Minimum operations required to make each row and column of matrix equals

Given a square matrix of size $n \times n$ . Find minimum number of operation are required such that sum of elements on each row and column becomes equals. In one operation, increment any value of cell of matrix by 1. In first line print minimum operation required and in next ‘n’ lines print ‘n’ integers representing the final matrix after operation.
Example:

Input: 
1 2
3 4
Output: 
4
4 3
3 4
Explanation
1. Increment value of cell(0, 0) by 3
2. Increment value of cell(0, 1) by 1
Hence total 4 operation are required

Input: 9
1 2 3
4 2 3
3 2 1
Output: 
6
2 4 3 
4 2 3 
3 3 3

Recommended: Please solve it on PRACTICE first, before moving on to the solution.

The approach is simple, let’s assume that maxSum is the maximum sum among all rows and columns. We just need to increment some cells such that the sum of any row or column becomes ‘maxSum’.
Let’s say X_i is the total number of operation needed to make the sum on row ‘i’ equals to maxSum and Y_j is the total number of operation needed to make the sum on column ‘j’ equals to maxSum. Since X_i = Y_j so we need to work at any one of them according to the condition.

In order to minimise X_i, we need to choose the maximum from rowSum_i and colSum_j as it will surely lead to minimum operation. After that, increment ‘i’ or ‘j’ according to the condition satisfied after increment.

Below is the implementation of the above approach.

C++

/* C++ Program to Find minimum number of 
operation required such that sum of 
elements on each row and column becomes same*/
#include <bits/stdc++.h> 
using namespace std; 
  
// Function to find minimum operation required 
// to make sum of each row and column equals 
int findMinOpeartion(int matrix[][2], int n) 
{ 
    // Initialize the sumRow[] and sumCol[] 
    // array to 0 
    int sumRow[n], sumCol[n]; 
    memset(sumRow, 0, sizeof(sumRow)); 
    memset(sumCol, 0, sizeof(sumCol)); 
  
    // Calculate sumRow[] and sumCol[] array 
    for (int i = 0; i < n; ++i) 
        for (int j = 0; j < n; ++j) { 
            sumRow[i] += matrix[i][j]; 
            sumCol[j] += matrix[i][j]; 
        } 
  
    // Find maximum sum value in either 
    // row or in column 
    int maxSum = 0; 
    for (int i = 0; i < n; ++i) { 
        maxSum = max(maxSum, sumRow[i]); 
        maxSum = max(maxSum, sumCol[i]); 
    } 
  
    int count = 0; 
    for (int i = 0, j = 0; i < n && j < n;) { 
  
        // Find minimum increment required in 
        // either row or column 
        int diff = min(maxSum - sumRow[i], 
                       maxSum - sumCol[j]); 
  
        // Add difference in corresponding cell, 
        // sumRow[] and sumCol[] array 
        matrix[i][j] += diff; 
        sumRow[i] += diff; 
        sumCol[j] += diff; 
  
        // Update the count variable 
        count += diff; 
  
        // If ith row satisfied, increment ith 
        // value for next iteration 
        if (sumRow[i] == maxSum) 
            ++i; 
  
        // If jth column satisfied, increment 
        // jth value for next iteration 
        if (sumCol[j] == maxSum) 
            ++j; 
    } 
    return count; 
} 
  
// Utility function to print matrix 
void printMatrix(int matrix[][2], int n) 
{ 
    for (int i = 0; i < n; ++i) { 
        for (int j = 0; j < n; ++j) 
            cout << matrix[i][j] << " "; 
        cout << "\n"; 
    } 
} 
  
// Driver code 
int main() 
{ 
    int matrix[][2] = { { 1, 2 }, 
                        { 3, 4 } }; 
    cout << findMinOpeartion(matrix, 2) << "\n"; 
    printMatrix(matrix, 2); 
    return 0; 
} 

Java

// Java Program to Find minimum  
// number of operation required  
// such that sum of elements on  
// each row and column becomes same 
import java.io.*; 
  
class GFG { 
  
    // Function to find minimum 
    // operation required 
    // to make sum of each row 
    // and column equals 
    static int findMinOpeartion(int matrix[][], 
                                         int n) 
    { 
        // Initialize the sumRow[] 
        // and sumCol[] array to 0 
        int[] sumRow = new int[n]; 
        int[] sumCol = new int[n]; 
          
        // Calculate sumRow[] and 
        // sumCol[] array 
        for (int i = 0; i < n; ++i) 
   
            for (int j = 0; j < n; ++j) 
            { 
                sumRow[i] += matrix[i][j]; 
                sumCol[j] += matrix[i][j]; 
            } 
      
        // Find maximum sum value  
        // in either row or in column 
        int maxSum = 0; 
        for (int i = 0; i < n; ++i)  
        { 
            maxSum = Math.max(maxSum, sumRow[i]); 
            maxSum = Math.max(maxSum, sumCol[i]); 
        } 
      
        int count = 0; 
        for (int i = 0, j = 0; i < n && j < n;)  
        { 
            // Find minimum increment 
            // required in either row 
            // or column 
            int diff = Math.min(maxSum - sumRow[i], 
                        maxSum - sumCol[j]); 
      
            // Add difference in  
            // corresponding cell, 
            // sumRow[] and sumCol[] 
            // array 
            matrix[i][j] += diff; 
            sumRow[i] += diff; 
            sumCol[j] += diff; 
      
            // Update the count  
            // variable 
            count += diff; 
      
            // If ith row satisfied, 
            // increment ith value  
            // for next iteration 
            if (sumRow[i] == maxSum) 
                ++i; 
      
            // If jth column satisfied,  
            // increment jth value for 
            // next iteration 
            if (sumCol[j] == maxSum) 
                ++j; 
        } 
        return count; 
    } 
  
    // Utility function to  
    // print matrix 
    static void printMatrix(int matrix[][], 
                                     int n) 
    { 
        for (int i = 0; i < n; ++i)  
        { 
            for (int j = 0; j < n; ++j) 
                System.out.print(matrix[i][j] + 
                                           " "); 
           
            System.out.println(); 
        } 
    } 
  
    /* Driver program */
    public static void main(String[] args) 
    { 
        int matrix[][] = {{1, 2}, 
                          {3, 4}}; 
          
        System.out.println(findMinOpeartion(matrix, 2)); 
        printMatrix(matrix, 2); 
  
    } 
} 
  
// This code is contributed by Gitanjali.  

Python 3

# Python 3 Program to Find minimum  
# number of operation required such  
# that sum of elements on each row 
# and column becomes same  
  
# Function to find minimum operation  
# required to make sum of each row  
# and column equals 
def findMinOpeartion(matrix, n): 
  
    # Initialize the sumRow[] and sumCol[] 
    # array to 0 
    sumRow = [0] * n 
    sumCol = [0] * n 
  
    # Calculate sumRow[] and sumCol[] array 
    for i in range(n): 
        for j in range(n) : 
            sumRow[i] += matrix[i][j] 
            sumCol[j] += matrix[i][j] 
  
    # Find maximum sum value in  
    # either row or in column 
    maxSum = 0
    for i in range(n) : 
        maxSum = max(maxSum, sumRow[i]) 
        maxSum = max(maxSum, sumCol[i]) 
  
    count = 0
    i = 0
    j = 0
    while i < n and j < n : 
  
        # Find minimum increment required  
        # in either row or column 
        diff = min(maxSum - sumRow[i],  
                   maxSum - sumCol[j]) 
  
        # Add difference in corresponding  
        # cell, sumRow[] and sumCol[] array 
        matrix[i][j] += diff 
        sumRow[i] += diff 
        sumCol[j] += diff 
  
        # Update the count variable 
        count += diff 
  
        # If ith row satisfied, increment  
        # ith value for next iteration 
        if (sumRow[i] == maxSum): 
            i += 1
  
        # If jth column satisfied, increment 
        # jth value for next iteration 
        if (sumCol[j] == maxSum): 
            j += 1
              
    return count 
  
# Utility function to print matrix 
def printMatrix(matrix, n): 
    for i in range(n) : 
        for j in range(n): 
            print(matrix[i][j], end = " ") 
        print() 
  
# Driver code 
if __name__ == "__main__": 
    matrix = [[ 1, 2 ], 
              [ 3, 4 ]] 
    print(findMinOpeartion(matrix, 2)) 
    printMatrix(matrix, 2) 
  
# This code is contributed 
# by ChitraNayal 

C#

// C# Program to Find minimum  
// number of operation required  
// such that sum of elements on  
// each row and column becomes same 
using System; 
  
class GFG { 
  
    // Function to find minimum 
    // operation required 
    // to make sum of each row 
    // and column equals 
    static int findMinOpeartion(int [,]matrix, 
                                        int n) 
    { 
        // Initialize the sumRow[] 
        // and sumCol[] array to 0 
        int[] sumRow = new int[n]; 
        int[] sumCol = new int[n]; 
          
        // Calculate sumRow[] and 
        // sumCol[] array 
        for (int i = 0; i < n; ++i) 
  
            for (int j = 0; j < n; ++j) 
            { 
                sumRow[i] += matrix[i,j]; 
                sumCol[j] += matrix[i,j]; 
            } 
      
        // Find maximum sum value  
        // in either row or in column 
        int maxSum = 0; 
        for (int i = 0; i < n; ++i)  
        { 
            maxSum = Math.Max(maxSum, sumRow[i]); 
            maxSum = Math.Max(maxSum, sumCol[i]); 
        } 
      
        int count = 0; 
        for (int i = 0, j = 0; i < n && j < n;)  
        { 
            // Find minimum increment 
            // required in either row 
            // or column 
            int diff = Math.Min(maxSum - sumRow[i], 
                        maxSum - sumCol[j]); 
      
            // Add difference in  
            // corresponding cell, 
            // sumRow[] and sumCol[] 
            // array 
            matrix[i,j] += diff; 
            sumRow[i] += diff; 
            sumCol[j] += diff; 
      
            // Update the count  
            // variable 
            count += diff; 
      
            // If ith row satisfied, 
            // increment ith value  
            // for next iteration 
            if (sumRow[i] == maxSum) 
                ++i; 
      
            // If jth column satisfied,  
            // increment jth value for 
            // next iteration 
            if (sumCol[j] == maxSum) 
                ++j; 
        } 
        return count; 
    } 
  
    // Utility function to  
    // print matrix 
    static void printMatrix(int [,]matrix, 
                                    int n) 
    { 
        for (int i = 0; i < n; ++i)  
        { 
            for (int j = 0; j < n; ++j) 
                Console.Write(matrix[i,j] + 
                                        " "); 
          
            Console.WriteLine(); 
        } 
    } 
  
    /* Driver program */
    public static void Main() 
    { 
        int [,]matrix = {{1, 2}, 
                        {3, 4}}; 
          
        Console.WriteLine(findMinOpeartion(matrix, 2)); 
        printMatrix(matrix, 2); 
  
    } 
} 
  
// This code is contributed by Vt_m.  

Output
4
4 3
3 4

Time complexity: O(n²)
Auxiliary space: O(n)

My Personal Notes

Recommended: Please solve it on PRACTICE first, before moving on to the solution.

C++

Java

Python 3

C#

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