Dadas dos arrays desordenadas de elementos distintos, la tarea es encontrar todos los pares de ambas arrays cuya suma sea igual a X .
Ejemplos:
C++
// C++ program to find all pairs in both arrays
// whose sum is equal to given value x
#include <bits/stdc++.h>
using namespace std;
// Function to print all pairs in both arrays
// whose sum is equal to given value x
void findPairs(int arr1[], int arr2[], int n, int m, int x)
{
for (int i = 0; i < n; i++)
for (int j = 0; j < m; j++)
if (arr1[i] + arr2[j] == x)
cout << arr1[i] << " " << arr2[j] << endl;
}
// Driver code
int main()
{
int arr1[] = { 1, 2, 3, 7, 5, 4 };
int arr2[] = { 0, 7, 4, 3, 2, 1 };
int n = sizeof(arr1) / sizeof(int);
int m = sizeof(arr2) / sizeof(int);
int x = 8;
findPairs(arr1, arr2, n, m, x);
return 0;
}
// This code is contributed by Aditya Kumar (adityakumar129)
C
// C program to find all pairs in both arrays
// whose sum is equal to given value x
#include <stdio.h>
// Function to print all pairs in both arrays
// whose sum is equal to given value x
void findPairs(int arr1[], int arr2[], int n, int m, int x)
{
for (int i = 0; i < n; i++)
for (int j = 0; j < m; j++)
if (arr1[i] + arr2[j] == x)
printf("%d %d\n", arr1[i], arr2[j]);
}
// Driver code
int main()
{
int arr1[] = { 1, 2, 3, 7, 5, 4 };
int arr2[] = { 0, 7, 4, 3, 2, 1 };
int n = sizeof(arr1) / sizeof(int);
int m = sizeof(arr2) / sizeof(int);
int x = 8;
findPairs(arr1, arr2, n, m, x);
return 0;
}
// This code is contributed by Aditya Kumar (adityakumar129)
Java
// Java program to find all pairs in both arrays
// whose sum is equal to given value x
import java.io.*;
class GFG {
// Function to print all pairs in both arrays
// whose sum is equal to given value x
static void findPairs(int arr1[], int arr2[], int n,
int m, int x)
{
for (int i = 0; i < n; i++)
for (int j = 0; j < m; j++)
if (arr1[i] + arr2[j] == x)
System.out.println(arr1[i] + " "
+ arr2[j]);
}
// Driver code
public static void main(String[] args)
{
int arr1[] = { 1, 2, 3, 7, 5, 4 };
int arr2[] = { 0, 7, 4, 3, 2, 1 };
int x = 8;
findPairs(arr1, arr2, arr1.length, arr2.length, x);
}
}
// This code is contributed
// by sunnysingh
Python3
# Python 3 program to find all # pairs in both arrays whose # sum is equal to given value x # Function to print all pairs # in both arrays whose sum is # equal to given value x def findPairs(arr1, arr2, n, m, x): for i in range(0, n): for j in range(0, m): if (arr1[i] + arr2[j] == x): print(arr1[i], arr2[j]) # Driver code arr1 = [1, 2, 3, 7, 5, 4] arr2 = [0, 7, 4, 3, 2, 1] n = len(arr1) m = len(arr2) x = 8 findPairs(arr1, arr2, n, m, x) # This code is contributed by Smitha Dinesh Semwal
C#
// C# program to find all
// pairs in both arrays
// whose sum is equal to
// given value x
using System;
class GFG {
// Function to print all
// pairs in both arrays
// whose sum is equal to
// given value x
static void findPairs(int[] arr1, int[] arr2,
int n, int m, int x)
{
for (int i = 0; i < n; i++)
for (int j = 0; j < m; j++)
if (arr1[i] + arr2[j] == x)
Console.WriteLine(arr1[i] + " " + arr2[j]);
}
// Driver code
static void Main()
{
int[] arr1 = { 1, 2, 3, 7, 5, 4 };
int[] arr2 = { 0, 7, 4, 3, 2, 1 };
int x = 8;
findPairs(arr1, arr2,
arr1.Length,
arr2.Length, x);
}
}
// This code is contributed
// by Sam007
PHP
<?php
// PHP program to find all pairs
// in both arrays whose sum is
// equal to given value x
// Function to print all pairs
// in both arrays whose sum is
// equal to given value x
function findPairs($arr1, $arr2,
$n, $m, $x)
{
for ($i = 0; $i < $n; $i++)
for ($j = 0; $j < $m; $j++)
if ($arr1[$i] + $arr2[$j] == $x)
echo $arr1[$i] . " " .
$arr2[$j] . "\n";
}
// Driver code
$arr1 = array(1, 2, 3, 7, 5, 4);
$arr2 = array(0, 7, 4, 3, 2, 1);
$n = count($arr1);
$m = count($arr2);
$x = 8;
findPairs($arr1, $arr2,
$n, $m, $x);
// This code is contributed
// by Sam007
?>
Javascript
<script>
// Javascript program to find all pairs in both arrays
// whose sum is equal to given value x
// Function to print all pairs in both arrays
// whose sum is equal to given value x
function findPairs(arr1, arr2, n, m, x)
{
for (let i = 0; i < n; i++)
for (let j = 0; j < m; j++)
if (arr1[i] + arr2[j] == x)
document.write(arr1[i] + " "
+ arr2[j] + "<br>");
}
// Driver code
let arr1 = [ 1, 2, 3, 7, 5, 4 ];
let arr2 = [ 0, 7, 4, 3, 2, 1 ];
let n = arr1.length;
let m = arr2.length;
let x = 8;
findPairs(arr1, arr2, n, m, x);
// This code is contributed by Surbhi Tyagi.
</script>
C++
#include<bits/stdc++.h>
using namespace std;
void heapify(int a[] , int n , int i)
{
int rootLargest = i;
int lchild = 2 * i;
int rchild = (2 * i) + 1;
if (lchild < n && a[lchild] > a[rootLargest])
rootLargest = lchild;
if (rchild < n && a[rchild] > a[rootLargest])
rootLargest = rchild;
if (rootLargest != i)
{
swap(a[i] , a[rootLargest]);
heapify(a , n , rootLargest);
}
}
int binarySearch(int a[] , int l , int r , int x)
{
while (l <= r)
{
int m = l + (r - l) / 2;
if (a[m] == x)
return m;
if (a[m] < x)
l = m + 1;
else
r = m - 1;
}
return -1;
}
int main()
{
int A[] = {1,2,1,3,4};
int B[] = {3,1,5,1,2};
int K = 8;
int n = sizeof(A) / sizeof(A[0]);
// Building the heap
for (int i = n / 2 - 1 ; i >= 1; i--)
heapify(A , n , i);
for(int i=0 ; i<n ; i++) //O(n)
{
int temp = K - B[i]; //O(1)
if(binarySearch(A , 0 , n-1 , temp)) //O(logn)
{
cout<<"\nFound the elements.\n";
break;
}
}
return 0;
}
// This code is contributed by Aditya Kumar (adityakumar129)
C
#include <stdio.h>
#include <stdlib.h>
void heapify(int a[], int n, int i)
{
int rootLargest = i;
int lchild = 2 * i;
int rchild = (2 * i) + 1;
if (lchild < n && a[lchild] > a[rootLargest])
rootLargest = lchild;
if (rchild < n && a[rchild] > a[rootLargest])
rootLargest = rchild;
if (rootLargest != i) {
int temp = a[i];
a[i] = a[rootLargest];
a[rootLargest] = temp;
heapify(a, n, rootLargest);
}
}
int binarySearch(int a[], int l, int r, int x)
{
while (l <= r) {
int m = l + (r - l) / 2;
if (a[m] == x)
return m;
if (a[m] < x)
l = m + 1;
else
r = m - 1;
}
return -1;
}
int main()
{
int A[] = { 1, 2, 1, 3, 4 };
int B[] = { 3, 1, 5, 1, 2 };
int K = 8;
int n = sizeof(A) / sizeof(A[0]);
// Building the heap
for (int i = n / 2 - 1; i >= 1; i--)
heapify(A, n, i);
for (int i = 0; i < n; i++) // O(n)
{
int temp = K - B[i]; // O(1)
if (binarySearch(A, 0, n - 1, temp)) // O(logn)
{
printf("\nFound the elements.\n");
break;
}
}
return 0;
}
// This code is contributed by Aditya Kumar (adityakumar129)
Java
import java.util.*;
class GFG{
static int A[] = {1,2,1,3,4};
static void heapify( int n , int i)
{
int rootLargest = i;
int lchild = 2 * i;
int rchild = (2 * i) + 1;
if (lchild < n && A[lchild] > A[rootLargest])
rootLargest = lchild;
if (rchild < n && A[rchild] > A[rootLargest])
rootLargest = rchild;
if (rootLargest != i)
{
int t = A[i];
A[i] = A[rootLargest];
A[rootLargest] = t;
//Recursion
heapify( n , rootLargest);
}
}
static int binarySearch( int l , int r , int x)
{
while (l <= r)
{
int m = l + (r - l) / 2;
if (A[m] == x)
return m;
if (A[m] < x)
l = m + 1;
else
r = m - 1;
}
return -1;
}
public static void main(String[] args)
{
int B[] = {3,1,5,1,2};
int K = 8;
int n = A.length;
// Building the heap
for (int i = n / 2 - 1 ; i >= 1; i--)
heapify( n , i);
for(int i = 0; i < n; i++) //O(n)
{
int temp = K - B[i]; //O(1)
if(binarySearch(0, n - 1, temp - 1) != -1) //O(logn)
{
System.out.print("\nFound the elements.\n");
break;
}
}
}
}
// This code is contributed by Rajput-Ji
Python3
A = [ 1, 2, 1, 3, 4 ];
def heapify(n, i):
rootLargest = i;
lchild = 2 * i;
rchild = (2 * i) + 1;
if (lchild < n and A[lchild] > A[rootLargest]):
rootLargest = lchild;
if (rchild < n and A[rchild] > A[rootLargest]):
rootLargest = rchild;
if (rootLargest != i):
t = A[i];
A[i] = A[rootLargest];
A[rootLargest] = t;
# Recursion
heapify(n, rootLargest);
def binarySearch(l, r, x):
while (l <= r):
m = l + (r - l) // 2;
if (A[m] == x):
return m;
if (A[m] < x):
l = m + 1;
else:
r = m - 1;
return -1;
if __name__ == '__main__':
B = [ 3, 1, 5, 1, 2 ];
K = 8;
n = len(A);
# Building the heap
for i in range(n// 2 - 1,0, -1):
heapify(n, i);
for i in range(n):
temp = K - B[i];
if (binarySearch(0, n - 1, temp - 1) != -1):
print("\nFound the elements.");
break;
# This code is contributed by Rajput-Ji
C#
using System;
public class GFG {
static int []A = { 1, 2, 1, 3, 4 };
static void heapify(int n, int i) {
int rootLargest = i;
int lchild = 2 * i;
int rchild = (2 * i) + 1;
if (lchild < n && A[lchild] > A[rootLargest])
rootLargest = lchild;
if (rchild < n && A[rchild] > A[rootLargest])
rootLargest = rchild;
if (rootLargest != i) {
int t = A[i];
A[i] = A[rootLargest];
A[rootLargest] = t;
// Recursion
heapify(n, rootLargest);
}
}
static int binarySearch(int l, int r, int x) {
while (l <= r) {
int m = l + (r - l) / 2;
if (A[m] == x)
return m;
if (A[m] < x)
l = m + 1;
else
r = m - 1;
}
return -1;
}
public static void Main(String[] args) {
int []B = { 3, 1, 5, 1, 2 };
int K = 8;
int n = A.Length;
// Building the heap
for (int i = n / 2 - 1; i >= 1; i--)
heapify(n, i);
for (int i = 0; i < n; i++) // O(n)
{
int temp = K - B[i]; // O(1)
if (binarySearch(0, n - 1, temp - 1) != -1) // O(logn)
{
Console.Write("\nFound the elements.\n");
break;
}
}
}
}
// This code is contributed by Rajput-Ji
Javascript
<script>
function heapify(a,n,i)
{
let rootLargest = i;
let lchild = 2 * i;
let rchild = (2 * i) + 1;
if (lchild < n && a[lchild] > a[rootLargest])
rootLargest = lchild;
if (rchild < n && a[rchild] > a[rootLargest])
rootLargest = rchild;
if (rootLargest != i)
{
swap(a[i] , a[rootLargest]);
//Recursion
heapify(a , n , rootLargest);
}
}
function binarySearch(a,l,r,x)
{
while (l <= r)
{
let m = l + (r - l) / 2;
if (a[m] == x)
return m;
if (a[m] < x)
l = m + 1;
else
r = m - 1;
}
return -1;
}
let A = [1,2,1,3,4];
let B = [3,1,5,1,2];
let K = 8;
let n = A.length;
// Building the heap
for (let i = n / 2 - 1 ; i >= 1; i--)
heapify(A , n , i);
for(let i=0 ; i<n ; i++) //O(n)
{
let temp = K - B[i]; //O(1)
if(binarySearch(A , 0 , n-1 , temp)) //O(logn)
{
document.write("\nFound the elements.\n");
break;
}
}
</script>
C++
// C++ program to find all pair in both arrays
// whose sum is equal to given value x
#include <bits/stdc++.h>
using namespace std;
// Function to find all pairs in both arrays
// whose sum is equal to given value x
void findPairs(int arr1[], int arr2[], int n,
int m, int x)
{
// Insert all elements of first array in a hash
unordered_set<int> s;
for (int i = 0; i < n; i++)
s.insert(arr1[i]);
// Subtract sum from second array elements one
// by one and check it's present in array first
// or not
for (int j = 0; j < m; j++)
if (s.find(x - arr2[j]) != s.end())
cout << x - arr2[j] << " "
<< arr2[j] << endl;
}
// Driver code
int main()
{
int arr1[] = { 1, 0, -4, 7, 6, 4 };
