# Find a pair of elements swapping which makes sum of two arrays same

Given two arrays of integers, find a pair of values (one value from each array) that you can swap to give the two arrays the same sum.
Examples:

Input: A[] = {4, 1, 2, 1, 1, 2}, B[] = (3, 6, 3, 3)
Output: {1, 3}
Sum of elements in A[] = 11
Sum of elements in B[] = 15
To get same sum from both arrays, we
can swap following values:
1 from A[] and 3 from B[]
Input: A[] = {5, 7, 4, 6}, B[] = {1, 2, 3, 8}
Output: 6 2

Method 1 (Naive Implementation)
Iterate through the arrays and check all pairs of values. Compare new sums or look for a pair with that difference.

## C++

 `// CPP code naive solution to find a pair swapping``// which makes sum of arrays sum.``#include ``using` `namespace` `std;` `// Function to calculate sum of elements of array``int` `getSum(``int` `X[], ``int` `n)``{``    ``int` `sum = 0;``    ``for` `(``int` `i = 0; i < n; i++)``        ``sum += X[i];``    ``return` `sum;``}` `void` `findSwapValues(``int` `A[], ``int` `n, ``int` `B[], ``int` `m)``{``    ``// Calculation of sums from both arrays``    ``int` `sum1 = getSum(A, n);``    ``int` `sum2 = getSum(B, m);` `    ``// Look for val1 and val2, such that``    ``// sumA - val1 + val2 = sumB - val2 + val1``    ``int` `newsum1, newsum2, val1, val2;``    ``for` `(``int` `i = 0; i < n; i++) {``        ``for` `(``int` `j = 0; j < m; j++) {``            ``newsum1 = sum1 - A[i] + B[j];``            ``newsum2 = sum2 - B[j] + A[i];``            ``if` `(newsum1 == newsum2) {``                ``val1 = A[i];``                ``val2 = B[j];``            ``}``        ``}``    ``}` `    ``cout << val1 << ``" "` `<< val2;``}` `// Driver code``int` `main()``{``    ``int` `A[] = { 4, 1, 2, 1, 1, 2 };``    ``int` `n = ``sizeof``(A) / ``sizeof``(A[0]);``    ``int` `B[] = { 3, 6, 3, 3 };``    ``int` `m = ``sizeof``(B) / ``sizeof``(B[0]);` `    ``// Call to function``    ``findSwapValues(A, n, B, m);``    ``return` `0;``}`

## Java

 `// Java program to find a pair swapping``// which makes sum of arrays sum``import` `java.io.*;` `class` `GFG``{``    ``// Function to calculate sum of elements of array``    ``static` `int` `getSum(``int` `X[], ``int` `n)``    ``{``        ``int` `sum = ``0``;``        ``for` `(``int` `i = ``0``; i < n; i++)``            ``sum += X[i];``        ``return` `sum;``    ``}``    ` `    ``// Function to prints elements to be swapped``    ``static` `void` `findSwapValues(``int` `A[], ``int` `n, ``int` `B[], ``int` `m)``    ``{``        ``// Calculation of sums from both arrays``        ``int` `sum1 = getSum(A, n);``        ``int` `sum2 = getSum(B, m);`` ` `        ``// Look for val1 and val2, such that``        ``// sumA - val1 + val2 = sumB - val2 + val1``        ``int` `newsum1, newsum2, val1 = ``0``, val2 = ``0``;``        ``for` `(``int` `i = ``0``; i < n; i++)``        ``{``            ``for` `(``int` `j = ``0``; j < m; j++)``            ``{``                ``newsum1 = sum1 - A[i] + B[j];``                ``newsum2 = sum2 - B[j] + A[i];``                ``if` `(newsum1 == newsum2)``                ``{``                    ``val1 = A[i];``                    ``val2 = B[j];``                ``}``            ``}``        ``}`` ` `        ``System.out.println(val1+``" "``+val2);``    ``}``    ` `    ``// driver program``    ``public` `static` `void` `main (String[] args)``    ``{``        ``int` `A[] = { ``4``, ``1``, ``2``, ``1``, ``1``, ``2` `};``        ``int` `n = A.length;``        ``int` `B[] = { ``3``, ``6``, ``3``, ``3` `};``        ``int` `m = B.length;`` ` `        ``// Call to function``        ``findSwapValues(A, n, B, m);``    ``}``}` `// Contributed by Pramod Kumar`

## Python3

 `# Python code naive solution to find a pair swapping``# which makes sum of lists sum.` `# Function to calculate sum of elements of list``def` `getSum(X):``    ``sum``=``0``    ``for` `i ``in` `X:``        ``sum``+``=``i``    ``return` `sum` `# Function to prints elements to be swapped``def` `findSwapValues(A,B):``    ``# Calculation if sums from both lists``    ``sum1``=``getSum(A)``    ``sum2``=``getSum(B)` `    ``# Boolean variable used to reduce further iterations``    ``# after the pair is found``    ``k``=``False` `    ``# Lool for val1 and val2, such that``    ``# sumA - val1 + val2 = sumB -val2 + val1``    ``val1,val2``=``0``,``0``    ``for` `i ``in` `A:``        ``for` `j ``in` `B:``            ``newsum1``=``sum1``-``i``+``j``            ``newsum2``=``sum2``-``j``+``i``            ` `            ``if` `newsum1 ``=``=``newsum2:``                ``val1``=``i``                ``val2``=``j``                ``# Set to True when pair is found``                ``k``=``True``                ``break``        ``# If k is True, it means pair is found.``        ``# So, no further iterations.``        ``if` `k``=``=``True``:``            ``break``    ``print` `(val1,val2)``    ``return`  `# Driver code``A``=``[``4``,``1``,``2``,``1``,``1``,``2``]``B``=``[``3``,``6``,``3``,``3``]` `# Call to function``findSwapValues(A,B)` `# code contributed by sachin bisht`

## C#

 `// C# program to find a pair swapping``// which makes sum of arrays sum``using` `System;` `class` `GFG``{``// Function to calculate sum``// of elements of array``static` `int` `getSum(``int``[] X, ``int` `n)``{``    ``int` `sum = 0;``    ``for` `(``int` `i = 0; i < n; i++)``        ``sum += X[i];``    ``return` `sum;``}` `// Function to prints elements``// to be swapped``static` `void` `findSwapValues(``int``[] A, ``int` `n,``                           ``int``[] B, ``int` `m)``{``    ``// Calculation of sums from``    ``// both arrays``    ``int` `sum1 = getSum(A, n);``    ``int` `sum2 = getSum(B, m);` `    ``// Look for val1 and val2, such that``    ``// sumA - val1 + val2 = sumB - val2 + val1``    ``int` `newsum1, newsum2,  ``        ``val1 = 0, val2 = 0;``    ``for` `(``int` `i = 0; i < n; i++)``    ``{``        ``for` `(``int` `j = 0; j < m; j++)``        ``{``            ``newsum1 = sum1 - A[i] + B[j];``            ``newsum2 = sum2 - B[j] + A[i];``            ``if` `(newsum1 == newsum2)``            ``{``                ``val1 = A[i];``                ``val2 = B[j];``            ``}``        ``}``    ``}` `    ``Console.Write(val1 + ``" "` `+ val2);``}` `// Driver Code``public` `static` `void` `Main ()``{``    ``int``[] A = { 4, 1, 2, 1, 1, 2 };``    ``int` `n = A.Length;``    ``int``[] B = { 3, 6, 3, 3 };``    ``int` `m = B.Length;` `    ``// Call to function``    ``findSwapValues(A, n, B, m);``}``}` `// This code is contributed``// by ChitraNayal`

## PHP

 ``

## Javascript

 ``

Output

`1 3`

Time Complexity :- O(n*m)
Space Complexity : O(1)

Method 2 -> Other Naive implementation

```We are looking for two values, a and b, such that:
sumA - a + b = sumB - b + a
2a - 2b  = sumA - sumB
a - b  = (sumA - sumB) / 2```

Therefore, we’re looking for two values that have a specific target difference: (sumA – sumB) / 2

## C++

 `// CPP code for naive implementation``#include ``using` `namespace` `std;` `// Function to calculate sum of elements of array``int` `getSum(``int` `X[], ``int` `n)``{``    ``int` `sum = 0;``    ``for` `(``int` `i = 0; i < n; i++)``        ``sum += X[i];``    ``return` `sum;``}` `// Function to calculate : a - b = (sumA - sumB) / 2``int` `getTarget(``int` `A[], ``int` `n, ``int` `B[], ``int` `m)``{``    ``// Calculation of sums from both arrays``    ``int` `sum1 = getSum(A, n);``    ``int` `sum2 = getSum(B, m);` `    ``// because that the target must be an integer``    ``if` `((sum1 - sum2) % 2 != 0)``        ``return` `0;``    ``return` `((sum1 - sum2) / 2);``}` `void` `findSwapValues(``int` `A[], ``int` `n, ``int` `B[], ``int` `m)``{``    ``int` `target = getTarget(A, n, B, m);``    ``if` `(target == 0)``        ``return``;` `    ``// Look for val1 and val2, such that``    ``// val1 - val2 = (sumA - sumB) / 2``    ``int` `val1, val2;``    ``for` `(``int` `i = 0; i < n; i++) {``        ``for` `(``int` `j = 0; j < m; j++) {``            ``if` `(A[i] - B[j] == target) {``                ``val1 = A[i];``                ``val2 = B[j];``            ``}``        ``}``    ``}` `    ``cout << val1 << ``" "` `<< val2;``}` `// Driver code``int` `main()``{``    ``int` `A[] = { 4, 1, 2, 1, 1, 2 };``    ``int` `n = ``sizeof``(A) / ``sizeof``(A[0]);``    ``int` `B[] = { 3, 6, 3, 3 };``    ``int` `m = ``sizeof``(B) / ``sizeof``(B[0]);` `    ``// Call to function``    ``findSwapValues(A, n, B, m);``    ``return` `0;``}`

## Java

 `// Java program to find a pair swapping``// which makes sum of arrays sum``import` `java.io.*;` `class` `GFG``{``    ``// Function to calculate sum of elements of array``    ``static` `int` `getSum(``int` `X[], ``int` `n)``    ``{``        ``int` `sum = ``0``;``        ``for` `(``int` `i = ``0``; i < n; i++)``            ``sum += X[i];``        ``return` `sum;``    ``}``    ` `    ``// Function to calculate : a - b = (sumA - sumB) / 2``    ``static` `int` `getTarget(``int` `A[], ``int` `n, ``int` `B[], ``int` `m)``    ``{``        ``// Calculation of sums from both arrays``        ``int` `sum1 = getSum(A, n);``        ``int` `sum2 = getSum(B, m);`` ` `        ``// because that the target must be an integer``        ``if` `((sum1 - sum2) % ``2` `!= ``0``)``            ``return` `0``;``        ``return` `((sum1 - sum2) / ``2``);``    ``}``    ` `    ``// Function to prints elements to be swapped``    ``static` `void` `findSwapValues(``int` `A[], ``int` `n, ``int` `B[], ``int` `m)``    ``{``        ``int` `target = getTarget(A, n, B, m);``        ``if` `(target == ``0``)``            ``return``;`` ` `        ``// Look for val1 and val2, such that``        ``// val1 - val2 = (sumA - sumB) / 2``        ``int` `val1 = ``0``, val2 = ``0``;``        ``for` `(``int` `i = ``0``; i < n; i++)``        ``{``            ``for` `(``int` `j = ``0``; j < m; j++)``            ``{``                ``if` `(A[i] - B[j] == target)``                ``{``                    ``val1 = A[i];``                    ``val2 = B[j];``                ``}``            ``}``        ``}``        ``System.out.println(val1+``" "``+val2);``    ``}``    ` `    ``// driver program``    ``public` `static` `void` `main (String[] args)``    ``{``        ``int` `A[] = { ``4``, ``1``, ``2``, ``1``, ``1``, ``2` `};``        ``int` `n = A.length;``        ``int` `B[] = { ``3``, ``6``, ``3``, ``3` `};``        ``int` `m = B.length;`` ` `        ``// Call to function``        ``findSwapValues(A, n, B, m);``    ``}``}` `// Contributed by Pramod Kumar`

## Python3

 `# Python Code for naive implementation` `# Function to calculate sum of elements of list``def` `getSum(X):``    ``sum``=``0``    ``for` `i ``in` `X:``        ``sum``+``=``i``    ``return` `sum` `# Function to calculate : a-b = (sumA - sumB) / 2``def` `getTarget(A,B):` `    ``#Calculations of sums from both lists``    ``sum1``=``getSum(A)``    ``sum2``=``getSum(B)` `    ``# Because the target must be an integer``    ``if``( (sum1``-``sum2)``%``2``!``=``0``):``        ``return` `0``    ``return` `(sum1``-``sum2)``/``/``2`  `def` `findSwapValues(A,B):``    ``target``=``getTarget(A,B)``    ``if` `target``=``=``0``:``        ``return` `    ``# Boolean variable used to reduce further iterations``    ``# after the pair is found``    ``flag``=``False` `    ``# Look for val1 and val2, such that``    ``# val1 - val2 = (sumA -sumB) /2``    ``val1,val2``=``0``,``0``    ``for` `i ``in` `A:``        ``for` `j ``in` `B:``            ` `            ``if` `i``-``j ``=``=` `target:``                ``val1``=``i``                ``val2``=``j``                ``# Set to True when pair is found``                ``flag``=``True``                ``break``        ``if` `flag``=``=``True``:``            ``break``    ``print` `(val1,val2)``    ``return`  `# Driver code``A``=``[``4``,``1``,``2``,``1``,``1``,``2``]``B``=``[``3``,``6``,``3``,``3``]` `# Call to function``findSwapValues(A,B)` `# code contributed by sachin bisht`

## C#

 `// C# program to find a pair swapping``// which makes sum of arrays sum``using` `System;` `class` `GFG``{` `    ``// Function to calculate sum of elements of array``    ``static` `int` `getSum(``int` `[]X, ``int` `n)``    ``{``        ``int` `sum = 0;``        ``for` `(``int` `i = 0; i < n; i++)``            ``sum += X[i];``        ``return` `sum;``    ``}``     ` `    ``// Function to calculate : a - b = (sumA - sumB) / 2``    ``static` `int` `getTarget(``int` `[]A, ``int` `n, ``int` `[]B, ``int` `m)``    ``{``        ``// Calculation of sums from both arrays``        ``int` `sum1 = getSum(A, n);``        ``int` `sum2 = getSum(B, m);``  ` `        ``// because that the target must be an integer``        ``if` `((sum1 - sum2) % 2 != 0)``            ``return` `0;``        ``return` `((sum1 - sum2) / 2);``    ``}``     ` `    ``// Function to prints elements to be swapped``    ``static` `void` `findSwapValues(``int` `[]A, ``int` `n, ``int` `[]B, ``int` `m)``    ``{``        ``int` `target = getTarget(A, n, B, m);``        ``if` `(target == 0)``            ``return``;``  ` `        ``// Look for val1 and val2, such that``        ``// val1 - val2 = (sumA - sumB) / 2``        ``int` `val1 = 0, val2 = 0;``        ``for` `(``int` `i = 0; i < n; i++)``        ``{``            ``for` `(``int` `j = 0; j < m; j++)``            ``{``                ``if` `(A[i] - B[j] == target)``                ``{``                    ``val1 = A[i];``                    ``val2 = B[j];``                ``}``            ``}``        ``}``        ``Console.Write(val1+``" "``+val2);``    ``}``     ` `    ``// Driver code``    ``public` `static` `void` `Main ()``    ``{``        ``int` `[]A = { 4, 1, 2, 1, 1, 2 };``        ``int` `n = A.Length;``        ``int` `[]B = { 3, 6, 3, 3 };``        ``int` `m = B.Length;``  ` `        ``// Call to function``        ``findSwapValues(A, n, B, m);``    ``}``}` `/*This code is contributed by 29AjayKumar*/`

## PHP

 ``

## Javascript

 ``

Output

`1 3`

Time Complexity :- O(n*m)
Space Complexity :- O(1)

Method 3 -> Optimized Solution :-

• Sort the arrays.
• Traverse both array simultaneously and do following for every pair.
1. If the difference is too small then, make it bigger by moving ‘a’ to a bigger value.
2. If it is too big then, make it smaller by moving b to a bigger value.
3. If it’s just right, return this pair.

Below image is a dry run of the above approach:

Below is the implementation of the above approach:

## C++

 `// CPP code for optimized implementation``#include ``using` `namespace` `std;` `// Returns sum of elements in X[]``int` `getSum(``int` `X[], ``int` `n)``{``    ``int` `sum = 0;``    ``for` `(``int` `i = 0; i < n; i++)``        ``sum += X[i];``    ``return` `sum;``}` `// Finds value of``// a - b = (sumA - sumB) / 2``int` `getTarget(``int` `A[], ``int` `n, ``int` `B[], ``int` `m)``{``    ``// Calculation of sums from both arrays``    ``int` `sum1 = getSum(A, n);``    ``int` `sum2 = getSum(B, m);` `    ``// because that the target must be an integer``    ``if` `((sum1 - sum2) % 2 != 0)``        ``return` `0;``    ``return` `((sum1 - sum2) / 2);``}` `// Prints elements to be swapped``void` `findSwapValues(``int` `A[], ``int` `n, ``int` `B[], ``int` `m)``{``    ``// Call for sorting the arrays``    ``sort(A, A + n);``    ``sort(B, B + m);` `    ``// Note that target can be negative``    ``int` `target = getTarget(A, n, B, m);` `    ``// target 0 means, answer is not possible``    ``if` `(target == 0)``        ``return``;` `    ``int` `i = 0, j = 0;``    ``while` `(i < n && j < m) {``        ``int` `diff = A[i] - B[j];``        ``if` `(diff == target) {``            ``cout << A[i] << ``" "` `<< B[j];``            ``return``;``        ``}` `        ``// Look for a greater value in A[]``        ``else` `if` `(diff < target)``            ``i++;` `        ``// Look for a greater value in B[]``        ``else``            ``j++;``    ``}``}` `// Driver code``int` `main()``{``    ``int` `A[] = { 4, 1, 2, 1, 1, 2 };``    ``int` `n = ``sizeof``(A) / ``sizeof``(A[0]);` `    ``int` `B[] = { 1, 6, 3, 3 };``    ``int` `m = ``sizeof``(B) / ``sizeof``(B[0]);` `    ``findSwapValues(A, n, B, m);``    ``return` `0;``}`

## Java

 `// Java code for optimized implementation``import` `java.io.*;``import` `java.util.*;` `class` `GFG``{``    ``// Function to calculate sum of elements of array``    ``static` `int` `getSum(``int` `X[], ``int` `n)``    ``{``        ``int` `sum = ``0``;``        ``for` `(``int` `i = ``0``; i < n; i++)``            ``sum += X[i];``        ``return` `sum;``    ``}``    ` `    ``// Function to calculate : a - b = (sumA - sumB) / 2``    ``static` `int` `getTarget(``int` `A[], ``int` `n, ``int` `B[], ``int` `m)``    ``{``        ``// Calculation of sums from both arrays``        ``int` `sum1 = getSum(A, n);``        ``int` `sum2 = getSum(B, m);`` ` `        ``// because that the target must be an integer``        ``if` `((sum1 - sum2) % ``2` `!= ``0``)``            ``return` `0``;``        ``return` `((sum1 - sum2) / ``2``);``    ``}``    ` `    ``// Function to prints elements to be swapped``    ``static` `void` `findSwapValues(``int` `A[], ``int` `n, ``int` `B[], ``int` `m)``    ``{``        ``// Call for sorting the arrays``        ``Arrays.sort(A);``        ``Arrays.sort(B);`` ` `        ``// Note that target can be negative``        ``int` `target = getTarget(A, n, B, m);`` ` `        ``// target 0 means, answer is not possible``        ``if` `(target == ``0``)``            ``return``;`` ` `        ``int` `i = ``0``, j = ``0``;``        ``while` `(i < n && j < m)``        ``{``            ``int` `diff = A[i] - B[j];``            ``if` `(diff == target)``            ``{``                ``System.out.println(A[i]+``" "``+B[i]);``                ``return``;``            ``}`` ` `            ``// Look for a greater value in A[]``            ``else` `if` `(diff < target)``                ``i++;`` ` `            ``// Look for a greater value in B[]``            ``else``                ``j++;``        ``}``    ``}``    ` `    ``// driver program``    ``public` `static` `void` `main (String[] args)``    ``{``        ``int` `A[] = { ``4``, ``1``, ``2``, ``1``, ``1``, ``2` `};``        ``int` `n = A.length;``        ``int` `B[] = { ``3``, ``6``, ``3``, ``3` `};``        ``int` `m = B.length;`` ` `        ``// Call to function``        ``findSwapValues(A, n, B, m);``    ``}``}` `// Contributed by Pramod Kumar`

