Dadas dos arrays desordenadas que representan dos conjuntos (los elementos de cada array son distintos), encuentre la unión y la intersección de dos arrays.
Por ejemplo, si las arrays de entrada son:
arr1[] = {7, 1, 5, 2, 3, 6}
arr2[] = {3, 8, 6, 20, 7}
Entonces su programa debería imprimir Unión como {1, 2, 3, 5, 6, 7, 8, 20} e Intersección como {3, 6, 7}. Tenga en cuenta que los elementos de unión e intersección se pueden imprimir en cualquier orden.
Método 1 (usando el conjunto) :
Unión de dos arrays que podemos obtener con la estructura de datos Set muy fácilmente. Set es una estructura de datos que permite solo los distintos elementos en él. Entonces, cuando colocamos los elementos de ambos arreglos en el conjunto, obtendremos solo los elementos distintos que son iguales a la operación de unión sobre los arreglos. Vamos a codificarlo ahora –>
C++
// C++ program for the union of two arrays using Set
#include <bits/stdc++.h>
using namespace std;
void getUnion(int a[], int n, int b[], int m)
{
// Defining set container s
set<int> s;
// Inserting array elements in s
for (int i = 0; i < n; i++)
s.insert(a[i]);
for (int i = 0; i < m; i++)
s.insert(b[i]);
cout << "Number of elements after union operation: " << s.size() << endl;
cout << "The union set of both arrays is :" << endl;
for (auto itr = s.begin(); itr != s.end(); itr++)
cout << *itr
<< " "; // s will contain only distinct
// elements from array a and b
}
// Driver Code
int main()
{
int a[9] = { 1, 2, 5, 6, 2, 3, 5, 7, 3 };
int b[10] = { 2, 4, 5, 6, 8, 9, 4, 6, 5, 4 };
getUnion(a, 9, b, 10);
}
// contributed by Anirban Chand
Java
// Java program for the union of two arrays using Set
import java.util.*;
class GFG{
static void getUnion(int a[], int n, int b[], int m)
{
// Defining set container s
HashSet<Integer> s = new HashSet<>();
// Inserting array elements in s
for (int i = 0; i < n; i++)
s.add(a[i]);
for (int i = 0; i < m; i++)
s.add(b[i]);
System.out.print("Number of elements after union operation: " + s.size() +"\n");
System.out.print("The union set of both arrays is :" +"\n");
System.out.print(s.toString()+ " "); // s will contain only distinct
// elements from array a and b
}
// Driver Code
public static void main(String[] args)
{
int a[] = { 1, 2, 5, 6, 2, 3, 5, 7, 3 };
int b[] = { 2, 4, 5, 6, 8, 9, 4, 6, 5, 4 };
getUnion(a, 9, b, 10);
}
}
// This code is contributed by gauravrajput1
Python3
# Python program for the union of two arrays using Set
def getUnion(a, n, b, m):
# Defining set container s
s =set();
# Inserting array elements in s
for i in range(n):
s.add(a[i]);
for i in range(m):
s.add(b[i]);
print("Number of elements after union operation: " , len(s) , "");
print("The union set of both arrays is :" + "");
print(s, end=""); # s will contain only distinct
# elements from array a and b
# Driver Code
if __name__ == '__main__':
a = [ 1, 2, 5, 6, 2, 3, 5, 7, 3 ];
b = [ 2, 4, 5, 6, 8, 9, 4, 6, 5, 4 ];
getUnion(a, 9, b, 10);
# This code is contributed by gauravrajput1
C#
// C# program for the union of two arrays using Set
using System;
using System.Collections.Generic;
public class GFG {
static void getUnion(int []a, int n, int []b, int m) {
// Defining set container s
HashSet<int> s = new HashSet<int>();
// Inserting array elements in s
for (int i = 0; i < n; i++)
s.Add(a[i]);
for (int i = 0; i < m; i++)
s.Add(b[i]);
Console.Write("Number of elements after union operation: " + s.Count + "\n");
Console.Write("The union set of both arrays is :" + "\n");
foreach(int i in s)
Console.Write(i + " "); // s will contain only distinct
// elements from array a and b
}
// Driver Code
public static void Main(String[] args) {
int []a = { 1, 2, 5, 6, 2, 3, 5, 7, 3 };
int []b = { 2, 4, 5, 6, 8, 9, 4, 6, 5, 4 };
getUnion(a, 9, b, 10);
}
}
// This code is contributed by umadevi9616
Javascript
<script>
// JavaScript program for the above
// approach using Set
function getUnion(a, n, b, m) {
// Defining set container s
var s = new Set();
// Inserting array elements in s
for (let i = 0; i < n; i++)
s.add(a[i]);
for (let i = 0; i < m; i++)
s.add(b[i]);
document.write(
"Number of elements after union operation: "
+ s.size + "<br>");
document.write("The union set of both arrays is :");
document.write("<br>");
var arr = [];
for (let itr of s)
arr.push(itr);
// s will contain only distinct
// elements from array a and b
arr.sort(function (a, b) { return a - b; })
for (let i = 0; i < arr.length; i++) {
document.write(arr[i] + " ");
}
}
// Driver Code
let a = [1, 2, 5, 6, 2, 3, 5, 7, 3];
let b = [2, 4, 5, 6, 8, 9, 4, 6, 5, 4];
getUnion(a, 9, b, 10);
// This code is contributed by Potta Lokesh
</script>
Producción
Number of elements after union operation: 9 The union set of both arrays is : 1 2 3 4 5 6 7 8 9
Complejidad de tiempo: O(m * log(m) + n * log(n))
Nota: O(m + n) en el caso de Python porque en python el método integrado establecido es bastante diferente al de C++ una vez, Python usa un mapa hash internamente.
Podemos mejorar el rendimiento del método getUnion iterando sobre ambas arrays para el índice de 0 a min(n, m)-1 agregando todos los elementos en ambas arrays al conjunto, y luego iterando sobre la otra array con los elementos restantes y agregarlos al conjunto.
C++
// C++ program for the union of two arrays using Set
#include <bits/stdc++.h>
using namespace std;
void getUnion(int a[], int n, int b[], int m)
{
int min=(n<m)? n : m;
// Defining set container s
set<int> s;
// add elements from both the arrays for
// index from 0 to min(n, m)-1
for (int i = 0; i < min; i++)
{
s.insert(a[i]);
s.insert(b[i]);
}
if (n > m)
{
for (int i = m; i < n; i++)
{
s.insert(a[i]);
}
}
else if (n < m)
{
for (int i = n; i < m; i++)
{
s.insert(b[i]);
}
}
cout << "Number of elements after union operation: " << s.size() << endl;
cout << "The union set of both arrays is :" << endl;
for (auto itr = s.begin(); itr != s.end(); itr++)
cout << *itr
<< " "; // s will contain only distinct
// elements from array a and b
}
// Driver Code
int main()
{
int a[9] = { 1, 2, 5, 6, 2, 3, 5, 7, 3 };
int b[10] = { 2, 4, 5, 6, 8, 9, 4, 6, 5, 4 };
getUnion(a, 9, b, 10);
}
// This code is contributed by Aarti_Rathi
Java
// Java program for the union of two arrays using Set
import java.util.*;
class GFG {
static void getUnion(int a[], int n, int b[], int m)
{
// find min of n and m
int min = (n < m) ? n : m;
// set container
Set<Integer> set = new HashSet<>();
// add elements from both the arrays for
// index from 0 to min(n, m)-1
for (int i = 0; i < min; i++) {
set.add(a[i]);
set.add(b[i]);
}
// add remiaining elements to the set from the other
// array (having greater length)
// note that only one of the loops will execute
if (n > m) {
for (int i = m; i < n; i++) {
set.add(a[i]);
}
}
else if (n < m) {
for (int i = n; i < m; i++) {
set.add(b[i]);
}
}
// driver code to print the output
System.out.println("Number of elements after union operation: " + set.size());
System.out.println("The union set of both arrays is :");
System.out.print(set.toString());
}
// Driver Code
public static void main(String[] args)
{
int a[] = { 1, 2, 5, 6, 2, 3, 5, 7, 3 };
int b[] = { 2, 4, 5, 6, 8, 9, 4, 6, 5, 4 };
getUnion(a, 9, b, 10);
}
}
// This code is contributed by Parth Malhotra
Python3
# Python program for the union of two arrays using Set
def getUnion(a, n, b, m):
# Defining set container s
hs = set()
if(n<m):
min=n
else:
min=m
# add elements from both the arrays for
# index from 0 to min(n, m)-1
for i in range(0, min):
hs.add(a[i])
hs.add(b[i])
if(n>m):
for i in range(m, n):
hs.add(a[i])
else:
if(n<m):
for i in range(m, n):
hs.add(b[i])
print("Number of elements after union operation: ",len(hs))
print("The union set of both arrays is :")
for i in hs:
print(i, end=" ")
print("\n")
# s will contain only distinct
# elements from array a and b
# Driver Program
a = [1, 2, 5, 6, 2, 3, 5, 7, 3]
b = [2, 4, 5, 6, 8, 9, 4, 6, 5, 4]
n1 = len(a)
n2 = len(b)
# Function call
getUnion(a, n1, b, n2)
# This code is contributed by Aarti_Rathi
C#
// C# program for the union of two arrays using Set
using System;
using System.Collections.Generic;
public class GFG {
static void getUnion(int []a, int n, int []b, int m)
{
// find min of n and m
int min = (n < m) ? n : m;
// set container
HashSet<int> set = new HashSet<int>();
// add elements from both the arrays for
// index from 0 to min(n, m)-1
for (int i = 0; i < min; i++) {
set.Add(a[i]);
set.Add(b[i]);
}
// add remiaining elements to the set from the other
// array (having greater length)
// note that only one of the loops will execute
if (n > m) {
for (int i = m; i < n; i++) {
set.Add(a[i]);
}
} else if (n < m) {
for (int i = n; i < m; i++) {
set.Add(b[i]);
}
}
// driver code to print the output
Console.WriteLine("Number of elements after union operation: " + set.Count);
Console.WriteLine("The union set of both arrays is :[");
foreach(int x in set)
Console.Write(x+", ");
Console.Write("]");
}
// Driver Code
public static void Main(String[] args) {
int []a = { 1, 2, 5, 6, 2, 3, 5, 7, 3 };
int []b = { 2, 4, 5, 6, 8, 9, 4, 6, 5, 4 };
getUnion(a, 9, b, 10);
}
}
// This code is contributed by Rajput-Ji
Javascript
<script>
// javascript program for the union of two arrays using Set
function getUnion(a , n , b , m) {
// find min of n and m
var min = (n < m) ? n : m;
// set container
var set = new Set();
// add elements from both the arrays for
// index from 0 to min(n, m)-1
for (i = 0; i < min; i++) {
set.add(a[i]);
set.add(b[i]);
}
// add remiaining elements to the set from the other
// array (having greater length)
// note that only one of the loops will execute
if (n > m) {
for (i = m; i < n; i++) {
set.add(a[i]);
}
} else if (n < m) {
for (i = n; i < m; i++) {
set.add(b[i]);
}
}
// driver code to print the output
document.write("Number of elements after union operation: " + set.size);
document.write("<br/>The union set of both arrays is :<br/>");
set.forEach (function(value) {
document.write(value+" ");
})
}
// Driver Code
var a = [ 1, 2, 5, 6, 2, 3, 5, 7, 3 ];
var b = [ 2, 4, 5, 6, 8, 9, 4, 6, 5, 4 ];
getUnion(a, 9, b, 10);
// This code contributed by Rajput-Ji
</script>
Producción
Number of elements after union operation: 9 The union set of both arrays is : [1, 2, 3, 4, 5, 6, 7, 8, 9]
Complejidad de tiempo : O (max (m, n))
Método 2: (usando la estructura de datos del mapa)
Por el conocimiento de las estructuras de datos, sabemos que el mapa almacena solo claves distintas. Entonces, si insertamos cualquier clave que aparece más de una vez, se almacena solo una vez. La idea es insertar ambas arrays en un mapa común que luego almacenaría los distintos elementos de ambas arrays (unión de ambas arrays).
A continuación se muestra la implementación del método anterior:
C++
// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
void printUnion(int* a, int n, int* b, int m)
{
// Defining map container mp
map<int, int> mp;
// Inserting array elements in mp
for (int i = 0; i < n; i++)
mp.insert({ a[i], i });
for (int i = 0; i < m; i++)
mp.insert({ b[i], i });
cout << "The union set of both arrays is :" << endl;
for (auto itr = mp.begin(); itr != mp.end(); itr++)
cout << itr->first
<< " "; // mp will contain only distinct
// elements from array a and b
}
// Driver Code
int main()
{
int a[7] = { 1, 2, 5, 6, 2, 3, 5 };
int b[9] = { 2, 4, 5, 6, 8, 9, 4, 6, 5 };
printUnion(a, 7, b, 9);
}
Java
// Java program for the above approach
import java.io.*;
import java.util.*;
class GFG{
static void printUnion(int[] a, int n,
int[] b, int m)
{
Map<Integer,
Integer> mp = new HashMap<Integer,
Integer>();
// Inserting array elements in mp
for(int i = 0; i < n; i++)
{
mp.put(a[i], i);
}
for(int i = 0; i < m; i++)
{
mp.put(b[i], i);
}
System.out.println("The union set of both arrays is :");
for(Map.Entry mapElement : mp.entrySet())
{
System.out.print(mapElement.getKey() + " ");
// mp will contain only distinct
// elements from array a and b
}
}
// Driver Code
public static void main (String[] args)
{
int a[] = { 1, 2, 5, 6, 2, 3, 5 };
int b[] = { 2, 4, 5, 6, 8, 9, 4, 6, 5 };
printUnion(a, 7, b, 9);
}
}
// This code is contributed by avanitrachhadiya2155
Python3
# Python program for the above approach
def printUnion(a , n, b , m):
mp = {}
# Inserting array elements in mp
for i in range(n):
mp[a[i]] = i
for i in range(m):
mp[b[i]] = i
print("The union set of both arrays is : ");
for key in mp.keys():
print(key,end=" ")
# Driver Code
a = [ 1, 2, 5, 6, 2, 3, 5 ];
b = [ 2, 4, 5, 6, 8, 9, 4, 6, 5 ];
printUnion(a, 7, b, 9)
# This code is contributed by shinjanpatra
C#
// C# program for the above approach
using System;
using System.Collections.Generic;
public class GFG{
static void printUnion(int[] a, int n,
int[] b, int m)
{
Dictionary<int,
int> mp = new Dictionary<int,
int>();
// Inserting array elements in mp
for(int i = 0; i < n; i++)
{
if(!mp.ContainsKey(a[i]))
mp.Add(a[i], i);
}
for(int i = 0; i < m; i++)
{
if(!mp.ContainsKey(b[i]))
mp.Add(b[i], i);
}
Console.WriteLine("The union set of both arrays is :");
foreach(KeyValuePair<int,int> mapElement in mp)
{
Console.Write(mapElement.Key + " ");
// mp will contain only distinct
// elements from array a and b
}
}
// Driver Code
public static void Main(String[] args)
{
int []a = { 1, 2, 5, 6, 2, 3, 5 };
int []b = { 2, 4, 5, 6, 8, 9, 4, 6, 5 };
printUnion(a, 7, b, 9);
}
}
// This code contributed by gauravrajput1
Javascript
<script>
// javascript program for the above approach
function printUnion(a , n, b , m)
{
var mp = new Map();
// Inserting array elements in mp
for(var i = 0; i < n; i++)
{
mp.set(a[i], i);
}
for(var i = 0; i < m; i++)
{
mp.set(b[i], i);
}
document.write("The union set of both arrays is :<br/>");
for(var key of mp.keys())
{
document.write(key + " ");
}
}
// Driver Code
var a = [ 1, 2, 5, 6, 2, 3, 5 ];
var b = [ 2, 4, 5, 6, 8, 9, 4, 6, 5 ];
printUnion(a, 7, b, 9);
// This code is contributed by gauravrajput1
</script>
Producción
The union set of both arrays is : 1 2 3 4 5 6 8 9
El método anterior tiene complejidad temporal O(m+n).
*Este método es sugerido por Vinay Verma
Método 3 (ingenuo)
Unión:
- Inicialice la unión U como vacía.
- Copie todos los elementos de la primera array en U.
- Haga lo siguiente para cada elemento x de la segunda array:
- Si x no está presente en la primera array, copie x en U.
- Vuelve U.
Intersección:
- Inicialice la intersección I como vacía.
- Haz lo siguiente para cada elemento x de la primera array
- Si x está presente en la segunda array, copie x en I.
- Regreso yo.
La complejidad temporal de este método es O(mn) para ambas operaciones. Aquí m y n son números de elementos en arr1[] y arr2[] respectivamente.
Sin embargo, el método anterior solo funciona para elementos distintos en arrays de entrada.
Método 4 (Uso de clasificación)
- Ordenar arr1[] y arr2[]. Este paso toma un tiempo O(mLogm + nLogn).
- Utilice algoritmos O(m + n) para encontrar la unión y la intersección de dos arrays ordenadas .
La complejidad temporal general de este método es O(mLogm + nLogn).
Método 5 (Uso de clasificación y búsqueda)
Unión:
- Inicialice la unión U como vacía.
- Encuentre m y n más pequeños y ordene la array más pequeña.
- Copie la array más pequeña en U.
- Para cada elemento x de una array más grande, haga lo siguiente
- Búsqueda binaria x en la array más pequeña. Si x no está presente, cópielo en U.
- Vuelve U.
Intersección:
- Inicialice la intersección I como vacía.
- Encuentre el menor de m y n y ordene la array más pequeña.
- Para cada elemento x de una array más grande, haga lo siguiente
- Búsqueda binaria x en la array más pequeña. Si x está presente, cópielo en I.
- Regreso yo.
La complejidad temporal de este método es min(mLogm + nLogm, mLogn + nLogn) que también se puede escribir como O((m+n)Logm, (m+n)Logn). Este enfoque funciona mucho mejor que el enfoque anterior cuando la diferencia entre los tamaños de dos arrays es significativa.
Gracias a use_the_force por sugerir este método en un comentario aquí .
A continuación se muestra la implementación de este método. Sin embargo, este método también funciona solo para elementos distintos en arrays de entrada.
C++
// A C++ program to print union and intersection
/// of two unsorted arrays
#include <algorithm>
#include <iostream>
using namespace std;
int binarySearch(int arr[], int l, int r, int x);
// Prints union of arr1[0..m-1] and arr2[0..n-1]
void printUnion(int arr1[], int arr2[], int m, int n)
{
// Before finding union, make sure arr1[0..m-1]
// is smaller
if (m > n) {
int* tempp = arr1;
arr1 = arr2;
arr2 = tempp;
int temp = m;
m = n;
n = temp;
}
// Now arr1[] is smaller
// Sort the first array and print its elements (these
// two steps can be swapped as order in output is not
// important)
sort(arr1, arr1 + m);
for (int i = 0; i < m; i++)
cout << arr1[i] << " ";
// Search every element of bigger array in smaller array
// and print the element if not found
for (int i = 0; i < n; i++)
if (binarySearch(arr1, 0, m - 1, arr2[i]) == -1)
cout << arr2[i] << " ";
}
// Prints intersection of arr1[0..m-1] and arr2[0..n-1]
void printIntersection(int arr1[], int arr2[], int m, int n)
{
// Before finding intersection, make sure arr1[0..m-1]
// is smaller
if (m > n) {
int* tempp = arr1;
arr1 = arr2;
arr2 = tempp;
int temp = m;
m = n;
n = temp;
}
// Now arr1[] is smaller
// Sort smaller array arr1[0..m-1]
sort(arr1, arr1 + m);
// Search every element of bigger array in smaller
// array and print the element if found
for (int i = 0; i < n; i++)
if (binarySearch(arr1, 0, m - 1, arr2[i]) != -1)
cout << arr2[i] << " ";
}
// A recursive binary search function. It returns
// location of x in given array arr[l..r] is present,
// otherwise -1
int binarySearch(int arr[], int l, int r, int x)
{
if (r >= l) {
int mid = l + (r - l) / 2;
// If the element is present at the middle itself
if (arr[mid] == x)
return mid;
// If element is smaller than mid, then it can only
// be present in left subarray
if (arr[mid] > x)
return binarySearch(arr, l, mid - 1, x);
// Else the element can only be present in right
// subarray
return binarySearch(arr, mid + 1, r, x);
}
// We reach here when element is not present in array
return -1;
}
/* Driver program to test above function */
int main()
{
int arr1[] = { 7, 1, 5, 2, 3, 6 };
int arr2[] = { 3, 8, 6, 20, 7 };
int m = sizeof(arr1) / sizeof(arr1[0]);
int n = sizeof(arr2) / sizeof(arr2[0]);
// Function call
cout << "Union of two arrays is n";
printUnion(arr1, arr2, m, n);
cout << "nIntersection of two arrays is n";
printIntersection(arr1, arr2, m, n);
return 0;
}
Java
// A Java program to print union and intersection
/// of two unsorted arrays
import java.util.Arrays;
class UnionAndIntersection {
// Prints union of arr1[0..m-1] and arr2[0..n-1]
void printUnion(int arr1[], int arr2[], int m, int n)
{
// Before finding union, make sure arr1[0..m-1]
// is smaller
if (m > n) {
int tempp[] = arr1;
arr1 = arr2;
arr2 = tempp;
int temp = m;
m = n;
n = temp;
}
// Now arr1[] is smaller
// Sort the first array and print its elements
// (these two steps can be swapped as order in
// output is not important)
Arrays.sort(arr1);
for (int i = 0; i < m; i++)
System.out.print(arr1[i] + " ");
// Search every element of bigger array in smaller
// array and print the element if not found
for (int i = 0; i < n; i++) {
if (binarySearch(arr1, 0, m - 1, arr2[i]) == -1)
System.out.print(arr2[i] + " ");
}
}
// Prints intersection of arr1[0..m-1] and arr2[0..n-1]
void printIntersection(int arr1[], int arr2[], int m,
int n)
{
// Before finding intersection, make sure
// arr1[0..m-1] is smaller
if (m > n) {
int tempp[] = arr1;
arr1 = arr2;
arr2 = tempp;
int temp = m;
m = n;
n = temp;
}
// Now arr1[] is smaller
// Sort smaller array arr1[0..m-1]
Arrays.sort(arr1);
// Search every element of bigger array in smaller
// array and print the element if found
for (int i = 0; i < n; i++) {
if (binarySearch(arr1, 0, m - 1, arr2[i]) != -1)
System.out.print(arr2[i] + " ");
}
}
// A recursive binary search function. It returns
// location of x in given array arr[l..r] is present,
// otherwise -1
int binarySearch(int arr[], int l, int r, int x)
{
if (r >= l) {
int mid = l + (r - l) / 2;
// If the element is present at the middle
// itself
if (arr[mid] == x)
return mid;
// If element is smaller than mid, then it can
// only be present in left subarray
if (arr[mid] > x)
return binarySearch(arr, l, mid - 1, x);
// Else the element can only be present in right
// subarray
return binarySearch(arr, mid + 1, r, x);
}
// We reach here when element is not present in
// array
return -1;
}
// Driver code
public static void main(String[] args)
{
UnionAndIntersection u_i
= new UnionAndIntersection();
int arr1[] = { 7, 1, 5, 2, 3, 6 };
int arr2[] = { 3, 8, 6, 20, 7 };
int m = arr1.length;
int n = arr2.length;
// Function call
System.out.println("Union of two arrays is ");
u_i.printUnion(arr1, arr2, m, n);
System.out.println("");
System.out.println(
"Intersection of two arrays is ");
u_i.printIntersection(arr1, arr2, m, n);
}
}
Python3
# A Python3 program to print union and intersection
# of two unsorted arrays
# Prints union of arr1[0..m-1] and arr2[0..n-1]
def printUnion(arr1, arr2, m, n):
# Before finding union, make sure arr1[0..m-1]
# is smaller
if (m > n):
tempp = arr1
arr1 = arr2
arr2 = tempp
temp = m
m = n
n = temp
# Now arr1[] is smaller
# Sort the first array and print its elements (these two
# steps can be swapped as order in output is not important)
arr1.sort()
for i in range(0, m):
print(arr1[i], end=" ")
# Search every element of bigger array in smaller array
# and print the element if not found
for i in range(0, n):
if (binarySearch(arr1, 0, m - 1, arr2[i]) == -1):
print(arr2[i], end=" ")
# Prints intersection of arr1[0..m-1] and arr2[0..n-1]
def printIntersection(arr1, arr2, m, n):
# Before finding intersection, make sure arr1[0..m-1]
# is smaller
if (m > n):
tempp = arr1
arr1 = arr2
arr2 = tempp
temp = m
m = n
n = temp
# Now arr1[] is smaller
# Sort smaller array arr1[0..m-1]
arr1.sort()
# Search every element of bigger array in smaller
# array and print the element if found
for i in range(0, n):
if (binarySearch(arr1, 0, m - 1, arr2[i]) != -1):
print(arr2[i], end=" ")
# A recursive binary search function. It returns
# location of x in given array arr[l..r] is present,
# otherwise -1
def binarySearch(arr, l, r, x):
if (r >= l):
mid = int(l + (r - l)/2)
# If the element is present at the middle itself
if (arr[mid] == x):
return mid
# If element is smaller than mid, then it can only
# be present in left subarray
if (arr[mid] > x):
return binarySearch(arr, l, mid - 1, x)
# Else the element can only be present in right subarray
return binarySearch(arr, mid + 1, r, x)
# We reach here when element is not present in array
return -1
# Driver code
arr1 = [7, 1, 5, 2, 3, 6]
arr2 = [3, 8, 6, 20, 7]
m = len(arr1)
n = len(arr2)
# Function call
print("Union of two arrays is ")
printUnion(arr1, arr2, m, n)
print("\nIntersection of two arrays is ")
printIntersection(arr1, arr2, m, n)
# This code is contributed by mits
C#
// A C# program to print union and
// intersection of two unsorted arrays
using System;
class GFG {
// Prints union of arr1[0..m-1] and arr2[0..n-1]
static void printUnion(int[] arr1, int[] arr2, int m,
int n)
{
// Before finding union, make
// sure arr1[0..m-1] is smaller
if (m > n) {
int[] tempp = arr1;
arr1 = arr2;
arr2 = tempp;
int temp = m;
m = n;
n = temp;
}
// Now arr1[] is smaller
// Sort the first array and print
// its elements (these two steps can
// be swapped as order in output is
// not important)
Array.Sort(arr1);
for (int i = 0; i < m; i++)
Console.Write(arr1[i] + " ");
// Search every element of bigger
// array in smaller array and print
// the element if not found
for (int i = 0; i < n; i++) {
if (binarySearch(arr1, 0, m - 1, arr2[i]) == -1)
Console.Write(arr2[i] + " ");
}
}
// Prints intersection of arr1[0..m-1]
// and arr2[0..n-1]
static void printIntersection(int[] arr1, int[] arr2,
int m, int n)
{
// Before finding intersection,
// make sure arr1[0..m-1] is smaller
if (m > n) {
int[] tempp = arr1;
arr1 = arr2;
arr2 = tempp;
int temp = m;
m = n;
n = temp;
}
// Now arr1[] is smaller
// Sort smaller array arr1[0..m-1]
Array.Sort(arr1);
// Search every element of bigger array in
// smaller array and print the element if found
for (int i = 0; i < n; i++) {
if (binarySearch(arr1, 0, m - 1, arr2[i]) != -1)
Console.Write(arr2[i] + " ");
}
}
// A recursive binary search function.