int arr2[] = { 0, 2, 4, -3, 2, 1 };
int x = 8;
int n = sizeof(arr1) / sizeof(int);
int m = sizeof(arr2) / sizeof(int);
findPairs(arr1, arr2, n, m, x);
return 0;
}
Java
// JAVA Code for Given two unsorted arrays,
// find all pairs whose sum is x
import java.util.*;
class GFG {
// Function to find all pairs in both arrays
// whose sum is equal to given value x
public static void findPairs(int arr1[], int arr2[],
int n, int m, int x)
{
// Insert all elements of first array in a hash
HashMap<Integer, Integer> s = new HashMap<Integer, Integer>();
for (int i = 0; i < n; i++)
s.put(arr1[i], 0);
// Subtract sum from second array elements one
// by one and check it's present in array first
// or not
for (int j = 0; j < m; j++)
if (s.containsKey(x - arr2[j]))
System.out.println(x - arr2[j] + " " + arr2[j]);
}
/* Driver program to test above function */
public static void main(String[] args)
{
int arr1[] = { 1, 0, -4, 7, 6, 4 };
int arr2[] = { 0, 2, 4, -3, 2, 1 };
int x = 8;
findPairs(arr1, arr2, arr1.length, arr2.length, x);
}
}
// This code is contributed by Arnav Kr. Mandal.
Python3
# Python3 program to find all # pair in both arrays whose # sum is equal to given value x # Function to find all pairs # in both arrays whose sum is # equal to given value x def findPairs(arr1, arr2, n, m, x): # Insert all elements of # first array in a hash s = set() for i in range (0, n): s.add(arr1[i]) # Subtract sum from second # array elements one by one # and check it's present in # array first or not for j in range(0, m): if ((x - arr2[j]) in s): print((x - arr2[j]), '', arr2[j]) # Driver code arr1 = [1, 0, -4, 7, 6, 4] arr2 = [0, 2, 4, -3, 2, 1] x = 8 n = len(arr1) m = len(arr2) findPairs(arr1, arr2, n, m, x) # This code is contributed # by ihritik
C#
// C# Code for Given two unsorted arrays,
// find all pairs whose sum is x
using System;
using System.Collections.Generic;
class GFG {
// Function to find all pairs in
// both arrays whose sum is equal
// to given value x
public static void findPairs(int[] arr1, int[] arr2,
int n, int m, int x)
{
// Insert all elements of first
// array in a hash
Dictionary<int,
int>
s = new Dictionary<int,
int>();
for (int i = 0; i < n; i++) {
s[arr1[i]] = 0;
}
// Subtract sum from second array
// elements one by one and check
// it's present in array first
// or not
for (int j = 0; j < m; j++) {
if (s.ContainsKey(x - arr2[j])) {
Console.WriteLine(x - arr2[j] + " " + arr2[j]);
}
}
}
// Driver Code
public static void Main(string[] args)
{
int[] arr1 = new int[] { 1, 0, -4, 7, 6, 4 };
int[] arr2 = new int[] { 0, 2, 4, -3, 2, 1 };
int x = 8;
findPairs(arr1, arr2, arr1.Length,
arr2.Length, x);
}
}
// This code is contributed by Shrikant13
Javascript
<script>
// Javascript Code for Given two unsorted arrays,
// find all pairs whose sum is x
// Function to find all pairs in both arrays
// whose sum is equal to given value x
function findPairs(arr1, arr2, n, m, x)
{
// Insert all elements of first array in a hash
let s = new Map();
for (let i = 0; i < n; i++)
s.set(arr1[i], 0);
// Subtract sum from second array elements one
// by one and check it's present in array first
// or not
for (let j = 0; j < m; j++)
if (s.has(x - arr2[j]))
document.write(x - arr2[j] + " " + arr2[j] + "<br/>");
}
// Driver code
let arr1 = [ 1, 0, -4, 7, 6, 4 ];
let arr2 = [ 0, 2, 4, -3, 2, 1 ];
let x = 8;
findPairs(arr1, arr2, arr1.length, arr2.length, x);
</script>
Publicación traducida automáticamente
Artículo escrito por GeeksforGeeks-1 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA