Find a pair of elements swapping which makes sum of two arrays same (original) (raw)

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
**Explanation: Sum of elements in A[] = 11 and Sum of elements in B[] = 15. To get same sum from both arrays, we can swap 1 from A[] with 3 from B[].

**Input: A[] = {5, 7, 4, 6}, B[] = {1, 2, 3, 8}
**Output: 6 2
**Explanation: Sum of elements in A[] = 22 and Sum of elements in B[] = 14. To get same sum from both arrays, we can swap 6 from A[] and 2 from B[].

Approaches to find a pair to swap which makes sum of two arrays same:

Table of Content

Approaches to find a pair to swap which makes sum of two arrays same

****[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.

Below is the implementation of the above algorithm:

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

Python

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

Javascript

PHP

i<i < i<n; $i++) sum+=sum += sum+=X[$i]; return $sum; } function findSwapValues($A, n,n, n,B, $m) { // Calculation of sums from both arrays sum1=getSum(sum1 = getSum(sum1=getSum(A, $n); sum2=getSum(sum2 = getSum(sum2=getSum(B, $m); // Look for val1 and val2, such that // sumA - val1 + val2 = sumB - val2 + val1 for ($i = 0; i<i < i<n; $i++) { for ($j = 0; j<j < j<m; $j++) { newsum1=newsum1 = newsum1=sum1 - A[A[A[i] + B[B[B[j]; newsum2=newsum2 = newsum2=sum2 - B[B[B[j] + A[A[A[i]; if ($newsum1 == $newsum2) { val1=val1 = val1=A[$i]; val2=val2 = val2=B[$j]; } } } echo val1."".val1 . " " . val1."".val2; } // Driver code $A = array(4, 1, 2, 1, 1, 2 ); n=sizeof(n = sizeof(n=sizeof(A); $B = array(3, 6, 3, 3 ); m=sizeof(m = sizeof(m=sizeof(B); // Call to function findSwapValues($A, n,n, n,B, $m); // This code is contributed // by Akanksha Rai ?>

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

[Better Approach] Sorting both the arrays - O(NlogN + MlogM) Time and O(1) Space:

Suppose the sum of array A[] is sumA and sum of array B[] us sumB, then we need to find a value in A[], say X and a value in B[], say Y, such that:

sumA - X + Y = sumB - Y + X
2X - 2Y = sumA - sumB
X - Y = (sumA - sumB) / 2

To find the elements X and Y, we can sort the array and traverse the array 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.

Below is the implementation of the above approach:

C++ `

// CPP code for optimized implementation #include <bits/stdc++.h> 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

Python

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 <target:
        i+=1
    # Look for a greater value in list B
    else:
        j+=1

A=[4,1,2,1,1,2] B=[3,6,3,3]

findSwapValues(A,B)

#code contributed by sachin bisht

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

**Time Complexity: O(nlog(n) + mlog(m))
**Auxiliary Space: O(1)

**[Expected Approach] Using Hashing - O(N + M) Time and O(N) Space:

As discussed in the previous approach, we need to find an element in A[], say X and an element in B[], say Y such that X - Y = (SumA - SumB)/2, where SumA is the sum of all elements in A[] and SumB is the sum of all elements in B[].

In order to find such a pair, we can use a hash set to store all the values of array A[]. Then, we can iterate on array B[], and for each value in B[] check if ((sumA - sumB)/2 + Y) is present in the hash set or not. If it is present, then print the current element as Y and the element present in the hashset as X.

Below is an implementation of the approach:

C++ `

#include <bits/stdc++.h> using namespace std;

// function to find the values to swap void findSwapValues(int A[], int n, int B[], int m) { // Find the sum of both the arrays int sumA = accumulate(A, A + n, 0); int sumB = accumulate(B, B + m, 0);

// Check if the difference between the sum of both the
// arrays is even or not
if ((sumA - sumB) % 2 != 0) {
    cout << "No Possible Pair exists" << endl;
    return;
}

// Unordered set to store all the elements of A
unordered_set<int> possibleX;
for (int i = 0; i < n; i++) {
    possibleX.insert(A[i]);
}

// Iterate over all the elements of B[] and check if an
// element with the value = X is present in A[] or not
for (int i = 0; i < m; i++) {
    int X = (sumA - sumB) / 2 + B[i];
    if (possibleX.find(X) != possibleX.end()) {
        cout << X << " " << B[i] << endl;
        return;
    }
}

cout << "No Possible Pair exists" << endl;

}

int main() {

// Sample Input
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]);

// Function call to print a valid pair
findSwapValues(A, n, B, m);

return 0;

}

Java

import java.util.*;

class GFG { // Function to find the values to swap public static void findSwapValues(int[] A, int n, int[] B, int m) { // Find the sum of both the arrays int sumA = Arrays.stream(A).sum(); int sumB = Arrays.stream(B).sum();

    // Check if the difference between the sum of both
    // the arrays is even or not
    if ((sumA - sumB) % 2 != 0) {
        System.out.println("No Possible Pair exists");
        return;
    }

    // Set to store all the elements of A
    Set<Integer> possibleX = new HashSet<>();
    for (int i = 0; i < n; i++) {
        possibleX.add(A[i]);
    }

    // Iterate over all the elements of B and check if
    // an element with the value = X is present in A or
    // not
    for (int i = 0; i < m; i++) {
        int X = (sumA - sumB) / 2 + B[i];
        if (possibleX.contains(X)) {
            System.out.println(X + " " + B[i]);
            return;
        }
    }

    System.out.println("No Possible Pair exists");
}

public static void main(String[] args)
{
    // Sample Input
    int[] A = { 4, 1, 2, 1, 1, 2 };
    int n = A.length;

    int[] B = { 3, 6, 3, 3 };
    int m = B.length;

    // Function call to print a valid pair
    findSwapValues(A, n, B, m);
}

}

Python

def find_swap_values(A, B): # Find the sum of both the arrays sumA = sum(A) sumB = sum(B)

# Check if the difference between the sum of both the arrays is even or not
if (sumA - sumB) % 2 != 0:
    print("No Possible Pair exists")
    return

# Set to store all the elements of A
possibleX = set(A)

# Iterate over all the elements of B and check if an
# element with the value = X is present in A or not
for Y in B:
    X = (sumA - sumB) // 2 + Y
    if X in possibleX:
        print(X, Y)
        return

print("No Possible Pair exists")

Sample Input

A = [4, 1, 2, 1, 1, 2] B = [3, 6, 3, 3]

Function call to print a valid pair

find_swap_values(A, B)

**Time Complexity: O(n + m)
**Auxiliary Space: O(n)

Find a pair of elements swapping which makes sum of two arrays same (original) (raw)

Approaches to find a pair to swap which makes sum of two arrays same:

**[Naive Approach] Check all possible pairs - O(N*M) Time and O(1) Space:

Python code naive solution to find a pair swapping

which makes sum of lists sum.

Function to calculate sum of elements of list

Function to prints elements to be swapped

Driver code

Call to function

code contributed by sachin bisht

[Better Approach] Sorting both the arrays - O(NlogN + MlogM) Time and O(1) Space:

Python code for optimized implementation

Finds value of

a - b = (sumA - sumB) / 2

Prints elements to be swapped

**[Expected Approach] Using Hashing - O(N + M) Time and O(N) Space:

Sample Input

Function call to print a valid pair

****[Naive Approach] Check all possible pairs - O(N*M) Time and O(1) Space**: