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

Difficulty Level : Medium
Last Updated : 05 May, 2021

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

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.

Javascript

<script>
// Javascript 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
    function findMinOpeartion(matrix,n)
    {
        // Initialize the sumRow[]
        // and sumCol[] array to 0
        let sumRow = new Array(n);
        let sumCol = new Array(n);
        for(let i=0;i<n;i++)
        {   
            sumRow[i]=0;
            sumCol[i]=0;
        }
           
        // Calculate sumRow[] and
        // sumCol[] array
        for (let i = 0; i < n; ++i)
    
            for (let 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
        let maxSum = 0;
        for (let i = 0; i < n; ++i)
        {
            maxSum = Math.max(maxSum, sumRow[i]);
            maxSum = Math.max(maxSum, sumCol[i]);
        }
       
        let count = 0;
        for (let i = 0, j = 0; i < n && j < n;)
        {
            // Find minimum increment
            // required in either row
            // or column
            let 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
    function printMatrix(matrix,n)
    {
         for (let i = 0; i < n; ++i)
        {
            for (let j = 0; j < n; ++j)
                document.write(matrix[i][j] +
                                           " ");
            
            document.write("<br>");
        }
    }
     
    /* Driver program */
    let matrix=[[1, 2],[3, 4]];
    document.write(findMinOpeartion(matrix, 2)+"<br>");
    printMatrix(matrix, 2);
    // This code is contributed by avanitrachhadiya2155
</script>

Output
4
4 3
3 4

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

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready. To complete your preparation from learning a language to DS Algo and many more, please refer Complete Interview Preparation Course.

In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.

My Personal Notes