Dado un arreglo arr[] , la tarea es imprimir el conteo de pares únicos (arr[i], arr[j]) tales que i < j .
Ejemplos:
Entrada: arr[] = {1, 2, 1, 4, 5, 2}
Salida: 11
Los pares posibles son (1, 2), (1, 1), (1, 4), (1, 5), (2, 1), (2, 4), (2, 5), (2, 2), (4, 5), (4, 2), (5, 2)Entrada: arr[] = {1, 2, 3, 4}
Salida: 6
Los pares posibles son (1, 2), (1, 3), (1, 4), (2, 3), (2, 4) ), (3, 4)
Enfoque ingenuo: la forma más fácil es iterar a través de cada par posible y, si cumple la condición, agregarlo a un conjunto. Luego, podemos devolver el tamaño del conjunto como nuestra respuesta.
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ implementation of the approach
#include <iostream>
#include <set>
using namespace std;
// Function to return the count
// of unique pairs in the array
int getPairs(int arr[], int n)
{
// Set to store unique pairs
set<pair<int, int>> h;
for(int i = 0; i < (n - 1); i++)
{
for (int j = i + 1; j < n; j++)
{
// Create pair of (arr[i], arr[j])
// and add it to the hashset
h.insert(make_pair(arr[i], arr[j]));
}
}
// Return the size of the HashSet
return h.size();
}
// Driver code
int main()
{
int arr[] = { 1, 2, 2, 4, 2, 5, 3, 5 };
int n = sizeof(arr) / sizeof(arr[0]);
printf("%d", getPairs(arr, n)) ;
return 0;
}
// This code is contributed by SHUBHAMSINGH10
Java
// Java implementation of the approach
import java.util.HashSet;
import javafx.util.Pair;
class GFG {
// Function to return the count
// of unique pairs in the array
static int getPairs(int arr[], int n)
{
// HashSet to store unique pairs
HashSet<Pair> h = new HashSet<Pair>();
for (int i = 0; i < n - 1; i++) {
for (int j = i + 1; j < n; j++) {
// Create pair of (a[i], a[j])
// and add it to the hashset
Pair<Integer, Integer> p
= new Pair<>(arr[i], arr[j]);
h.add(p);
}
}
// Return the size of the HashSet
return h.size();
}
// Driver code
public static void main(String[] args)
{
int arr[] = { 1, 2, 2, 4, 2, 5, 3, 5 };
int n = arr.length;
System.out.println(getPairs(arr, n));
}
}
Python3
# Python3 implementation of the approach # Function to return the count # of unique pairs in the array def getPairs(arr, n) : # Set to store unique pairs h = set() for i in range(n - 1) : for j in range(i + 1, n) : # Create pair of (a[i], a[j]) # and add it to the hashset h.add((arr[i], arr[j])); # Return the size of the HashSet return len(h); # Driver code if __name__ == "__main__" : arr = [ 1, 2, 2, 4, 2, 5, 3, 5 ] n = len(arr) print(getPairs(arr, n)) # This code is contributed by Ryuga
C#
// C# implementation of the approach
using System;
using System.Collections.Generic;
class GFG{
// Function to return the count
// of unique pairs in the array
static int getPairs(int []arr, int n)
{
// HashSet to store unique pairs
HashSet<Tuple<int,
int>> h = new HashSet<Tuple<int,
int>>();
for(int i = 0; i < n - 1; i++)
{
for(int j = i + 1; j < n; j++)
{
// Create pair of (a[i], a[j])
// and add it to the hashset
Tuple<int,
int> p = new Tuple<int,
int>(arr[i],
arr[j]);
h.Add(p);
}
}
// Return the size of the HashSet
return h.Count;
}
// Driver code
public static void Main(String[] args)
{
int []arr = { 1, 2, 2, 4, 2, 5, 3, 5 };
int n = arr.Length;
Console.WriteLine(getPairs(arr, n));
}
}
// This code is contributed by Amit Katiyar
Javascript
<script>
// JavaScript implementation of the approach
// Function to return the count
// of unique pairs in the array
function getPairs(arr, n)
{
// Set to store unique pairs
var h = new Set();
for(var i = 0; i < (n - 1); i++)
{
for (var j = i + 1; j < n; j++)
{
// Create pair of (arr[i], arr[j])
// and add it to the hashset
h.add([arr[i], arr[j]]);
}
}
// Return the size of the HashSet
return h.size/2;
}
// Driver code
var arr = [ 1, 2, 2, 4, 2, 5, 3, 5 ];
var n = arr.length;
document.write(getPairs(arr, n));
// This code is contributed by SHUBHAMSINGH10
</script>
Producción:
14
Complejidad temporal: O(n 2 )
Nota: utilice un IDE sin conexión para compilar el código anterior. Es posible que los compiladores en línea no admitan JavaFX.
