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/C++
// 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 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 |
PHP
<?php // PHP code naive solution to find // a pair swapping which makes sum // of arrays sum. // Function to calculate sum of // elements of array function getSum( $X , $n ) { $sum = 0; for ( $i = 0; $i < $n ; $i ++) $sum += $X [ $i ]; return $sum ; } function findSwapValues( $A , $n , $B , $m ) { // Calculation of sums from both arrays $sum1 = getSum( $A , $n ); $sum2 = getSum( $B , $m ); // Look for val1 and val2, such that // sumA - val1 + val2 = sumB - val2 + val1 for ( $i = 0; $i < $n ; $i ++) { for ( $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 ]; } } } echo $val1 . " " . $val2 ; } // Driver code $A = array (4, 1, 2, 1, 1, 2 ); $n = sizeof( $A ); $B = array (3, 6, 3, 3 ); $m = sizeof( $B ); // Call to function findSwapValues( $A , $n , $B , $m ); // This code is contributed // by Akanksha Rai ?> |
Output :
1 3
Time Complexity :- O(n*m)
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/C++
// CPP code for naive implementation #include <iostream> 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 |
Python
# 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
<?php // PHP code for naive implementation // Function to calculate sum // of elements of array function getSum( $X , $n ) { $sum = 0; for ( $i = 0; $i < $n ; $i ++) $sum += $X [ $i ]; return $sum ; } // Function to calculate : // a - b = (sumA - sumB) / 2 function getTarget( $A , $n , $B , $m ) { // Calculation of sums from // both arrays $sum1 = getSum( $A , $n ); $sum2 = getSum( $B , $m ); // because that the target // must be an integer if (( $sum1 - $sum2 ) % 2 != 0) return 0; return (( $sum1 - $sum2 ) / 2); } function findSwapValues( $A , $n , $B , $m ) { $target = getTarget( $A , $n , $B , $m ); if ( $target == 0) return ; // Look for val1 and val2, such that // val1 - val2 = (sumA - sumB) / 2 for ( $i = 0; $i < $n ; $i ++) { for ( $j = 0; $j < $m ; $j ++) { if ( $A [ $i ] - $B [ $j ] == $target ) { $val1 = $A [ $i ]; $val2 = $B [ $j ]; } } } echo $val1 . " " . $val2 ; } // Driver code $A = array (4, 1, 2, 1, 1, 2); $n = sizeof( $A ); $B = array (3, 6, 3, 3); $m = sizeof( $B ); // Call to function findSwapValues( $A , $n , $B , $m ); // This code is cotnributed // by Akanksha Rai ?> |
Output:
1 3
Time Complexity :- O(n*m)
Method 3 -> Optimized Solution :-
- Sort the arrays.
- Traverse both array simultaneously and do following for every pair.
- If the difference is too small then, make it bigger by moving ‘a’ to a bigger value.
- If it is too big then, make it smaller by moving b to a bigger value.
- 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/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 contibuted 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 |
Output:
2 3
Time Complexity :-
If arrays are sorted : O(n + m)
If arrays aren’t sorted : O(nlog(n) + mlog(m))
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) ans 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.
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