// It returns location of x in given
// array arr[l..r] is present, otherwise -1
static int binarySearch(int[] arr, int l, int r, int x)
{
if (r >= l) {
int mid = l + (r - l) / 2;
// If the element is present at
// the middle itself
if (arr[mid] == x)
return mid;
// If element is smaller than mid, then it
// can only be present in left subarray
if (arr[mid] > x)
return binarySearch(arr, l, mid - 1, x);
// Else the element can only be
// present in right subarray
return binarySearch(arr, mid + 1, r, x);
}
// We reach here when element is
// not present in array
return -1;
}
// Driver Code
static public void Main()
{
int[] arr1 = { 7, 1, 5, 2, 3, 6 };
int[] arr2 = { 3, 8, 6, 20, 7 };
int m = arr1.Length;
int n = arr2.Length;
// Function call
Console.WriteLine("Union of two arrays is ");
printUnion(arr1, arr2, m, n);
Console.WriteLine("");
Console.WriteLine("Intersection of two arrays is ");
printIntersection(arr1, arr2, m, n);
}
}
// This code is contributed
// by Sach_Code
PHP
<?php
// A PHP program to print union and intersection
/// of two unsorted arrays
// Prints union of arr1[0..m-1] and arr2[0..n-1]
function printUnion($arr1, $arr2, $m, $n)
{
// Before finding union, make sure arr1[0..m-1]
// is smaller
if ($m > $n)
{
$tempp = $arr1;
$arr1 = $arr2;
$arr2 = $tempp;
$temp = $m;
$m = $n;
$n = $temp;
}
// Now arr1[] is smaller
// Sort the first array and print its elements (these two
// steps can be swapped as order in output is not important)
sort($arr1);
for ($i = 0; $i < $m; $i++)
echo $arr1[$i]." ";
// Search every element of bigger array in smaller array
// and print the element if not found
for ($i = 0; $i < $n; $i++)
if (binarySearch($arr1, 0, $m - 1, $arr2[$i]) == -1)
echo $arr2[$i]." ";
}
// Prints intersection of arr1[0..m-1] and arr2[0..n-1]
function printIntersection($arr1, $arr2, $m, $n)
{
// Before finding intersection, make sure arr1[0..m-1]
// is smaller
if ($m > $n)
{
$tempp = $arr1;
$arr1 = $arr2;
$arr2 = $tempp;
$temp = $m;
$m = $n;
$n = $temp;
}
// Now arr1[] is smaller
// Sort smaller array arr1[0..m-1]
sort($arr1);
// Search every element of bigger array in smaller
// array and print the element if found
for ($i = 0; $i < $n; $i++)
if (binarySearch($arr1, 0, $m - 1, $arr2[$i]) != -1)
echo $arr2[$i]." ";
}
// A recursive binary search function. It returns
// location of x in given array arr[l..r] is present,
// otherwise -1
function binarySearch($arr, $l, $r,$x)
{
if ($r >= $l)
{
$mid = (int)($l + ($r - $l)/2);
// If the element is present at the middle itself
if ($arr[$mid] == $x) return $mid;
// If element is smaller than mid, then it can only
// be present in left subarray
if ($arr[$mid] > $x)
return binarySearch($arr, $l, $mid - 1, $x);
// Else the element can only be present in right subarray
return binarySearch($arr, $mid + 1, $r, $x);
}
// We reach here when element is not present in array
return -1;
}
/* Driver program to test above function */
$arr1 = array(7, 1, 5, 2, 3, 6);
$arr2 = array(3, 8, 6, 20, 7);
$m = count($arr1);
$n = count($arr2);
echo "Union of two arrays is \n";
printUnion($arr1, $arr2, $m, $n);
echo "\nIntersection of two arrays is \n";
printIntersection($arr1, $arr2, $m, $n);
// This code is contributed by mits
?>
Javascript
<script>
// A JavaScript program to
// print union and intersection
// of two unsorted arrays
// A recursive binary search function. It returns
// location of x in given array arr[l..r] is present,
// otherwise -1
function binarySearch(arr, l, r, x)
{
if (r >= l) {
let mid = l + Math.floor((r - l) / 2);
// If the element is present at the middle itself
if (arr[mid] == x)
return mid;
// If element is smaller than mid, then it can only
// be present in left subarray
if (arr[mid] > x)
return binarySearch(arr, l, mid - 1, x);
// Else the element can only be present in right
// subarray
return binarySearch(arr, mid + 1, r, x);
}
// We reach here when element is not present in array
return -1;
}
// Prints union of arr1[0..m-1] and arr2[0..n-1]
function printUnion(arr1, arr2, m, n)
{
// Before finding union, make sure arr1[0..m-1]
// is smaller
if (m > n) {
let tempp = arr1;
arr1 = arr2;
arr2 = tempp;
let temp = m;
m = n;
n = temp;
}
// Now arr1[] is smaller
// Sort the first array and print its elements (these
// two steps can be swapped as order in output is not
// important)
arr1.sort((a, b) => a - b);
for (let i = 0; i < m; i++)
document.write(arr1[i] + " ");
// Search every element of bigger array in smaller array
// and print the element if not found
for (let i = 0; i < n; i++)
if (binarySearch(arr1, 0, m - 1, arr2[i]) == -1)
document.write(arr2[i] + " ");
}
// Prints intersection of arr1[0..m-1] and arr2[0..n-1]
function printIntersection(arr1, arr2, m, n)
{
// Before finding intersection, make sure arr1[0..m-1]
// is smaller
if (m > n) {
let tempp = arr1;
arr1 = arr2;
arr2 = tempp;
let temp = m;
m = n;
n = temp;
}
// Now arr1[] is smaller
// Sort smaller array arr1[0..m-1]
arr1.sort((a, b) => a - b);
// Search every element of bigger array in smaller
// array and print the element if found
for (let i = 0; i < n; i++)
if (binarySearch(arr1, 0, m - 1, arr2[i]) != -1)
document.write(arr2[i] + " ");
}
/* Driver program to test above function */
let arr1 = [ 7, 1, 5, 2, 3, 6 ];
let arr2 = [ 3, 8, 6, 20, 7 ];
let m = arr1.length;
let n = arr2.length;
// Function call
document.write("Union of two arrays is <br>");
printUnion(arr1, arr2, m, n);
document.write("<br>Intersection of two arrays is<br>");
printIntersection(arr1, arr2, m, n);
// This code is contributed by Surbhi Tyagi.
</script>
Producción
Union of two arrays is n3 6 7 8 20 1 5 2 nIntersection of two arrays is n7 3 6
Otro enfoque (cuando los elementos de la array pueden no ser distintos):
C++
// C++ code to find intersection when
// elements may not be distinct
#include <bits/stdc++.h>
using namespace std;
// Function to find intersection
void intersection(int a[], int b[], int n, int m)
{
int i = 0, j = 0;
while (i < n && j < m) {
if (a[i] > b[j]) {
j++;
}
else if (b[j] > a[i]) {
i++;
}
else {
// when both are equal
cout << a[i] << " ";
i++;
j++;
}
}
}
// Driver Code
int main()
{
int a[] = { 1, 3, 2, 3, 3, 4, 5, 5, 6 };
int b[] = { 3, 3, 5 };
int n = sizeof(a) / sizeof(a[0]);
int m = sizeof(b) / sizeof(b[0]);
// sort
sort(a, a + n);
sort(b, b + m);
// Function call
intersection(a, b, n, m);
}
Java
// Java code to find intersection when
// elements may not be distinct
import java.io.*;
import java.util.Arrays;
class GFG {
// Function to find intersection
static void intersection(int a[], int b[], int n, int m)
{
int i = 0, j = 0;
while (i < n && j < m) {
if (a[i] > b[j]) {
j++;
}
else if (b[j] > a[i]) {
i++;
}
else {
// when both are equal
System.out.print(a[i] + " ");
i++;
j++;
}
}
}
// Driver Code
public static void main(String[] args)
{
int a[] = { 1, 3, 2, 3, 4, 5, 5, 6 };
int b[] = { 3, 3, 5 };
int n = a.length;
int m = b.length;
// sort
Arrays.sort(a);
Arrays.sort(b);
// Function call
intersection(a, b, n, m);
}
}
Python3
# Python 3 code to find intersection # when elements may not be distinct # Function to find intersection def intersection(a, b, n, m): i = 0 j = 0 while (i < n and j < m): if (a[i] > b[j]): j += 1 else: if (b[j] > a[i]): i += 1 else: # when both are equal print(a[i], end=" ") i += 1 j += 1 # Driver Code if __name__ == "__main__": a = [1, 3, 2, 3, 4, 5, 5, 6] b = [3, 3, 5] n = len(a) m = len(b) # sort a.sort() b.sort() # function call intersection(a, b, n, m) # This code is contributed by Ita_c
C#
// C# code to find intersection when
// elements may not be distinct
using System;
class GFG {
// Function to find intersection
static void intersection(int[] a, int[] b, int n, int m)
{
int i = 0, j = 0;
while (i < n && j < m) {
if (a[i] > b[j]) {
j++;
}
else if (b[j] > a[i]) {
i++;
}
else {
// when both are equal
Console.Write(a[i] + " ");
i++;
j++;
}
}
}
// Driver Code
public static void Main()
{
int[] a = { 1, 3, 2, 3, 4, 5, 5, 6 };
int[] b = { 3, 3, 5 };
int n = a.Length;
int m = b.Length;
// sort
Array.Sort(a);
Array.Sort(b);
// Function call
intersection(a, b, n, m);
}
}
// this code is contributed by mukul singh
PHP
<?php
// PHP code to find intersection when
// elements may not be distinct
// Function to find intersection
function intersection($a, $b, $n, $m)
{
$i = 0; $j = 0;
while ($i < $n && $j < $m)
{
if ($a[$i] > $b[$j])
{
$j++;
}
else
if ($b[$j] > $a[$i])
{
$i++;
}
else
{
// when both are equal
echo($a[$i] . " ");
$i++;
$j++;
}
}
}
// Driver Code
$a = array(1, 3, 2, 3, 4, 5, 5, 6);
$b = array(3, 3, 5);
$n = sizeof($a);
$m = sizeof($b);
// sort
sort($a);
sort($b);
// Function call
intersection($a, $b, $n, $m);
// This code is contributed
// by Mukul Singh
?>
Javascript
<script>
// Javascript code to find intersection when
// elements may not be distinct
// Function to find intersection
function intersection(a,b,n,m)
{
let i = 0, j = 0;
while (i < n && j < m)
{
if (a[i] > b[j])
{
j++;
}
else if (b[j] > a[i])
{
i++;
}
else
{
// when both are equal
document.write(a[i] + " ");
i++;
j++;
}
}
}
// Driver Code
let a = [1, 3, 2, 3, 4, 5, 5, 6 ];
let b = [3, 3, 5 ]
let n = a.length;
let m = b.length;
// sort
a.sort();
b.sort();
// Function call
intersection(a, b, n, m);
// This code is contributed by rag2127
</script>
Producción
3 3 5
Gracias, Sanny Kumar por sugerir el método anterior.
Método 6 (sin usar hashing o cualquier biblioteca predefinida como conjuntos o mapas y funciona incluso para elementos repetidos y distantes)
En primer lugar, ordenamos ambas arrays y procedemos de la siguiente manera:
Unión
- Iterar en el bucle while hasta que finalice cualquier array.
- En cada iteración buscamos el más pequeño en ambas arrays y lo imprimimos e incrementamos su puntero solo si no es el mismo que el último elemento impreso en unión.
- Después de que terminemos, iteramos el resto de dos arrays de manera similar a la anterior e imprimimos la unión.
Intersección
- Iterar en el bucle while hasta que finalice cualquiera de los arreglos.
- En cada iteración, buscamos el más pequeño de los dos elementos de la array y aumentamos su puntero porque no estará en otra lista, por lo tanto, no formará parte de la intersección.
- Para la intersección, si ambos elementos son iguales, lo imprimimos e incrementamos ambos punteros solo si no es el mismo que el último elemento impreso en la intersección.
C++
#include <bits/stdc++.h>
using namespace std;
// Function to find union
void Union(int a[], int b[], int n, int m) {
int *result=new int[n+m];
int index=0;
int left=0,right=0;
while(left<n && right<m){
if(a[left]<b[right]){
if(index!=0 && a[left]==result[index-1]){
left++;
}else{
result[index]=a[left];
left++;
index++;
}
}else{
if(index!=0 && b[right]==result[index-1]){
right++;
}else{
result[index]=b[right];
right++;
index++;
}
}
}
while(left<n){
if(index!=0 && a[left]==result[index-1]){
left++;
}else{
result[index]=a[left];
left++;
index++;
}
}
while(right<m){
if(index!=0 && b[right]==result[index-1]){
right++;
}else{
result[index]=b[right];
right++;
index++;
}
}
cout<<"Union: ";
for(int k=0;k<index;k++)
cout<<result[k]<<" ";
cout<<endl;
};
// Function to find intersection
void intersection(int a[], int b[], int n, int m) {
int i=0,j=0,k=0;
int *result=new int[n+m];
while(i<n && j<m){
if(a[i]<b[j])
i++;
else if(a[i]>b[j])
j++;
else{
if(k!=0 && a[i]==result[k-1]){
i++;
j++;
}
else{
result[k]=a[i];
i++;
j++;
k++;
}
}
}
cout<<"Intersection: ";
for(int x=0;x<k;x++)
cout<<result[x]<<" ";
cout<<endl;
}
// Driver Code
int main()
{
int a[] = { 1, 3, 2, 3, 3, 4, 5, 5, 6 };
int b[] = { 3, 3, 5 };
int n = sizeof(a) / sizeof(a[0]);
int m = sizeof(b) / sizeof(b[0]);
// sort
sort(a, a + n);
sort(b, b + m);
// Function call
Union(a,b,n,m);
intersection(a, b, n, m);
}
Python3
# Python3 code to implement the approach
# Function to find union
def Union(a, b, n, m):
result = [0 for _ in range(n + m)]
index, left, right = 0, 0, 0
while left<n and right<m:
if (a[left] < b[right]):
if(index != 0 and a[left] == result[index-1]):
left += 1
else:
result[index] = a[left];
left += 1
index += 1
else:
if (index !=0 and b[right] == result[index-1]):
right += 1
else:
result[index] = b[right];
right += 1
index += 1
while(left < n):
if(index != 0 and a[left] == result[index-1]):
left += 1
else:
result[index]=a[left];
left += 1;
index += 1;
while(right < m):
if(index != 0 and b[right] == result[index - 1]):
right += 1
else:
result[index] = b[right];
right += 1;
index += 1;
print("Union:", *result[:index])
# Function to find intersection
def intersection(a, b, n, m):
i, j, k = 0, 0, 0
result = [0 for _ in range(n + m)]
while i < n and j < m:
if a[i] < b[j]:
i += 1
elif a[i] > b[j]:
j += 1
else:
if k != 0 and a[i] == result[k - 1]:
i += 1
j += 1
else:
result[k] = a[i]
i += 1
j += 1
k += 1
print("Intersection:", *result[:k])
# Driver Code
a = [ 1, 3, 2, 3, 3, 4, 5, 5, 6 ];
b = [ 3, 3, 5 ];
n, m = len(a), len(b)
# sort
a.sort()
b.sort()
# Function call
Union(a,b,n,m);
intersection(a, b, n, m);
# This code is contributed by phasing17
Javascript
// JavaScript code to implement the approach
// Function to find union
function Union(a, b, n, m)
{
let result = new Array(n + m);
let index = 0;
let left = 0, right = 0;
while (left < n && right < m) {
if (a[left] < b[right]) {
if (index != 0
&& a[left] == result[index - 1]) {
left++;
}
else {
result[index] = a[left];
left++;
index++;
}
}
else {
if (index != 0
&& b[right] == result[index - 1]) {
right++;
}
else {
result[index] = b[right];
right++;
index++;
}
}
}
while (left < n) {
if (index != 0 && a[left] == result[index - 1]) {
left++;
}
else {
result[index] = a[left];
left++;
index++;
}
}
while (right < m) {
if (index != 0 && b[right] == result[index - 1]) {
right++;
}
else {
result[index] = b[right];
right++;
index++;
}
}
process.stdout.write("Union: ");
for (var k = 0; k < index; k++)
process.stdout.write(result[k] + " ");
process.stdout.write("\n");
}
// Function to find intersection
function intersection(a, b, n, m)
{
let i = 0, j = 0, k = 0;
let result = new Array(n + m);
while (i < n && j < m) {
if (a[i] < b[j])
i++;
else if (a[i] > b[j])
j++;
else {
if (k != 0 && a[i] == result[k - 1]) {
i++;
j++;
}
else {
result[k] = a[i];
i++;
j++;
k++;
}
}
}
process.stdout.write("Intersection: ");
for (var x = 0; x < k; x++)
process.stdout.write(result[x] + " ");
process.stdout.write("\n");
}
// Driver Code
let a = [ 1, 3, 2, 3, 3, 4, 5, 5, 6 ];
let b = [ 3, 3, 5 ];
let n = a.length;
let m = b.length;
// sort
a.sort();
b.sort();
// Function call
Union(a, b, n, m);
intersection(a, b, n, m);
// This code is contributed by phasing17
Método 7 (Usar hash)
Unión
- Inicialice un conjunto hash vacío hs .
- Iterar a través de la primera array y colocar cada elemento de la primera array en el conjunto S.
- Repita el proceso para la segunda array.
- Imprima el conjunto hs .
Intersección
- Inicializar un conjunto vacío hs.
- Iterar a través de la primera array y colocar cada elemento de la primera array en el conjunto S.
- Para cada elemento x de la segunda array, haga lo siguiente:
Buscar x en el conjunto hs. Si x está presente, entonces imprímalo. La complejidad de tiempo de este método es ?(m+n) bajo el supuesto de que las operaciones de búsqueda e inserción de tablas hash toman ?(1) tiempo.
A continuación se muestra la implementación de la idea anterior:
C++
// CPP program to find union and intersection
// using sets
#include <bits/stdc++.h>
using namespace std;
// Prints union of arr1[0..n1-1] and arr2[0..n2-1]
void printUnion(int arr1[], int arr2[], int n1, int n2)
{
set<int> hs;
// Insert the elements of arr1[] to set hs
for (int i = 0; i < n1; i++)
hs.insert(arr1[i]);
// Insert the elements of arr2[] to set hs
for (int i = 0; i < n2; i++)
hs.insert(arr2[i]);
// Print the content of set hs
for (auto it = hs.begin(); it != hs.end(); it++)
cout << *it << " ";
cout << endl;
}
// Prints intersection of arr1[0..n1-1] and
// arr2[0..n2-1]
void printIntersection(int arr1[], int arr2[], int n1,
int n2)
{
set<int> hs;
// Insert the elements of arr1[] to set S
for (int i = 0; i < n1; i++)
hs.insert(arr1[i]);
for (int i = 0; i < n2; i++)
// If element is present in set then
// push it to vector V
if (hs.find(arr2[i]) != hs.end())
cout << arr2[i] << " ";
}
// Driver Program
int main()
{
int arr1[] = { 7, 1, 5, 2, 3, 6 };
int arr2[] = { 3, 8, 6, 20, 7 };
int n1 = sizeof(arr1) / sizeof(arr1[0]);
int n2 = sizeof(arr2) / sizeof(arr2[0]);
// Function call
printUnion(arr1, arr2, n1, n2);
printIntersection(arr1, arr2, n1, n2);
return 0;
}
Java
// Java program to find union and intersection
// using Hashing
import java.util.HashSet;
class Test {
// Prints union of arr1[0..m-1] and arr2[0..n-1]
static void printUnion(int arr1[], int arr2[])
{
HashSet<Integer> hs = new HashSet<>();
for (int i = 0; i < arr1.length; i++)
hs.add(arr1[i]);
for (int i = 0; i < arr2.length; i++)
hs.add(arr2[i]);
System.out.println(hs);
}
// Prints intersection of arr1[0..m-1] and arr2[0..n-1]
static void printIntersection(int arr1[], int arr2[])
{
HashSet<Integer> hs = new HashSet<>();
HashSet<Integer> hs1 = new HashSet<>();
for (int i = 0; i < arr1.length; i++)
hs.add(arr1[i]);
for (int i = 0; i < arr2.length; i++)
if (hs.contains(arr2[i]))
System.out.print(arr2[i] + " ");
}
// Driver code
public static void main(String[] args)
{
int arr1[] = { 7, 1, 5, 2, 3, 6 };
int arr2[] = { 3, 8, 6, 20, 7 };
// Function call
System.out.println("Union of two arrays is : ");
printUnion(arr1, arr2);
System.out.println(
"Intersection of two arrays is : ");
printIntersection(arr1, arr2);
}
}
Python
# Python program to find union and intersection
# using sets
def printUnion(arr1, arr2, n1, n2):
hs = set()
# Insert the elements of arr1[] to set hs
for i in range(0, n1):
hs.add(arr1[i])
# Insert the elements of arr1[] to set hs
for i in range(0, n2):
hs.add(arr2[i])
print("Union:")
for i in hs:
print(i, end=" ")
print("\n")
# Prints intersection of arr1[0..n1-1] and
# arr2[0..n2-1]
def printIntersection(arr1, arr2, n1, n2):
hs = set()
# Insert the elements of arr1[] to set S
for i in range(0, n1):
hs.add(arr1[i])
print("Intersection:")
for i in range(0, n2):
# If element is present in set then
# push it to vector V
if arr2[i] in hs:
print(arr2[i], end=" ")
# Driver Program
arr1 = [7, 1, 5, 2, 3, 6]
arr2 = [3, 8, 6, 20, 7]
n1 = len(arr1)
n2 = len(arr2)
# Function call
printUnion(arr1, arr2, n1, n2)
printIntersection(arr1, arr2, n1, n2)
# This article is contributed by Kumar Suman .
C#
// C# program to find union and intersection
// using Hashing
using System;
using System.Linq;
using System.Collections.Generic;
class GFG
{
// Prints union of arr1[0..m-1] and arr2[0..n-1]
static void printUnion(int []arr1, int []arr2)
{
HashSet<int> hs = new HashSet<int>();
for (int i = 0; i < arr1.Length; i++)
hs.Add(arr1[i]);
for (int i = 0; i < arr2.Length; i++)
hs.Add(arr2[i]);
Console.WriteLine(string.Join(", ", hs));
}
// Prints intersection of arr1[0..m-1] and arr2[0..n-1]
static void printIntersection(int []arr1, int []arr2)
{
HashSet<int> hs = new HashSet<int>();
for (int i = 0; i < arr1.Length; i++)
hs.Add(arr1[i]);
for (int i = 0; i < arr2.Length; i++)
if (hs.Contains(arr2[i]))
Console.Write(arr2[i] + " ");
}
// Driver Code
static void Main()
{
int []arr1 = {7, 1, 5, 2, 3, 6};
int []arr2 = {3, 8, 6, 20, 7};
Console.WriteLine("Union of two arrays is : ");
printUnion(arr1, arr2);
Console.WriteLine("\nIntersection of two arrays is : ");
printIntersection(arr1, arr2);
}
}
// This code is contributed by mits
Javascript
<script>
// javascript program to find union and intersection
// using Hashing
// Prints union of arr1[0..m-1] and arr2[0..n-1]
function printUnion(arr1 , arr2) {
var hs = new Set();
for (i = 0; i < arr1.length; i++)
hs.add(arr1[i]);
for (i = 0; i < arr2.length; i++)
hs.add(arr2[i]);
for(var k of hs)
document.write(k+" ");
}
// Prints intersection of arr1[0..m-1] and arr2[0..n-1]
function printIntersection(arr1 , arr2) {
var hs = new Set();
var hs1 = new Set();
for (i = 0; i < arr1.length; i++)
hs.add(arr1[i]);
for (var i = 0; i < arr2.length; i++)
if (hs.has(arr2[i]))
document.write(arr2[i] + " ");
}
// Driver code
var arr1 = [ 7, 1, 5, 2, 3, 6 ];
var arr2 = [ 3, 8, 6, 20, 7 ];
// Function call
document.write("Union of two arrays is :<br/> ");
printUnion(arr1, arr2);
document.write("<br/>Intersection of two arrays is : <br/>");
printIntersection(arr1, arr2);
// This code is contributed by gauravrajput1
</script>
Producción
1 2 3 5 6 7 8 20 3 6 7
Este método es aportado por Ankur Singh .
La complejidad de tiempo de este método es O (m + n) bajo el supuesto de que las operaciones de búsqueda e inserción de tablas hash toman O (1) tiempo.
Método 8 (Tipo de técnica de hashing sin usar ninguna colección de Java predefinida)
- Inicializar la array con un tamaño de m+n
- Rellene el primer valor de array en una array resultante haciendo hash (para encontrar la posición adecuada)
- Repita para la segunda array
- Mientras hace hash si ocurre una colisión, incremente la posición de forma recursiva
A continuación se muestra la implementación del código anterior:
C++
// CPP program to find union and intersection
#include <bits/stdc++.h>
using namespace std;
// Prints union of arr1[0..n1-1] and arr2[0..n2-1]
void printUnion(int v, int ans[], int zero)
{
int zero1 = 0;
cout<<"\nUnion : ";
for (int i = 0; i < v; i++) {
if ((zero == 0 && ans[i] == 0)
|| (ans[i] == 0 && zero1 > 0))
continue;
if (ans[i] == 0)
zero1++;
cout<<ans[i] << ",";
}
}
void placeValue(int a[], int ans[], int i, int p, int v)
{
p = p % v;
if (ans[p] == 0)
ans[p] = a[i];
else {
if (ans[p] == a[i])
cout<<a[i] << ",";
else {
// Hashing collision happened increment
// position and do recursive call
p = p + 1;
placeValue(a, ans, i, p, v);
}
}
}
void placeZeros(int v, int ans[], int zero)
{
if (zero == 2) {
cout<<"0"<<endl;
int d[] = { 0 };
placeValue(d, ans, 0, 0, v);
}
if (zero == 1) {
int d[] = { 0 };
placeValue(d, ans, 0, 0, v);
}
}
// Function to iterate array
int iterateArray(int a[], int v, int ans[], int i)
{
if (a[i] != 0) {
int p = a[i] % v;
placeValue(a, ans, i, p, v);
}
else
return 1;
return 0;
}
// Prints intersection of arr1[0..n1-1] and
// arr2[0..n2-1]
void findPosition(int a[], int b[], int n1, int n2)
{
int v = (n1+n2);
int ans[v];
for(int i=0;i<v;i++)
{
ans[i]=0;
}
int zero1 = 0;
int zero2 = 0;
cout<<"Intersection : ";
// Iterate first array
for (int i = 0; i < n1; i++)
zero1 = iterateArray(a, v, ans, i);
// Iterate second array
for (int j = 0; j < n2; j++)
zero2 = iterateArray(b, v, ans, j);
int zero = zero1 + zero2;
placeZeros(v, ans, zero);
printUnion(v, ans, zero);
}
// Driver Program
int main()
{
int arr1[] = { 7, 1, 5, 2, 3, 6 };
int arr2[] = { 3, 8, 6, 20, 7 };
int n1 = sizeof(arr1) / sizeof(arr1[0]);
int n2 = sizeof(arr2) / sizeof(arr2[0]);
// Function call
findPosition(arr1, arr2, n1, n2);
return 0;
}
// This code is contributed by Aarti_Rathi
Java
// Java program to find union and intersection
// using similar Hashing Technique
// without using any predefined Java Collections
// Time Complexity best case & avg case = O(m+n)
// Worst case = O(nlogn)
// package com.arrays.math;
public class UnsortedIntersectionUnion {
// Prints intersection of arr1[0..n1-1] and
// arr2[0..n2-1]
public void findPosition(int a[], int b[])
{
int v = (a.length + b.length);
int ans[] = new int[v];
int zero1 = 0;
int zero2 = 0;
System.out.print("Intersection : ");
// Iterate first array
for (int i = 0; i < a.length; i++)
zero1 = iterateArray(a, v, ans, i);
// Iterate second array
for (int j = 0; j < b.length; j++)
zero2 = iterateArray(b, v, ans, j);
int zero = zero1 + zero2;
placeZeros(v, ans, zero);
printUnion(v, ans, zero);
}
// Prints union of arr1[0..n1-1] and arr2[0..n2-1]
private void printUnion(int v, int[] ans, int zero)
{
int zero1 = 0;
System.out.print("\nUnion : ");
for (int i = 0; i < v; i++) {
if ((zero == 0 && ans[i] == 0)
|| (ans[i] == 0 && zero1 > 0))
continue;
if (ans[i] == 0)
zero1++;
System.out.print(ans[i] + ",");
}
}
private void placeZeros(int v, int[] ans, int zero)
{
if (zero == 2) {
System.out.println("0");
int d[] = { 0 };
placeValue(d, ans, 0, 0, v);
}
if (zero == 1) {
int d[] = { 0 };
placeValue(d, ans, 0, 0, v);
}
}
// Function to iterate array
private int iterateArray(int[] a, int v, int[] ans,
int i)
{
if (a[i] != 0) {
int p = a[i] % v;
placeValue(a, ans, i, p, v);
}
else
return 1;
return 0;
}
private void placeValue(int[] a, int[] ans, int i,
int p, int v)
{
p = p % v;
if (ans[p] == 0)
ans[p] = a[i];
else {
if (ans[p] == a[i])
System.out.print(a[i] + ",");
else {
// Hashing collision happened increment
// position and do recursive call
p = p + 1;
placeValue(a, ans, i, p, v);
}
}
}
// Driver code
public static void main(String args[])
{
int a[] = { 7, 1, 5, 2, 3, 6 };
int b[] = { 3, 8, 6, 20, 7 };
// Function call
UnsortedIntersectionUnion uiu
= new UnsortedIntersectionUnion();
uiu.findPosition(a, b);
}
}
// This code is contributed by Mohanakrishnan S.
Python3
# Python3 program to find union and intersection
# using similar Hashing Technique
# without using any predefined Java Collections
# Time Complexity best case & avg case = O(m+n)
# Worst case = O(nlogn)
# Prints intersection of arr1[0..n1-1] and
# arr2[0..n2-1]
def findPosition(a, b):
v = len(a) + len(b);
ans = [0]*v;
zero1 = zero2 = 0;
print("Intersection :",end=" ");
# Iterate first array
for i in range(len(a)):
zero1 = iterateArray(a, v, ans, i);
# Iterate second array
for j in range(len(b)):
zero2 = iterateArray(b, v, ans, j);
zero = zero1 + zero2;
placeZeros(v, ans, zero);
printUnion(v, ans, zero);
# Prints union of arr1[0..n1-1] and arr2[0..n2-1]
def printUnion(v, ans,zero):
zero1 = 0;
print("\nUnion :",end=" ");
for i in range(v):
if ((zero == 0 and ans[i] == 0) or
(ans[i] == 0 and zero1 > 0)):
continue;
if (ans[i] == 0):
zero1+=1;
print(ans[i],end=",");
def placeZeros(v, ans, zero):
if (zero == 2):
print("0");
d = [0];
placeValue(d, ans, 0, 0, v);
if (zero == 1):
d=[0];
placeValue(d, ans, 0, 0, v);
# Function to iterate array
def iterateArray(a,v,ans,i):
if (a[i] != 0):
p = a[i] % v;
placeValue(a, ans, i, p, v);
else:
return 1;
return 0;
def placeValue(a,ans,i,p,v):
p = p % v;
if (ans[p] == 0):
ans[p] = a[i];
else:
if (ans[p] == a[i]):
print(a[i],end=",");
else:
# Hashing collision happened increment
# position and do recursive call
p = p + 1;
placeValue(a, ans, i, p, v);
# Driver code
a = [ 7, 1, 5, 2, 3, 6 ];
b = [ 3, 8, 6, 20, 7 ];
findPosition(a, b);
# This code is contributed by mits
C#
// C# program to find union and intersection
// using similar Hashing Technique
// without using any predefined Java Collections
// Time Complexity best case & avg case = O(m+n)
// Worst case = O(nlogn)
//package com.arrays.math;
using System;
class UnsortedIntersectionUnion
{
// Prints intersection of arr1[0..n1-1] and
// arr2[0..n2-1]
public void findPosition(int []a, int []b)
{
int v = (a.Length + b.Length);
int []ans = new int[v];
int zero1 = 0;
int zero2 = 0;
Console.Write("Intersection : ");
// Iterate first array
for (int i = 0; i < a.Length; i++)
zero1 = iterateArray(a, v, ans, i);
// Iterate second array
for (int j = 0; j < b.Length; j++)
zero2 = iterateArray(b, v, ans, j);
int zero = zero1 + zero2;
placeZeros(v, ans, zero);
printUnion(v, ans, zero);
}
// Prints union of arr1[0..n1-1]
// and arr2[0..n2-1]
private void printUnion(int v, int[] ans, int zero)
{
int zero1 = 0;
Console.Write("\nUnion : ");
for (int i = 0; i < v; i++)
{
if ((zero == 0 && ans[i] == 0) ||
(ans[i] == 0 && zero1 > 0))
continue;
if (ans[i] == 0)
zero1++;
Console.Write(ans[i] + ",");
}
}
private void placeZeros(int v, int[] ans, int zero)
{
if (zero == 2)
{
Console.WriteLine("0");
int []d = { 0 };
placeValue(d, ans, 0, 0, v);
}
if (zero == 1)
{
int []d = { 0 };
placeValue(d, ans, 0, 0, v);
}
}
// Function to iterate array
private int iterateArray(int[] a, int v,
int[] ans, int i)
{
if (a[i] != 0)
{
int p = a[i] % v;
placeValue(a, ans, i, p, v);
} else
return 1;
return 0;
}
private void placeValue(int[] a, int[] ans,
int i, int p, int v)
{
p = p % v;
if (ans[p] == 0)
ans[p] = a[i];
else {
if (ans[p] == a[i])
Console.Write(a[i] + ",");
else
{
//Hashing collision happened increment
// position and do recursive call
p = p + 1;
placeValue(a, ans, i, p, v);
}
}
}
// Driver code
public static void Main()
{
int []a = { 7, 1, 5, 2, 3, 6 };
int []b = { 3, 8, 6, 20, 7 };
UnsortedIntersectionUnion uiu = new UnsortedIntersectionUnion();
uiu.findPosition(a, b);
}
}
// This code is contributed by PrinciRaj1992
Javascript
<script>
// javascript program to find union and intersection
// using similar Hashing Technique
// without using any predefined javascript Collections
// Time Complexity best case & avg case = O(m+n)
// Worst case = O(nlogn)
// package com.arrays.math;
// Prints intersection of arr1[0..n1-1] and
// arr2[0..n2-1]
function findPosition(a , b) {
var v = (a.length + b.length);
var ans = Array(v).fill(0);
var zero1 = 0;
var zero2 = 0;
document.write("Intersection : ");
// Iterate first array
for (var i = 0; i < a.length; i++)
zero1 = iterateArray(a, v, ans, i);
// Iterate second array
for (var j = 0; j < b.length; j++)
zero2 = iterateArray(b, v, ans, j);
var zero = zero1 + zero2;
placeZeros(v, ans, zero);
printUnion(v, ans, zero);
}
// Prints union of arr1[0..n1-1] and arr2[0..n2-1]
function printUnion(v, ans , zero) {
var zero1 = 0;
document.write("<br/>Union : ");
for (i = 0; i < v; i++) {
if ((zero == 0 && ans[i] == 0) || (ans[i] == 0 && zero1 > 0))
continue;
if (ans[i] == 0)
zero1++;
document.write(ans[i] + ",");
}
}
function placeZeros(v, ans , zero) {
if (zero == 2) {
document.write("0");
var d = [ 0 ];
placeValue(d, ans, 0, 0, v);
}
if (zero == 1) {
var d = [ 0 ];
placeValue(d, ans, 0, 0, v);
}
}
// Function to iterate array
function iterateArray(a , v, ans , i) {
if (a[i] != 0) {
var p = a[i] % v;
placeValue(a, ans, i, p, v);
} else
return 1;
return 0;
}
function placeValue(a, ans , i , p , v) {
p = p % v;
if (ans[p] == 0)
ans[p] = a[i];
else {
if (ans[p] == a[i])
document.write(a[i] + ",");
else {
// Hashing collision happened increment
// position and do recursive call
p = p + 1;
placeValue(a, ans, i, p, v);
}
}
}
// Driver code
var a = [ 7, 1, 5, 2, 3, 6 ];
var b = [ 3, 8, 6, 20, 7 ];
// Function call
findPosition(a, b);
// This code is contributed by gauravrajput1
</script>
Producción
Intersection : 3,6,7, Union : 1,2,3,5,6,7,8,20,
C++
// C++ program to find union and intersection
// using sets
#include <bits/stdc++.h>
using namespace std;
void printUnion(int arr1[], int arr2[], int n1, int n2)
{
// Defining set container s
set<int> s;
// Insert the elements of arr1[] to set s
for (int i = 0; i < n1; i++)
{
s.insert(arr1[i]);
}
// Insert the elements of arr2[] to set s
for (int i = 0; i < n2; i++)
{
s.insert(arr2[i]);
}
cout << "Union:" << endl;
for (auto itr = s.begin(); itr != s.end(); itr++)
// s will contain only distinct
// elements from array a and b
cout << *itr<< " ";
cout <<endl;
// Prints intersection of arr1[0..n1-1] and
// arr2[0..n2-1]
}
void printIntersection(int arr1[], int arr2[], int n1, int n2)
{
// Defining set container s
set<int> s;
// Insert the elements of arr1[] to set s
for (int i = 0; i < n1; i++)
{
s.insert(arr1[i]);
}
cout << "Intersection:" << endl;
for (int i = 0; i < n2; i++)
{
// If element is present in set then
if(s.count(arr2[i]))
{
cout<<arr2[i]<<" ";
}
}
cout <<endl;
}
// Driver Code
int main()
{
int arr1[] = { 7, 1, 5, 2, 3, 6 };
int arr2[] = { 3, 8, 6, 20, 7 };
int n1 = sizeof(arr1) / sizeof(arr1[0]);
int n2 = sizeof(arr2) / sizeof(arr2[0]);
printUnion(arr1, arr2, n1, n2);
printIntersection(arr1, arr2, n1, n2);
}
// This code is contributed by Aarti_Rathi
Java
// Java program to find union and intersection
// using sets
import java.util.*;
public class GFG {
static void printUnion(int arr1[], int arr2[], int n1,
int n2)
{
// Defining set container s
Set<Integer> s = new HashSet<Integer>();
// Insert the elements of arr1[] to set s
for (int i = 0; i < n1; i++) {
s.add(arr1[i]);
}
// Insert the elements of arr2[] to set s
for (int i = 0; i < n2; i++) {
s.add(arr2[i]);
}
System.out.println("Union");
for (int itr : s)
// s will contain only distinct
// elements from array a and b
System.out.print(itr + " ");
System.out.println();
}
static void printIntersection(int arr1[], int arr2[],
int n1, int n2)
{
// Defining set container s
Set<Integer> s = new HashSet<Integer>();
// Insert the elements of arr1[] to set s
for (int i = 0; i < n1; i++) {
s.add(arr1[i]);
}
System.out.println("Intersection");
for (int i = 0; i < n2; i++) {
// If element is present in set then
if (s.contains(arr2[i])) {
System.out.print(arr2[i] + " ");
}
}
System.out.println();
}
// Driver code
public static void main(String args[])
{
int arr1[] = { 7, 1, 5, 2, 3, 6 };
int arr2[] = { 3, 8, 6, 20, 7 };
int n1 = arr1.length;
int n2 = arr2.length;
// Function call
printUnion(arr1, arr2, n1, n2);
printIntersection(arr1, arr2, n1, n2);
}
}
// This code is contributed by Aarti_Rathi
Python3
# Python program to find union and intersection
# using sets
def printUnion(arr1, arr2, n1, n2):
hs = set()
# Insert the elements of arr1[] to set hs
for i in range(0, n1):
hs.add(arr1[i])
# Insert the elements of arr1[] to set hs
for i in range(0, n2):
hs.add(arr2[i])
print("Union:")
for i in hs:
print(i, end=" ")
print("\n")
# Prints intersection of arr1[0..n1-1] and
# arr2[0..n2-1]
def printIntersection(arr1, arr2, n1, n2):
hs = set()
# Insert the elements of arr1[] to set S
for i in range(0, n1):
hs.add(arr1[i])
print("Intersection:")
for i in range(0, n2):
# If element is present in set then
# push it to vector V
if arr2[i] in hs:
print(arr2[i], end=" ")
# Driver Program
arr1 = [7, 1, 5, 2, 3, 6]
arr2 = [3, 8, 6, 20, 7]
n1 = len(arr1)
n2 = len(arr2)
# Function call
printUnion(arr1, arr2, n1, n2)
printIntersection(arr1, arr2, n1, n2)
# This code is contributed by Kumar Suman .
C#
// C# program to find union and intersection
// using sets
using System;
using System.Collections.Generic;
public class GFG {
static void printUnion(int[] arr1, int[] arr2, int n1,
int n2)
{
// Defining set container s
SortedSet<int> s = new SortedSet<int>();
// Insert the elements of arr1[] to set s
for (int i = 0; i < n1; i++) {
s.Add(arr1[i]);
}
// Insert the elements of arr2[] to set s
for (int i = 0; i < n2; i++) {
s.Add(arr2[i]);
}
Console.WriteLine("Union");
foreach(var itr in s)
// s will contain only distinct
// elements from array a and b
Console.Write(itr + " ");
Console.WriteLine();
}
static void printIntersection(int[] arr1, int[] arr2,
int n1, int n2)
{
// Defining set container s
SortedSet<int> s = new SortedSet<int>();
// Insert the elements of arr1[] to set s
for (int i = 0; i < n1; i++) {
s.Add(arr1[i]);
}
Console.WriteLine("Intersection");
for (int i = 0; i < n2; i++) {
// If element is present in set then
if (s.Contains(arr2[i])) {
Console.Write(arr2[i] + " ");
}
}
Console.WriteLine();
}
// Driver code
public static void Main(string[] args)
{
int[] arr1 = { 7, 1, 5, 2, 3, 6 };
int[] arr2 = { 3, 8, 6, 20, 7 };
int n1 = arr1.Length;
int n2 = arr2.Length;
// Function call
printUnion(arr1, arr2, n1, n2);
printIntersection(arr1, arr2, n1, n2);
}
}
// This code is contributed by phasing17
Producción
Union: 1 2 3 5 6 7 8 20 Intersection: 3 6 7
Consulte la siguiente publicación para arrays ordenadas. Encuentra la unión y la intersección de dos arrays ordenadas
Publicación traducida automáticamente
Artículo escrito por GeeksforGeeks-1 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA