# 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).
1 more than abs(arr) in the second row
and 1 more than abs(arr) 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

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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++

 `// C++ implementation of the apporach ` `#include ` `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; ` `} `

## Java

 `// 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 `

## Python3

 `# 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 `

## C#

 `// 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. `

Output:

```0
```

Time complexity: O(N2)

My Personal Notes arrow_drop_up Discovering ways to develop a plane for soaring career goals

If you like GeeksforGeeks 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.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.

Article Tags :
Practice Tags :

1

Please write to us at contribute@geeksforgeeks.org to report any issue with the above content.