Enfoque eficiente: cada elemento arr[i] puede formar un par con el elemento arr[j] si i < j . Pero (arr[i], arr[j]) debe ser único, por lo tanto, para cada único arr[i] , los posibles pares serán iguales al número de números distintos en el subarreglo arr[i + 1], arr[i + 2], …, arr[n – 1] . Entonces, para cada arr[i] , encontraremos los elementos únicos de derecha a izquierda. Para esta tarea, es fácil realizar un seguimiento de los elementos visitados mediante una tabla Hash. De esta forma, tendremos un único arr[i] para cada único arr[j] . Ahora, sumaremos estos valores para cada arr[i] únicoque es el número deseado de pares.
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ implementation of the approach
#include<bits/stdc++.h>
using namespace std;
// Function to return the count
// of unique pairs in the array
int getPairs(int a[], int n)
{
set<int> visited1;
// un[i] stores number of unique elements
// from un[i + 1] to un[n - 1]
int un[n] ;
// Last element will have no unique elements
// after it
un[n - 1] = 0;
// To count unique elements after every a[i]
int count = 0;
// auto pos = s.find(3);
// prints the set elements
// cout << "The set elements after 3 are: ";
// for (auto it = pos; it != s.end(); it++)
// cout << *it << " "
for (int i = n - 1; i > 0; i--)
{
// If current element has already been used
// i.e. not unique
auto pos = visited1.find(a[i]);
if (pos != visited1.end())
un[i - 1] = count;
else
un[i - 1] = ++count;
// Set to true if a[i] is visited
visited1.insert(a[i]);
}
set<int>visited2;
// To know which a[i] is already visited
int answer = 0;
for (int i = 0; i < n - 1; i++)
{
// If visited, then the pair would
// not be unique
auto pos = visited2.find(a[i]);
if (pos != visited2.end())
continue;
// Calculating total unique pairs
answer += un[i];
// Set to true if a[i] is visited
visited2.insert(a[i]);
}
return answer;
}
// Driver code
int main()
{
int a[] = { 1, 2, 2, 4, 2, 5, 3, 5 };
int n = sizeof(a)/sizeof(a[0]);
// Print the count of unique pairs
cout<<(getPairs(a, n));
}
// This code is contributed by Rajput-Ji
Java
// Java implementation of the approach
import java.util.HashSet;
public class GFG {
// Function to return the count
// of unique pairs in the array
static int getPairs(int a[], int n)
{
HashSet<Integer> visited1 = new HashSet<Integer>();
// un[i] stores number of unique elements
// from un[i + 1] to un[n - 1]
int un[] = new int[n];
// Last element will have no unique elements
// after it
un[n - 1] = 0;
// To count unique elements after every a[i]
int count = 0;
for (int i = n - 1; i > 0; i--) {
// If current element has already been used
// i.e. not unique
if (visited1.contains(a[i]))
un[i - 1] = count;
else
un[i - 1] = ++count;
// Set to true if a[i] is visited
visited1.add(a[i]);
}
HashSet<Integer> visited2 = new HashSet<Integer>();
// To know which a[i] is already visited
int answer = 0;
for (int i = 0; i < n - 1; i++) {
// If visited, then the pair would
// not be unique
if (visited2.contains(a[i]))
continue;
// Calculating total unique pairs
answer += un[i];
// Set to true if a[i] is visited
visited2.add(a[i]);
}
return answer;
}
// Driver code
public static void main(String[] args)
{
int a[] = { 1, 2, 2, 4, 2, 5, 3, 5 };
int n = a.length;
// Print the count of unique pairs
System.out.println(getPairs(a, n));
}
}
Python3
# Python3 implementation of the approach # Function to return the count # of unique pairs in the array def getPairs(a, n): visited1 = set() # un[i] stores number of unique elements # from un[i + 1] to un[n - 1] un = [0] * n # Last element will have no unique elements # after it un[n - 1] = 0 # To count unique elements after every a[i] count = 0 for i in range(n - 1, -1, -1): # If current element has already been used # i.e. not unique if (a[i] in visited1): un[i - 1] = count else: count += 1 un[i - 1] = count # Set to true if a[i] is visited visited1.add(a[i]) visited2 = set() # To know which a[i] is already visited answer = 0 for i in range(n - 1): # If visited, then the pair would # not be unique if (a[i] in visited2): continue # Calculating total unique pairs answer += un[i] # Set to true if a[i] is visited visited2.add(a[i]) return answer # Driver code a = [1, 2, 2, 4, 2, 5, 3, 5] n = len(a) # Print the count of unique pairs print(getPairs(a, n)) # This code is contributed by SHUBHAMSINGH10
C#
// C# implementation of the approach
using System;
using System.Collections.Generic;
class GFG
{
// Function to return the count
// of unique pairs in the array
static int getPairs(int []a, int n)
{
HashSet<int> visited1 = new HashSet<int>();
// un[i] stores number of unique elements
// from un[i + 1] to un[n - 1]
int []un = new int[n];
// Last element will have no unique elements
// after it
un[n - 1] = 0;
// To count unique elements after every a[i]
int count = 0;
for (int i = n - 1; i > 0; i--)
{
// If current element has already been used
// i.e. not unique
if (visited1.Contains(a[i]))
un[i - 1] = count;
else
un[i - 1] = ++count;
// Set to true if a[i] is visited
visited1.Add(a[i]);
}
HashSet<int> visited2 = new HashSet<int>();
// To know which a[i] is already visited
int answer = 0;
for (int i = 0; i < n - 1; i++)
{
// If visited, then the pair would
// not be unique
if (visited2.Contains(a[i]))
continue;
// Calculating total unique pairs
answer += un[i];
// Set to true if a[i] is visited
visited2.Add(a[i]);
}
return answer;
}
// Driver code
public static void Main(String[] args)
{
int []a = { 1, 2, 2, 4, 2, 5, 3, 5 };
int n = a.Length;
// Print the count of unique pairs
Console.WriteLine(getPairs(a, n));
}
}
/* This code contributed by PrinciRaj1992 */
Javascript
<script>
// JavaScript Program to implement
// the above approach
// Function to return the count
// of unique pairs in the array
function getPairs(a, n)
{
let visited1 = new Set();
// un[i] stores number of unique elements
// from un[i + 1] to un[n - 1]
let un = Array.from({length: n}, (_, i) => 0);
// Last element will have no unique elements
// after it
un[n - 1] = 0;
// To count unique elements after every a[i]
let count = 0;
for (let i = n - 1; i > 0; i--) {
// If current element has already been used
// i.e. not unique
if (visited1.has(a[i]))
un[i - 1] = count;
else
un[i - 1] = ++count;
// Set to true if a[i] is visited
visited1.add(a[i]);
}
let visited2 = new Set();
// To know which a[i] is already visited
let answer = 0;
for (let i = 0; i < n - 1; i++) {
// If visited, then the pair would
// not be unique
if (visited2.has(a[i]))
continue;
// Calculating total unique pairs
answer += un[i];
// Set to true if a[i] is visited
visited2.add(a[i]);
}
return answer;
}
// Driver Code
let a = [ 1, 2, 2, 4, 2, 5, 3, 5 ];
let n = a.length;
// Print the count of unique pairs
document.write(getPairs(a, n));
</script>
Producción:
14
Complejidad de tiempo: O (nlogn)