## Python3

 `# Python code for optimized implementation` `#Returns sum of elements in list``def` `getSum(X):``    ``sum``=``0``    ``for` `i ``in` `X:``        ``sum``+``=``i``    ``return` `sum` `# Finds value of``# a - b = (sumA - sumB) / 2``def` `getTarget(A,B):``    ``# Calculations of sumd from both lists``    ``sum1``=``getSum(A)``    ``sum2``=``getSum(B)` `    ``# Because that target must be an integer``    ``if``( (sum1``-``sum2)``%``2``!``=``0``):``        ``return` `0``    ``return` `(sum1``-``sum2)``/``/``2` `# Prints elements to be swapped``def` `findSwapValues(A,B):``    ``# Call for sorting the lists``    ``A.sort()``    ``B.sort()` `    ``#Note that target can be negative``    ``target``=``getTarget(A,B)` `    ``# target 0 means, answer is not possible``    ``if``(target``=``=``0``):``        ``return``    ``i,j``=``0``,``0``    ``while``(i<``len``(A) ``and` `j<``len``(B)):``        ``diff``=``A[i]``-``B[j]``        ``if` `diff ``=``=` `target:``            ``print` `(A[i],B[j])``            ``return``        ``# Look for a greater value in list A``        ``elif` `diff

## C#

 `// C# code for optimized implementation``using` `System;` `class` `GFG``{``    ``// Function to calculate sum of elements of array``    ``static` `int` `getSum(``int` `[]X, ``int` `n)``    ``{``        ``int` `sum = 0;``        ``for` `(``int` `i = 0; i < n; i++)``            ``sum += X[i];``        ``return` `sum;``    ``}``    ` `    ``// Function to calculate : a - b = (sumA - sumB) / 2``    ``static` `int` `getTarget(``int` `[]A, ``int` `n, ``int` `[]B, ``int` `m)``    ``{``        ``// Calculation of sums from both arrays``        ``int` `sum1 = getSum(A, n);``        ``int` `sum2 = getSum(B, m);` `        ``// because that the target must be an integer``        ``if` `((sum1 - sum2) % 2 != 0)``            ``return` `0;``        ``return` `((sum1 - sum2) / 2);``    ``}``    ` `    ``// Function to prints elements to be swapped``    ``static` `void` `findSwapValues(``int` `[]A, ``int` `n, ``int` `[]B, ``int` `m)``    ``{``        ``// Call for sorting the arrays``        ``Array.Sort(A);``        ``Array.Sort(B);` `        ``// Note that target can be negative``        ``int` `target = getTarget(A, n, B, m);` `        ``// target 0 means, answer is not possible``        ``if` `(target == 0)``            ``return``;` `        ``int` `i = 0, j = 0;``        ``while` `(i < n && j < m)``        ``{``            ``int` `diff = A[i] - B[j];``            ``if` `(diff == target)``            ``{``                ``Console.WriteLine(A[i]+``" "``+B[i]);``                ``return``;``            ``}` `            ``// Look for a greater value in A[]``            ``else` `if` `(diff < target)``                ``i++;` `            ``// Look for a greater value in B[]``            ``else``                ``j++;``        ``}``    ``}``    ` `    ``// Driver code``    ``public` `static` `void` `Main (String[] args)``    ``{``        ``int` `[]A = { 4, 1, 2, 1, 1, 2 };``        ``int` `n = A.Length;``        ``int` `[]B = { 3, 6, 3, 3 };``        ``int` `m = B.Length;` `        ``// Call to function``        ``findSwapValues(A, n, B, m);``    ``}``}` `// This code has been contributed by 29AjayKumar`

## Javascript

 ``

Output

`2 3`

Time Complexity :-
If arrays are sorted : O(n + m)
If arrays aren’t sorted : O(nlog(n) + mlog(m))

Space Complexity : O(1)

Method 4 (Hashing)
We can solve this problem in O(m+n) time and O(m) auxiliary space. Below are algorithmic steps.

```// assume array1 is small i.e. (m < n)
// where m is array1.length and n is array2.length
1. Find sum1(sum of small array elements) and sum2
(sum of larger array elements). // time O(m+n)
2. Make a hashset for small array(here array1).
3. Calculate diff as (sum1-sum2)/2.
4. Run a loop for array2
for (int i equal to 0 to n-1)
if (hashset contains (array2[i]+diff))
print array2[i]+diff and array2[i]
set flag  and break;
5. If flag is unset then there is no such kind of
pair.```

Thanks to nicky khan for suggesting method 4.
This article is contributed by Sakshi Tiwari. If you like GeeksforGeeks (We know you do!) and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Another Approach:

We can also solve this problem in linear time using hashing. Let us assume that sum of the elements of the first array a[] is s1, and of the second array b[] is s2. Also suppose that a pair to be swapped is (p, q), where p belongs to a[] and q belongs to b[]. Therefore, we have the equation s1 – p + q = s2 – q + p, i.e. 2q = s2 – s1 + 2p. Since both 2p and 2q are even integers, the difference s2 – s1 must be an even integer too. So, given any p, our aim is to find an appropriate q satisfying the above conditions.

Below is an implementation of the said approach:

## C++

 `#include ``using` `namespace` `std;``void` `findSwapValues(``int` `a[], ``int` `m, ``int` `b[], ``int` `n);``int` `main()``{``    ``int` `a[] = { 4, 1, 2, 1, 1, 2 }, b[] = { 1, 6, 3, 3 };``    ``int` `m, n;``    ``m = ``sizeof``(a) / ``sizeof``(``int``),``    ``n = ``sizeof``(b) / ``sizeof``(``int``);``    ``findSwapValues(a, m, b, n);``    ``return` `0;``}``void` `findSwapValues(``int` `a[], ``int` `m, ``int` `b[], ``int` `n)``{``    ``unordered_set<``int``> x,``        ``y; ``/* Unordered sets (and unordered maps) are``              ``implemented internally using hash tables; they``              ``support dictionary operations (i.e. search,``              ``insert, delete) in O(1) time on an average. */``    ``unordered_set<``int``>::iterator p, q;``    ``int` `s1, s2;``    ``int` `i;``    ``s1 = 0;``    ``for` `(i = 0; i < m;``         ``i++) ``/* Determining sum s1 of the elements of array``                 ``a[], and simultaneously inserting the array``                 ``elements in the unordered set. */``        ``s1 += a[i], x.insert(a[i]);``    ``s2 = 0;``    ``for` `(i = 0; i < n; i++)``        ``s2 += b[i], y.insert(b[i]);``    ``if` `((s1 - s2) % 2) ``/* Checking if difference between the``                                              ``two array sums``                          ``is even or not. */``    ``{``        ``printf``(``"No such values exist.\n"``);``        ``return``;``    ``}``    ``for` `(p = x.begin(); p != x.end(); p++) {``        ``q = y.find(``            ``((s2 - s1) + 2 * *p)``            ``/ 2); ``// Finding q for a given p in O(1) time.``        ``if` `(q != y.end()) {``            ``printf``(``"%d %d\n"``, *p, *q);``            ``return``;``        ``}``    ``}``    ``printf``(``"No such values exist.\n"``);``}`

## Java

 `import` `java.util.*;` `public` `class` `Solution {``  ``static` `void` `findSwapValues(``int` `a[], ``int` `m, ``int` `b[],``                             ``int` `n)``  ``{``    ``HashSet x = ``new` `HashSet<>();``    ``HashSet y = ``new` `HashSet<>();``    ` `    ``/* Unordered sets (and unordered maps) are``    ``implemented internally using hash tables; they``    ``support dictionary operations (i.e. search,``    ``insert, delete) in O(1) time on an average. */` `    ``int` `s1, s2;``    ``int` `i;``    ``s1 = ``0``;``    ``for` `(i = ``0``; i < m; i++) {``      ``/* Determining sum s1 of the elements of array``            ``a[], and simultaneously inserting the array``            ``elements in the unordered set. */``      ``s1 += a[i];``      ``x.add(a[i]);``    ``}` `    ``s2 = ``0``;``    ``for` `(i = ``0``; i < n; i++) {``      ``s2 += b[i];``      ``y.add(b[i]);``    ``}` `    ``if` `((s1 - s2) % ``2` `!= ``0``)``      ``/* Checking if difference between the``        ``two array sums is even or not. */``    ``{``      ``System.out.println(``"No such values exist."``);``      ``return``;``    ``}` `    ``for` `(Integer p : x) {``      ``int` `q = ((s2 - s1) + ``2` `* p) / ``2``;``      ``if` `(y.contains(q)) {``        ``System.out.println(p + ``" "` `+ q);``        ``return``;``      ``}``    ``}``    ``System.out.println(``"No such values exist."``);``  ``}` `  ``// Driver Code``  ``public` `static` `void` `main(String[] args)``  ``{``    ``int` `a[] = { ``4``, ``1``, ``2``, ``1``, ``1``, ``2` `};``    ``int` `b[] = { ``1``, ``6``, ``3``, ``3` `};``    ``int` `m = a.length;``    ``int` `n = b.length;``    ``findSwapValues(a, m, b, n);``  ``}``}` `// This code is contributed by karandeep1234.`

## Python3

 `# Python Code for the above approach``def` `findSwapValues(a, m, b, n):` `    ``# initialize 2 dictionary. dictionary takes operations (i.e. search,``    ``# insert, delete) in O(1) time on an average.``    ``x, y ``=` `{}, {}` `    ``s1, s2 ``=` `0``, ``0` `    ``# Determining sum s1 of the elements of array``    ``# a[], and simultaneously inserting the array in the dictionary x``    ``for` `i ``in` `range``(m):``        ``s1 ``+``=` `a[i]``        ``x[a[i]] ``=` `x.get(a[i], ``0``) ``+` `1` `    ``# Determining sum s2 of the elements of array``    ``# b[], and simultaneously inserting the array in the dictionary y``    ``for` `i ``in` `range``(n):``        ``s2 ``+``=` `b[i]``        ``y[b[i]] ``=` `y.get(b[i], ``0``) ``+` `1` `    ``if` `(s1 ``-` `s2) ``%` `2``:  ``# Checking if difference between the two arrays sums is even or not``        ``print``(``"No such values exist."``)``        ``return` `    ``for` `p ``in` `x:``        ``q ``=` `((s2 ``-` `s1)``/``/``2``) ``+` `p``        ``if` `q ``in` `y:  ``# Finding q for a given p in O(1) time.``            ``print``(p, q)``            ``return``    ``print``(``"No such values exist."``)`  `if` `__name__ ``=``=` `"__main__"``:``    ``a ``=` `[``4``, ``1``, ``2``, ``1``, ``1``, ``2``]``    ``b ``=` `[``1``, ``6``, ``3``, ``3``]``    ``m ``=` `len``(a)``    ``n ``=` `len``(b)``    ``findSwapValues(a, m, b, n)``# This Code is Contributed by Vivek Maddeshiya`

## C#

 `using` `System;``using` `System.Collections.Generic;``class` `GFG {``  ``static` `void` `findSwapValues(``int``[] a, ``int` `m, ``int``[] b,``                             ``int` `n)``  ``{``    ``HashSet<``int``> x = ``new` `HashSet<``int``>();``    ``HashSet<``int``> y = ``new` `HashSet<``int``>();``    ``int` `s1, s2;``    ``int` `i;``    ``s1 = 0;``    ` `    ``/* Determining sum s1 of the elements of array``                     ``a[], and simultaneously inserting the``           ``array elements in the unordered set. */``    ``for` `(i = 0; i < m; i++) {``      ``s1 += a[i];``      ``x.Add(a[i]);``    ``}``    ``s2 = 0;``    ``for` `(i = 0; i < n; i++) {``      ``s2 += b[i];``      ``y.Add(b[i]);``    ``}``    ``if` `((s1 - s2) % 2``        ``!= 0) ``/* Checking if difference between the two``                     ``array sums is even or not. */``    ``{``      ``Console.Write(``"No such values exist"``);``      ``return``;``    ``}` `    ``foreach``(``int` `p ``in` `x)``    ``{``      ``int` `q = ((s2 - s1) + 2 * p)``        ``/ 2; ``// Finding q for a given p in O(1)``      ``// time.``      ``if` `(y.Contains(q)) {``        ``Console.Write(p + ``" "` `+ q);``        ``return``;``      ``}``    ``}``    ``Console.Write(``"No such values exist"``);``  ``}``  ``public` `static` `void` `Main(``string``[] args)``  ``{``    ``int``[] a = { 4, 1, 2, 1, 1, 2 };``    ``int``[] b = { 1, 6, 3, 3 };``    ``int` `m = 6;``    ``int` `n = 4;``    ``findSwapValues(a, m, b, n);``  ``}``}` `// This code is contributed by garg28harsh.`

Output

`2 3`

Time complexity of the above code is O(m + n), where m and n respectively represent the sizes of the two input arrays, and space complexity O(s + t), where s and t respectively represent the number of distinct elements present in the two input arrays.

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