Given two arrays of integers a[] and b[], the task is to check if a pair of values (one value from each array) exists such that swapping the elements of the pair will make the sum of two arrays equal.
Examples:
Input: a[] = [4, 1, 2, 1, 1, 2], b[] = [3, 6, 3, 3]
Output: true
Explanation: 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: true
Explanation: We can swap 6 from array a[] and 2 from array b[]
Try It Yourself
Table of Content
[Naive Approach] Check all possible pairs - O(n * m) Time and O(1) Space
Iterate through the arrays and check all pairs of values. For each element in A[], iterate over all the elements of B[], and check if swapping these two elements will make the sum equal.
// CPP code naive solution to find a pair swapping
// which makes sum of arrays sum.
#include <iostream>
using namespace std;
// Function to calculate sum of elements of array
bool findSwapValues(vector<int> &a, vector<int> &b)
{
// getting sizes of both arrays
int n = a.size();
int m = b.size();
// calculating sum of elements of both arrays
int sum1 = 0, sum2 = 0;
for (int i = 0; i < n; i++)
sum1 += a[i];
for (int j = 0; j < m; j++)
sum2 += b[j];
// variables to store new sums after swapping
int newsum1, newsum2;
// traversing each element of first array
for (int i = 0; i < n; i++)
{
// traversing each element of second array
for (int j = 0; j < m; j++)
{
// calculating new sum if a[i] and b[j] are swapped
newsum1 = sum1 - a[i] + b[j];
newsum2 = sum2 - b[j] + a[i];
// checking if both sums become equal
if (newsum1 == newsum2)
{
return true; // valid pair found
}
}
}
// if no such pair exists
return false;
}
// Driver code
int main() {
// initializing first array
vector<int> a = {4, 1, 2, 1, 1, 2};
// initializing second array
vector<int> b = {3, 6, 3, 3};
// calling function and printing result
if(findSwapValues(a, b))
cout << "True";
else
cout << "False";
return 0;
}
// Java code naive solution to find a pair swapping which makes sum of arrays sum.
import java.util.*;
public class GFG {
// Function to calculate sum of elements of array
public static boolean findSwapValues(int[] a, int[] b) {
// getting sizes of both arrays
int n = a.length;
int m = b.length;
// calculating sum of elements of both arrays
int sum1 = 0, sum2 = 0;
for (int i = 0; i < n; i++)
sum1 += a[i];
for (int j = 0; j < m; j++)
sum2 += b[j];
// variables to store new sums after swapping
int newsum1, newsum2;
// traversing each element of first array
for (int i = 0; i < n; i++) {
// traversing each element of second array
for (int j = 0; j < m; j++) {
// calculating new sum if a[i] and b[j] are swapped
newsum1 = sum1 - a[i] + b[j];
newsum2 = sum2 - b[j] + a[i];
// checking if both sums become equal
if (newsum1 == newsum2) {
return true; // valid pair found
}
}
}
// if no such pair exists
return false;
}
public static void main(String[] args) {
// initializing first array
int[] a = {4, 1, 2, 1, 1, 2};
// initializing second array
int[] b = {3, 6, 3, 3};
// calling function and printing result
if (findSwapValues(a, b))
System.out.println("True");
else
System.out.println("False");
}
}
# Python code naive solution to find a pair swapping which makes sum of arrays sum.
# Function to calculate sum of elements of array
def findSwapValues(a, b):
# getting sizes of both arrays
n = len(a)
m = len(b)
# calculating sum of elements of both arrays
sum1 = sum(a)
sum2 = sum(b)
# variables to store new sums after swapping
newsum1, newsum2 = 0, 0
# traversing each element of first array
for i in range(n):
# traversing each element of second array
for j in range(m):
# calculating new sum if a[i] and b[j] are swapped
newsum1 = sum1 - a[i] + b[j]
newsum2 = sum2 - b[j] + a[i]
# checking if both sums become equal
if newsum1 == newsum2:
return True # valid pair found
# if no such pair exists
return False
# Driver code
if __name__ == '__main__':
# initializing first array
a = [4, 1, 2, 1, 1, 2]
# initializing second array
b = [3, 6, 3, 3]
# calling function and printing result
if findSwapValues(a, b):
print("True")
else:
print("False")
// C# code naive solution to find a pair swapping which makes sum of arrays sum.
using System;
using System.Collections.Generic;
public class GFG {
// Function to calculate sum of elements of array
public static bool FindSwapValues(int[] a, int[] b) {
// getting sizes of both arrays
int n = a.Length;
int m = b.Length;
// calculating sum of elements of both arrays
int sum1 = 0, sum2 = 0;
for (int i = 0; i < n; i++)
sum1 += a[i];
for (int j = 0; j < m; j++)
sum2 += b[j];
// variables to store new sums after swapping
int newsum1, newsum2;
// traversing each element of first array
for (int i = 0; i < n; i++) {
// traversing each element of second array
for (int j = 0; j < m; j++) {
// calculating new sum if a[i] and b[j] are swapped
newsum1 = sum1 - a[i] + b[j];
newsum2 = sum2 - b[j] + a[i];
// checking if both sums become equal
if (newsum1 == newsum2) {
return true; // valid pair found
}
}
}
// if no such pair exists
return false;
}
public static void Main() {
// initializing first array
int[] a = {4, 1, 2, 1, 1, 2};
// initializing second array
int[] b = {3, 6, 3, 3};
// calling function and printing result
if (FindSwapValues(a, b))
Console.WriteLine("True");
else
Console.WriteLine("False");
}
}
// JavaScript code naive solution to find a pair swapping which makes sum of arrays sum.
// Function to calculate sum of elements of array
function findSwapValues(a, b) {
// getting sizes of both arrays
let n = a.length;
let m = b.length;
// calculating sum of elements of both arrays
let sum1 = a.reduce((acc, val) => acc + val, 0);
let sum2 = b.reduce((acc, val) => acc + val, 0);
// variables to store new sums after swapping
let newsum1, newsum2;
// traversing each element of first array
for (let i = 0; i < n; i++) {
// traversing each element of second array
for (let j = 0; j < m; j++) {
// calculating new sum if a[i] and b[j] are swapped
newsum1 = sum1 - a[i] + b[j];
newsum2 = sum2 - b[j] + a[i];
// checking if both sums become equal
if (newsum1 === newsum2) {
return true; // valid pair found
}
}
}
// if no such pair exists
return false;
}
// Driver code
let a = [4, 1, 2, 1, 1, 2];
let b = [3, 6, 3, 3];
if (findSwapValues(a, b)) {
console.log("True");
} else {
console.log("False");
}
Output
True
[Better Approach] Using Sorting + Two Pointer Technique - O(n log n) Time and O(1) Space
Let the sum of array a[] be sumA and sum of array b[] be sumB, then we need to find a value x in a[] and a y in b[] such that:
sumA - x + y = sumB - y + x
2x - 2y = sumA - sumB
x - y = (sumA - sumB) / 2To find the elements x and y, we sort the arrays and traverse simultaneously using two pointers,
- If the difference of x and y is too small then, make it bigger by moving x to a bigger value.
- If the difference of x and y is too big then, make it smaller by moving y to a bigger value.
- If the difference of x and y is equal to (sumA - sumB)/2, return this pair.
Working of Approach:
- Calculate the sum of both arrays sum1 and sum2, and check if their difference is even.
- Compute target = (sum1 - sum2) / 2.
- Sort both arrays to apply the two-pointer technique.
- Initialize two pointers i and j at the start of both arrays.
- Traverse and compare a[i] - b[j] with target to find a valid pair.
- If a match is found return true, otherwise move pointers accordingly; if none found, return false.
#include <bits/stdc++.h>
using namespace std;
// function to calculate sum of elements of array
int getSum(vector<int> &arr)
{
int sum = 0;
for (int i = 0; i < arr.size(); i++)
sum += arr[i];
return sum;
}
// function to get target value
// a - b = (sum1 - sum2) / 2
int getTarget(vector<int> &a, vector<int> &b)
{
int sum1 = getSum(a);
int sum2 = getSum(b);
// if difference is odd, equal sum not possible
if ((sum1 - sum2) % 2 != 0)
return INT_MIN;
return (sum1 - sum2) / 2;
}
bool findSwapValues(vector<int> &a, vector<int> &b)
{
int n = a.size();
int m = b.size();
// sorting both arrays
sort(a.begin(), a.end());
sort(b.begin(), b.end());
// getting target difference
int target = getTarget(a, b);
// check only invalid case
if (target == INT_MIN)
return false;
int i = 0, j = 0;
// using two pointer approach
while (i < n && j < m)
{
int diff = a[i] - b[j];
if (diff == target)
return true;
else if (diff < target)
i++;
else
j++;
}
return false;
}
// Driver code
int main() {
vector<int> a = {4, 1, 2, 1, 1, 2};
vector<int> b = {3, 6, 3, 3};
bool ans = findSwapValues(a, b);
if(ans)
cout << "True";
else
cout << "False";
return 0;
}
import java.util.Arrays;
public class GFG {
// function to calculate sum of elements of array
public static int getSum(int[] arr) {
int sum = 0;
for (int i = 0; i < arr.length; i++)
sum += arr[i];
return sum;
}
// function to get target value
// a - b = (sum1 - sum2) / 2
public static int getTarget(int[] a, int[] b) {
int sum1 = getSum(a);
int sum2 = getSum(b);
// if difference is odd, equal sum not possible
if ((sum1 - sum2) % 2!= 0)
return Integer.MIN_VALUE;
return (sum1 - sum2) / 2;
}
public static boolean findSwapValues(int[] a, int[] b) {
int n = a.length;
int m = b.length;
// sorting both arrays
Arrays.sort(a);
Arrays.sort(b);
// getting target difference
int target = getTarget(a, b);
// check only invalid case
if (target == Integer.MIN_VALUE)
return false;
int i = 0, j = 0;
// using two pointer approach
while (i < n && j < m) {
int diff = a[i] - b[j];
if (diff == target)
return true;
else if (diff < target)
i++;
else
j++;
}
return false;
}
public static void main(String[] args) {
int[] a = {4, 1, 2, 1, 1, 2};
int[] b = {3, 6, 3, 3};
boolean ans = findSwapValues(a, b);
if(ans)
System.out.println("True");
else
System.out.println("False");
}
}
def getSum(arr):
# function to calculate sum of elements of array
sum = 0
for i in range(len(arr)):
sum += arr[i]
return sum
def getTarget(a, b):
# function to get target value
# a - b = (sum1 - sum2) / 2
sum1 = getSum(a)
sum2 = getSum(b)
# if difference is odd, equal sum not possible
if (sum1 - sum2) % 2!= 0:
return float('-inf')
return (sum1 - sum2) // 2
def findSwapValues(a, b):
n = len(a)
m = len(b)
# sorting both arrays
a.sort()
b.sort()
# getting target difference
target = getTarget(a, b)
# check only invalid case
if target == float('-inf'):
return False
i = 0
j = 0
# using two pointer approach
while i < n and j < m:
diff = a[i] - b[j]
if diff == target:
return True
elif diff < target:
i += 1
else:
j += 1
return False
# Driver code
a = [4, 1, 2, 1, 1, 2]
b = [3, 6, 3, 3]
ans = findSwapValues(a, b)
if ans:
print('True')
else:
print('False')
using System;
using System.Linq;
public class GFG
{
// function to calculate sum of elements of array
public static int GetSum(int[] arr)
{
int sum = 0;
for (int i = 0; i < arr.Length; i++)
sum += arr[i];
return sum;
}
// function to get target value
// a - b = (sum1 - sum2) / 2
public static int GetTarget(int[] a, int[] b)
{
int sum1 = GetSum(a);
int sum2 = GetSum(b);
// if difference is odd, equal sum not possible
if ((sum1 - sum2) % 2!= 0)
return int.MinValue;
return (sum1 - sum2) / 2;
}
public static bool FindSwapValues(int[] a, int[] b)
{
int n = a.Length;
int m = b.Length;
// sorting both arrays
Array.Sort(a);
Array.Sort(b);
// getting target difference
int target = GetTarget(a, b);
// check only invalid case
if (target == int.MinValue)
return false;
int i = 0, j = 0;
// using two pointer approach
while (i < n && j < m)
{
int diff = a[i] - b[j];
if (diff == target)
return true;
else if (diff < target)
i++;
else
j++;
}
return false;
}
public static void Main(string[] args)
{
int[] a = { 4, 1, 2, 1, 1, 2 };
int[] b = { 3, 6, 3, 3 };
bool ans = FindSwapValues(a, b);
if (ans)
Console.WriteLine("True");
else
Console.WriteLine("False");
}
}
function getSum(arr) {
// function to calculate sum of elements of array
let sum = 0;
for (let i = 0; i < arr.length; i++) {
sum += arr[i];
}
return sum;
}
function getTarget(a, b) {
// function to get target value
// a - b = (sum1 - sum2) / 2
let sum1 = getSum(a);
let sum2 = getSum(b);
// if difference is odd, equal sum not possible
if ((sum1 - sum2) % 2!= 0) {
return Number.NEGATIVE_INFINITY;
}
return Math.floor((sum1 - sum2) / 2);
}
function findSwapValues(a, b) {
let n = a.length;
let m = b.length;
// sorting both arrays
a.sort((x, y) => x - y);
b.sort((x, y) => x - y);
// getting target difference
let target = getTarget(a, b);
// check only invalid case
if (target === Number.NEGATIVE_INFINITY) {
return false;
}
let i = 0, j = 0;
// using two pointer approach
while (i < n && j < m) {
let diff = a[i] - b[j];
if (diff === target) {
return true;
} else if (diff < target) {
i++;
} else {
j++;
}
}
return false;
}
// Driver code
let a = [4, 1, 2, 1, 1, 2];
let b = [3, 6, 3, 3];
let ans = findSwapValues(a, b);
if (ans) {
console.log('True');
} else {
console.log('False');
}
Output
True
[Expected Approach] Using Hashing (Unordered Set) - O(n + m) Time and O(n) Space
Let sumA and sumB be the sums of arrays a[] and b[]. To make both sums equal after swapping elements x and y, the condition becomes sumA - x + y = sumB - y + x. Simplifying this, we get x - y = (sumA - sumB) / 2. Let this value be the target. Now, the problem reduces to finding a pair (x, y) such that x = target + y. We store all elements of a[] in a hash set and check for each element of b[] whether the required value exists in the set.
Working of Approach:
- Calculate sums of both arrays sumA and sumB.
- If their difference is odd, return false.
- Compute target = (sumA - sumB) / 2.
- Store elements of a[] in a hash set.
- For each element in b[], check if target + b[j] exists in the set.
- If found, return true, otherwise return false.
#include <bits/stdc++.h>
using namespace std;
bool findSwapValues(vector<int> &a, vector<int> &b)
{
// getting sizes of both arrays
int n = a.size();
int m = b.size();
// calculating sum of elements of both arrays
int sumA = 0, sumB = 0;
for (int i = 0; i < n; i++)
sumA += a[i];
for (int j = 0; j < m; j++)
sumB += b[j];
// if difference of sums is odd, equal sum is not possible
if ((sumA - sumB) % 2 != 0)
return false;
// creating a hash set to store elements of array a[]
unordered_set<int> possibleX;
// inserting all elements of a[] into hash set
for (int i = 0; i < n; i++)
{
possibleX.insert(a[i]);
}
// traversing array b[] to find required pair
for (int j = 0; j < m; j++)
{
// calculating value of x using formula
int X = (sumA - sumB) / 2 + b[j];
// checking if this value exists in array a[]
if (possibleX.find(X) != possibleX.end())
{
return true; // valid pair found
}
}
// if no such pair exists
return false;
}
// Driver Code
int main() {
vector<int> a = {4, 1, 2, 1, 1, 2};
vector<int> b = {3, 6, 3, 3};
bool ans = findSwapValues(a, b);
if(ans)
cout << "True";
else
cout << "False";
return 0;
}
import java.util.HashSet;
public class GFG {
public static boolean findSwapValues(int[] a, int[] b) {
// getting sizes of both arrays
int n = a.length;
int m = b.length;
// calculating sum of elements of both arrays
int sumA = 0, sumB = 0;
for (int i = 0; i < n; i++)
sumA += a[i];
for (int j = 0; j < m; j++)
sumB += b[j];
// if difference of sums is odd, equal sum is not possible
if ((sumA - sumB) % 2!= 0)
return false;
// creating a hash set to store elements of array a[]
HashSet<Integer> possibleX = new HashSet<>();
// inserting all elements of a[] into hash set
for (int i = 0; i < n; i++) {
possibleX.add(a[i]);
}
// traversing array b[] to find required pair
for (int j = 0; j < m; j++) {
// calculating value of x using formula
int X = (sumA - sumB) / 2 + b[j];
// checking if this value exists in array a[]
if (possibleX.contains(X)) {
return true; // valid pair found
}
}
// if no such pair exists
return false;
}
public static void main(String[] args) {
int[] a = {4, 1, 2, 1, 1, 2};
int[] b = {3, 6, 3, 3};
boolean ans = findSwapValues(a, b);
if (ans)
System.out.println("True");
else
System.out.println("False");
}
}
def findSwapValues(a, b):
# getting sizes of both arrays
n = len(a)
m = len(b)
# calculating sum of elements of both arrays
sumA = sum(a)
sumB = sum(b)
# if difference of sums is odd, equal sum is not possible
if (sumA - sumB) % 2!= 0:
return False
# creating a set to store elements of array a[]
possibleX = set(a)
# traversing array b[] to find required pair
for j in range(m):
# calculating value of x using formula
X = (sumA - sumB) // 2 + b[j]
# checking if this value exists in array a[]
if X in possibleX:
return True # valid pair found
# if no such pair exists
return False
# Driver Code
a = [4, 1, 2, 1, 1, 2]
b = [3, 6, 3, 3]
ans = findSwapValues(a, b)
if ans:
print('True')
else:
print('False')
using System;
using System.Collections.Generic;
class GFG
{
static bool findSwapValues(List<int> a, List<int> b)
{
// getting sizes of both arrays
int n = a.Count;
int m = b.Count;
// calculating sum of elements of both arrays
int sumA = 0, sumB = 0;
for (int i = 0; i < n; i++)
sumA += a[i];
for (int j = 0; j < m; j++)
sumB += b[j];
// if difference of sums is odd, equal sum is not possible
if ((sumA - sumB) % 2!= 0)
return false;
// creating a hash set to store elements of array a[]
HashSet<int> possibleX = new HashSet<int>();
// inserting all elements of a[] into hash set
for (int i = 0; i < n; i++)
{
possibleX.Add(a[i]);
}
// traversing array b[] to find required pair
for (int j = 0; j < m; j++)
{
// calculating value of x using formula
int X = (sumA - sumB) / 2 + b[j];
// checking if this value exists in array a[]
if (possibleX.Contains(X))
{
return true; // valid pair found
}
}
// if no such pair exists
return false;
}
// Driver Code
static void Main(string[] args)
{
List<int> a = new List<int> { 4, 1, 2, 1, 1, 2 };
List<int> b = new List<int> { 3, 6, 3, 3 };
bool ans = findSwapValues(a, b);
Console.WriteLine(ans ? "True" : "False");
}
}
function findSwapValues(a, b) {
// getting sizes of both arrays
const n = a.length;
const m = b.length;
// calculating sum of elements of both arrays
let sumA = 0, sumB = 0;
for (let i = 0; i < n; i++)
sumA += a[i];
for (let j = 0; j < m; j++)
sumB += b[j];
// if difference of sums is odd, equal sum is not possible
if ((sumA - sumB) % 2!== 0)
return false;
// creating a set to store elements of array a[]
const possibleX = new Set();
// inserting all elements of a[] into set
for (let i = 0; i < n; i++)
possibleX.add(a[i]);
// traversing array b[] to find required pair
for (let j = 0; j < m; j++)
{
// calculating value of x using formula
const X = (sumA - sumB) / 2 + b[j];
// checking if this value exists in array a[]
if (possibleX.has(X))
{
return true; // valid pair found
}
}
// if no such pair exists
return false;
}
// Driver Code
const a = [4, 1, 2, 1, 1, 2];
const b = [3, 6, 3, 3];
const ans = findSwapValues(a, b);
console.log(ans ? 'True' : 'False');
Output
True