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.
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.
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