Maximum increase in value of Matrix to keep maximum rows and columns unchanged

Given a matrix mat[][] of dimensionM*N, the task is to find the total maximum possible value must be increased to each cell(say mat[i][j]) such that maximum element of row i and column j remains unchanged.

Examples:

Input: N = 3, M = 3, mat[][] = { {4, 1, 3}, {3, 1, 1}, {2, 2, 0} }
Output: 6
Explanation:


The given matrix is:
4 1 3
3 1 1
2 2 0
Row-wise maximum values are: {4, 3, 2}
Column-wise maximum values are: {4, 2, 3}
After increasing the elements without affecting the row-wise and column-wise maximum values:
4 2 3
3 2 3
2 2 2
The total increase in corresponding cells of the two matrix is ((0 + 1 + 0) + (0 + 1 + 2) + (0 + 0 + 2)) = 6.

Input: M = 4, N = 4, mat[][] = { {3, 0, 8, 4}, {2, 4, 5, 7}, {9, 2, 6, 3}, {0, 3, 1, 0} }
Output: 35
Explanation:
The given matrix is:
3 0 8 4
2 4 5 7
9 2 6 3
0 3 1 0
Row-wise maximum values are: {8, 7, 9, 3}
Column-wise maximum values are: {9, 4, 8, 7}
After increasing the elements without affecting the row-wise and column-wise maximum values:
8 4 8 7
7 4 7 7
9 4 8 7
3 3 3 3
The total increase in corresponding cells of the two matrix is ((5 + 4 + 0 + 3) + (5 + 0 + 2 + 0) + (0 + 2 + 2 + 4) + (3 + 0 + 2 + 3)) = 35.

Approach:



  1. Traverse along each row and column of the given matrix to store the maximum values for each row and column.
  2. Traverse the given matrix mat[][] and for each cell(say mat[i][j]) and add the difference between the current element and the minimum value of the maximum values along its corresponding row and column as:

    difference = min(row[i], cols[j]) – mat[i][j]

  3. The total values added in the above operations is the required result.

Below is the implementation of the above approach:

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// C++ program for the above approach
  
#include <bits/stdc++.h>
using namespace std;
  
// Function to find the maximum increment
// in each cell of the given matrix such
// that maximum and minimum value remains
// unaffected
int maxIncrease(vector<vector<int> >& matrix)
{
    // Get the row of matrix as M
    int M = matrix.size();
  
    // Get the column of matrix as N
    int N = matrix[0].size();
  
    // To store the row-wise
    // maximum values
    vector<int> maxRowVal(M, 0);
  
    // To store the column-wise
    // maximum values
    vector<int> maxColVal(N, 0);
  
    // Traverse along the matrix
    // to store the maximum values
    // of each row and column
    for (int i = 0; i < M; i++) {
  
        for (int j = 0; j < N; j++) {
  
            // Store the row-wise
            // maximum values
            maxRowVal[i] = max(maxRowVal[i],
                               matrix[i][j]);
  
            // Store the column-wise
            // maximum values
            maxColVal[j] = max(maxColVal[j],
                               matrix[i][j]);
        }
    }
  
    // Calculate the total amount
    // of increment
    int totalIncrease = 0;
  
    // Traverse matrix mat[][]
    for (int i = 0; i < M; i++) {
  
        for (int j = 0; j < N; j++) {
  
            // The maximum possible
            // amount to increase
            int needToIncrease
                = min(maxRowVal[i],
                      maxColVal[j])
                  - matrix[i][j];
  
            // Total increased value
            totalIncrease += needToIncrease;
        }
    }
  
    // Return the total
    // increased value
    return totalIncrease;
}
  
// Driver Code
int main()
{
    // Given matrix
    vector<vector<int> > matrix{ { 3, 0, 8, 4 },
                                 { 2, 4, 5, 7 },
                                 { 9, 2, 6, 3 },
                                 { 0, 3, 1, 0 } };
  
    // Function Call
    cout << maxIncrease(matrix)
         << endl;
    return 0;
}
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// Java program for the above approach
import java.util.*;
class GFG{
  
// Function to find the maximum increment
// in each cell of the given matrix such
// that maximum and minimum value remains
// unaffected
static int maxIncrease(int [][] matrix)
{
    // Get the row of matrix as M
    int M = matrix.length;
  
    // Get the column of matrix as N
    int N = matrix[0].length;
  
    // To store the row-wise
    // maximum values
    int []maxRowVal = new int[M];
  
    // To store the column-wise
    // maximum values
    int []maxColVal = new int[N];
      
    // Traverse along the matrix
    // to store the maximum values
    // of each row and column
    for (int i = 0; i < M; i++) 
    {
        for (int j = 0; j < N; j++) 
        {
  
            // Store the row-wise
            // maximum values
            maxRowVal[i] = Math.max(maxRowVal[i],
                                    matrix[i][j]);
  
            // Store the column-wise
            // maximum values
            maxColVal[j] = Math.max(maxColVal[j],
                                    matrix[i][j]);
        }
    }
  
    // Calculate the total amount
    // of increment
    int totalIncrease = 0;
  
    // Traverse matrix mat[][]
    for (int i = 0; i < M; i++) 
    {
        for (int j = 0; j < N; j++) 
        {
  
            // The maximum possible
            // amount to increase
            int needToIncrease = Math.min(maxRowVal[i],
                                          maxColVal[j]) - 
                                          matrix[i][j];
  
            // Total increased value
            totalIncrease += needToIncrease;
        }
    }
  
    // Return the total
    // increased value
    return totalIncrease;
}
  
// Driver Code
public static void main(String[] args)
{
    // Given matrix
    int [][] matrix = { { 3, 0, 8, 4 },
                        { 2, 4, 5, 7 },
                        { 9, 2, 6, 3 },
                        { 0, 3, 1, 0 } };
  
    // Function Call
    System.out.print(maxIncrease(matrix) + "\n");
}
}
  
// This code is contributed by gauravrajput1
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// C# program for the above approach
using System;
  
class GFG{
  
// Function to find the maximum increment
// in each cell of the given matrix such
// that maximum and minimum value remains
// unaffected
static int maxIncrease(int [,] matrix)
{
      
    // Get the row of matrix as M
    int M = matrix.GetLength(0);
  
    // Get the column of matrix as N
    int N = matrix.GetLength(1);
  
    // To store the row-wise
    // maximum values
    int []maxRowVal = new int[M];
  
    // To store the column-wise
    // maximum values
    int []maxColVal = new int[N];
      
    // Traverse along the matrix
    // to store the maximum values
    // of each row and column
    for(int i = 0; i < M; i++) 
    {
       for(int j = 0; j < N; j++)
       {
            
          // Store the row-wise
          // maximum values
          maxRowVal[i] = Math.Max(maxRowVal[i],
                                  matrix[i, j]);
            
          // Store the column-wise
          // maximum values
          maxColVal[j] = Math.Max(maxColVal[j],
                                  matrix[i, j]);
       }
    }
  
    // Calculate the total amount
    // of increment
    int totalIncrease = 0;
  
    // Traverse matrix [,]mat
    for(int i = 0; i < M; i++) 
    {
       for(int j = 0; j < N; j++)
       {
             
          // The maximum possible
          // amount to increase
          int needToIncrease = Math.Min(maxRowVal[i],
                                        maxColVal[j]) - 
                                        matrix[i, j];
            
          // Total increased value
          totalIncrease += needToIncrease;
       }
    }
      
    // Return the total
    // increased value
    return totalIncrease;
}
  
// Driver Code
public static void Main(String[] args)
{
      
    // Given matrix
    int [,] matrix = { { 3, 0, 8, 4 },
                       { 2, 4, 5, 7 },
                       { 9, 2, 6, 3 },
                       { 0, 3, 1, 0 } };
  
    // Function call
    Console.Write(maxIncrease(matrix) + "\n");
}
}
  
// This code is contributed by 29AjayKumar
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Output:
35

Time Complexity: O(M*N)





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Improved By : GauravRajput1, 29AjayKumar

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