Minimum number of steps to convert a given matrix into Diagonally Dominant Matrix

Given a matrix of order NxN, the task is to find the minimum number of steps to convert given matrix into Diagonally Dominant Matrix. In each step, the only operation allowed is to decrease or increase any element by 1.

Examples:

Input: mat[][] = {{3, 2, 4}, {1, 4, 4}, {2, 3, 4}}
Output: 5
Sum of the absolute values of elements of row 1 except
the diagonal element is 3 more than abs(arr[0][0]).
1 more than abs(arr[1][1]) in the second row
and 1 more than abs(arr[2][2]) in the third row.
Hence, 3 + 1 + 1 = 5

Input: mat[][] = {{1, 2, 4, 0}, {1, 3, 4, 2}, {3, 3, 4, 2}, {-1, 0, 1, 4}}
Output: 13

Approach:

  • A square matrix is said to be diagonally dominant matrix if for every row of the matrix, the magnitude of the diagonal entry in a row is larger than or equal to the sum of the magnitudes of all the other (non-diagonal) entries in that row.
  • The minimum number of steps required to convert a given matrix into the diagonally dominant matrix can be calculated depending upon two case:
    1. If the sum of the absolute value of all elements of a row except diagonal element is greater than the absolute value of diagonal element then the difference between these two values will be added to the result.
    2. Else no need to add anything in the result as in that case row satisfies the condition for a diagonally dominant matrix.

Below is the implementation of the above approach:

C++

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// C++ implementation of the apporach
#include <bits/stdc++.h>
using namespace std;
#define N 3
  
// Function to return the minimum steps
// required to convert the given matrix
// to a Diagonally Dominant Matrix
int findStepsForDDM(int arr[][N])
{
    int result = 0;
  
    // For each row
    for (int i = 0; i < N; i++) {
  
        // To store the sum of the current row
        int sum = 0;
        for (int j = 0; j < N; j++)
            sum += abs(arr[i][j]);
  
        // Remove the element of the current row
        // which lies on the main diagonal
        sum -= abs(arr[i][i]);
  
        // Checking if the diagonal element is less
        // than the sum of non-diagonal element
        // then add their difference to the result
        if (abs(arr[i][i]) < abs(sum))
            result += abs(abs(arr[i][i]) - abs(sum));
    }
  
    return result;
}
  
// Driven code
int main()
{
    int arr[N][N] = { { 3, -2, 1 },
                      { 1, -3, 2 },
                      { -1, 2, 4 } };
  
    cout << findStepsForDDM(arr);
  
    return 0;
}

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Java

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// Java implementation of the apporach 
class GFG 
{
      
    final static int N = 3 ;
      
    // Function to return the minimum steps 
    // required to convert the given matrix 
    // to a Diagonally Dominant Matrix 
    static int findStepsForDDM(int arr[][]) 
    
        int result = 0
      
        // For each row 
        for (int i = 0; i < N; i++) 
        
      
            // To store the sum of the current row 
            int sum = 0
            for (int j = 0; j < N; j++) 
                sum += Math.abs(arr[i][j]); 
      
            // Remove the element of the current row 
            // which lies on the main diagonal 
            sum -= Math.abs(arr[i][i]); 
      
            // Checking if the diagonal element is less 
            // than the sum of non-diagonal element 
            // then add their difference to the result 
            if (Math.abs(arr[i][i]) < Math.abs(sum)) 
                result += Math.abs(Math.abs(arr[i][i]) - Math.abs(sum)); 
        
      
        return result; 
    
      
    // Driven code 
    public static void main (String[] args) 
    {
          
        int arr[][] = { { 3, -2, 1 }, 
                        { 1, -3, 2 }, 
                        { -1, 2, 4 } }; 
      
        System.out.println(findStepsForDDM(arr)); 
    }
}
  
// This code is contributed by AnkitRai01

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Python3

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# Python3 implementation of the apporach
  
N = 3
  
# Function to return the minimum steps
# required to convert the given matrix
# to a Diagonally Dominant Matrix
def findStepsForDDM(arr):
  
    result = 0
  
    # For each row
    for i in range(N):
  
        # To store the sum of the current row
        sum = 0
        for j in range(N):
            sum += abs(arr[i][j])
  
        # Remove the element of the current row
        # which lies on the main diagonal
        sum -= abs(arr[i][i])
  
        # Checking if the diagonal element is less
        # than the sum of non-diagonal element
        # then add their difference to the result
        if (abs(arr[i][i]) < abs(sum)):
            result += abs(abs(arr[i][i]) - abs(sum))
  
    return result
  
# Driver code
  
arr= [ [ 3, -2, 1 ],
    [ 1, -3, 2 ],
    [ -1, 2, 4 ] ]
  
print(findStepsForDDM(arr))
  
# This code is contributed by mohit kumar 29

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C#

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// C# implementation of the apporach
using System;
  
class GFG
{
          
    static int N = 3 ;
      
    // Function to return the minimum steps 
    // required to convert the given matrix 
    // to a Diagonally Dominant Matrix 
    static int findStepsForDDM(int [,]arr) 
    
        int result = 0; 
      
        // For each row 
        for (int i = 0; i < N; i++) 
        
      
            // To store the sum of the current row 
            int sum = 0; 
            for (int j = 0; j < N; j++) 
                sum += Math.Abs(arr[i,j]); 
      
            // Remove the element of the current row 
            // which lies on the main diagonal 
            sum -= Math.Abs(arr[i,i]); 
      
            // Checking if the diagonal element is less 
            // than the sum of non-diagonal element 
            // then add their difference to the result 
            if (Math.Abs(arr[i,i]) < Math.Abs(sum)) 
                result += Math.Abs(Math.Abs(arr[i,i]) - Math.Abs(sum)); 
        
      
        return result; 
    
      
    // Driven code 
    static public void Main ()
    {
      
        int [,]arr = { { 3, -2, 1 }, 
                        { 1, -3, 2 }, 
                        { -1, 2, 4 } }; 
      
        Console.WriteLine(findStepsForDDM(arr)); 
    }
}
  
// This code is contributed by ajit.

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Output:

0

Time complexity: O(N2)



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