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# Program to convert the diagonal elements of the matrix to 0

• Last Updated : 27 Apr, 2021

Given an N*N matrix. The task is to convert the elements of diagonal of a matrix to 0.
Examples

```Input: mat[][] =
{{ 2, 1, 7 },
{ 3, 7, 2 },
{ 5, 4, 9 }}
Output:
{ {0, 1, 0},
{3, 0, 2},
{0, 4, 0}}

Input:  mat[][] =
{{1, 3, 5, 6, 7},
{3, 5, 3, 2, 1},
{1, 2, 3, 4, 5},
{7, 9, 2, 1, 6},
{9, 1, 5, 3, 2}}
Output:
{{0, 3, 5, 6, 0},
{3, 0, 3, 0, 1},
{1, 2, 0, 4, 5},
{7, 0, 2, 0, 6},
{0, 1, 5, 3, 0}}```

Approach: Run two loops i.e. outer loop for no. of rows and inner loop for no. of columns. Check for the below condition:

if ((i == j ) || (i + j + 1) == n)
mat[i][j] = 0

Below is the implementation of the above approach:

## C++

 `// C++ program to change value of``// diagonal elements of a matrix to 0.``#include ``using` `namespace` `std;``const` `int` `MAX = 100;` `// to print the resultant matrix``void` `print(``int` `mat[][MAX], ``int` `n, ``int` `m)``{``    ``for` `(``int` `i = 0; i < n; i++) {``        ``for` `(``int` `j = 0; j < m; j++)``            ``cout << mat[i][j] << ``" "``;` `        ``cout << endl;``    ``}``}` `// function to change the values``// of diagonal elements to 0``void` `makediagonalzero(``int` `mat[][MAX], ``int` `n, ``int` `m)``{``    ``for` `(``int` `i = 0; i < n; i++) {``        ``for` `(``int` `j = 0; j < m; j++) {` `            ``// right and left diagonal condition``            ``if` `(i == j || (i + j + 1) == n)``                ``mat[i][j] = 0;``        ``}``    ``}` `    ``// print resultant matrix``    ``print(mat, n, m);``}` `// Driver code``int` `main()``{` `    ``int` `n = 3, m = 3;``    ``int` `mat[][MAX] = { { 2, 1, 7 },``                       ``{ 3, 7, 2 },``                       ``{ 5, 4, 9 } };` `    ``makediagonalzero(mat, n, m);` `    ``return` `0;``}`

## Java

 `// Java program to change value of``// diagonal elements of a matrix to 0.``class` `GFG``{` `static` `final` `int` `MAX = ``100``;` `// to print the resultant matrix``static` `void` `print(``int` `mat[][], ``int` `n, ``int` `m)``{``    ``for` `(``int` `i = ``0``; i < n; i++)``    ``{``        ``for` `(``int` `j = ``0``; j < m; j++)``        ``{``            ``System.out.print(mat[i][j] + ``" "``);``        ``}` `        ``System.out.println();``    ``}``}` `// function to change the values``// of diagonal elements to 0``static` `void` `makediagonalzero(``int` `mat[][],``                             ``int` `n, ``int` `m)``{``    ``for` `(``int` `i = ``0``; i < n; i++)``    ``{``        ``for` `(``int` `j = ``0``; j < m; j++)``        ``{` `            ``// right and left diagonal condition``            ``if` `(i == j || (i + j + ``1``) == n)``            ``{``                ``mat[i][j] = ``0``;``            ``}``        ``}``    ``}` `    ``// print resultant matrix``    ``print(mat, n, m);``}` `// Driver code``public` `static` `void` `main(String args[])``{``    ``int` `n = ``3``, m = ``3``;``    ``int` `mat[][] = {{``2``, ``1``, ``7``},``                   ``{``3``, ``7``, ``2``},``                   ``{``5``, ``4``, ``9``}};` `    ``makediagonalzero(mat, n, m);``}``}` `// This code is contributed``// by PrinciRaj1992`

## Python 3

 `# Python 3 program to change value of``# diagonal elements of a matrix to 0.``MAX` `=` `100` `# to print the resultant matrix``def` `print_1(mat, n, m):``    ` `    ``for` `i ``in` `range``(n):``        ``for` `j ``in` `range``(m):``            ``print``( mat[i][j], end ``=` `" "``)` `        ``print``()` `# function to change the values``# of diagonal elements to 0``def` `makediagonalzero(mat, n, m):` `    ``for` `i ``in` `range``(n):``        ``for` `j ``in` `range``(m):` `            ``# right and left diagonal condition``            ``if` `(i ``=``=` `j ``or` `(i ``+` `j ``+` `1``) ``=``=` `n):``                ``mat[i][j] ``=` `0` `    ``# print resultant matrix``    ``print_1(mat, n, m)` `# Driver code``if` `__name__ ``=``=` `"__main__"``:` `    ``n ``=` `3``    ``m ``=` `3``    ``mat ``=` `[[ ``2``, ``1``, ``7` `],``           ``[ ``3``, ``7``, ``2` `],``           ``[ ``5``, ``4``, ``9` `]]` `    ``makediagonalzero(mat, n, m)` `# This code is contributed by ChitraNayal`

## C#

 `// C# program to change value of``// diagonal elements of a matrix to 0.``using` `System;` `class` `GFG``{` `static` `int` `MAX = 100;` `// to print the resultant matrix``static` `void` `print(``int``[,] mat, ``int` `n, ``int` `m)``{``    ``for` `(``int` `i = 0; i < n; i++)``    ``{``        ``for` `(``int` `j = 0; j < m; j++)``        ``{``            ``Console.Write(mat[i, j] + ``" "``);``        ``}` `        ``Console.WriteLine();``    ``}``}` `// function to change the values``// of diagonal elements to 0``static` `void` `makediagonalzero(``int``[,] mat,``                            ``int` `n, ``int` `m)``{``    ``for` `(``int` `i = 0; i < n; i++)``    ``{``        ``for` `(``int` `j = 0; j < m; j++)``        ``{` `            ``// right and left diagonal condition``            ``if` `(i == j || (i + j + 1) == n)``            ``{``                ``mat[i, j] = 0;``            ``}``        ``}``    ``}` `    ``// print resultant matrix``    ``print(mat, n, m);``}` `// Driver code``public` `static` `void` `Main()``{``    ``int` `n = 3, m = 3;``    ``int``[,] mat = {{2, 1, 7},``                  ``{3, 7, 2},``                  ``{5, 4, 9}};` `    ``makediagonalzero(mat, n, m);``}``}` `// This code is contributed``// by Akanksha Rai`

## PHP

 ``

## Javascript

 ``
Output:
```0 1 0
3 0 2
0 4 0```

Another approach :

## C++

 `// C++ program to change value of``// diagonal elements of a matrix to 0.``#include ``#define COL 100``using` `namespace` `std;` `// Method to replace the diagonal matrix with zeros``void` `diagonalMat( ``int` `m[][COL] ,``int` `row, ``int` `col)``{` `    ``// l is the left iterator which is``    ``// iterationg from 0 to col-1 here``    ``//k is the right iterator which is``    ``// iterating from col-1 to 0``    ``int` `i = 0, l = 0, k = col - 1;` `    ``// i used to iterate over rows of the matrix``    ``while` `(i < row)``    ``{``        ``int` `j = 0;``        ` `        ``// condition to check if it is``        ``// the centre of the matrix``        ``if` `(l == k)``        ``{``            ``m[l][k] = 0;``            ``l++;``            ``k--;``        ``}         ``//otherwize the diagonal will be equivalaent to l or k``            ``//increment l because l is traversing from left``            ``//to right and decrement k for vice-cersa``        ``else``        ``{``            ``m[i][l] = 0;``            ``l++;``            ``m[i][k] = 0;``            ``k--;``        ``}``        ` `        ``// print every element after replacing from the column``        ``while` `(j < col)``        ``{``            ``cout << ``" "``<< m[i][j];``            ``j++;``        ``}``        ``i++;``        ``cout << ``"\n"``;``    ``}` `}` `// Driver code``int` `main()``{``    ``int` `m[][COL] ={{ 2, 1, 7},``                    ``{ 3, 7, 2},``                    ``{ 5, 4, 9}};``    ``int` `row = 3, col = 3;``    ``diagonalMat( m, row, col);``    ``return` `0;``}` `// This code has been contributed by 29AjayKumar`

## Java

 `// Java program to change value of``// diagonal elements of a matrix to 0.``import` `java.io.*;` `class` `GFG {``    ``public` `static` `void` `main (String[] args) {``        ``int` `m[][] = {{ ``2``, ``1``, ``7` `},``                    ``{ ``3``, ``7``, ``2` `},``                    ``{ ``5``, ``4``, ``9` `}};``        ``int` `row = ``3``, col = ``3``;``        ``GFG.diagonalMat(row,col,m);``    ``}``    ``// method to replace the diagonal matrix with zeros``    ``public` `static` `void` `diagonalMat(``int` `row, ``int` `col, ``int` `m[][]){``        ` `        ``// l is the left iterator which is``        ``// iterationg from 0 to col-1 here``        ``//k is the right iterator which is``        ``// iterating from col-1 to 0``        ``int` `i =``0``,l=``0``,k=col-``1``;``        ` `        ``// i used to iterate over rows of the matrix``        ``while``(i

## Python3

 `# Python3 program to change value of``# diagonal elements of a matrix to 0.` `# method to replace the diagonal``# matrix with zeros``def` `diagonalMat(row, col, m):` `    ``# l is the left iterator which is``    ``# iterationg from 0 to col-1 here``    ``# k is the right iterator which is``    ``# iterating from col-1 to 0``    ``i, l, k ``=` `0``, ``0``, col ``-` `1``;` `    ``# i used to iterate over rows of the matrix``    ``while``(i < row):``        ``j ``=` `0``;``        ` `        ``# condition to check if it is``        ``# the centre of the matrix``        ``if``(l ``=``=` `k):``            ``m[l][k] ``=` `0``;``            ``l ``+``=` `1``;``            ``k ``-``=` `1``;``            ` `        ``# otherwize the diagonal will be equivalaent to l or k``        ``# increment l because l is traversing from left``        ``# to right and decrement k for vice-cersa``        ``else``:``            ``m[i][l] ``=` `0``;``            ``l ``+``=` `1``;``            ``m[i][k] ``=` `0``;``            ``k ``-``=` `1``;``            ` `        ``# print every element``        ``# after replacing from the column``        ``while``(j < col):``            ``print``(``" "``, m[i][j], end ``=` `"");``            ``j ``+``=` `1``;``        ``i ``+``=` `1``;``        ``print``("");` `# Driver Code``if` `__name__ ``=``=` `'__main__'``:``    ``m ``=` `[[``2``, ``1``, ``7` `],``        ``[ ``3``, ``7``, ``2` `],``        ``[ ``5``, ``4``, ``9` `]];``    ``row, col ``=` `3``, ``3``;``    ``diagonalMat(row, col, m);``    ` `# This code contributed by Rajput-Ji`

## C#

 `// C# program to change value of``// diagonal elements of a matrix to 0.``using` `System;` `class` `GFG {``    ``public` `static` `void` `Main () {``        ``int``[,]  m = {{ 2, 1, 7 },``                    ``{ 3, 7, 2 },``                    ``{ 5, 4, 9 }};``        ``int` `row = 3, col = 3;``        ``GFG.diagonalMat(row,col,m);``    ``}``    ``// method to replace the diagonal matrix with zeros``    ``public` `static` `void` `diagonalMat(``int` `row, ``int` `col, ``int``[,] m){``        ` `        ``// l is the left iterator which is``        ``// iterationg from 0 to col-1 here``        ``//k is the right iterator which is``        ``// iterating from col-1 to 0``        ``int` `i =0,l=0,k=col-1;``        ` `        ``// i used to iterate over rows of the matrix``        ``while``(i < row){``            ``int` `j=0;``            ``// condition to check if it is``            ``// the centre of the matrix``            ``if``(l==k){``                ``m[l,k] = 0;``                ``l++;``                ``k--;``            ``}``            ``//otherwize the diagonal will be equivalaent to l or k``            ``//increment l because l is traversing from left``            ``//to right and decrement k for vice-cersa``            ``else``{``                ``m[i,l] = 0;``                ``l++;``                ``m[i,k]=0;``                ``k--;``            ``}``            ``// print every element after replacing from the column``            ``while``(j < col){``                ``Console.Write(``" "` `+ m[i,j]);``                ``j++;``            ``}``            ``i++;``            ``Console.WriteLine();``        ``}``        ` `    ``}` `}``      ``//This code is contributed by Mukul singh`

## Javascript

 ``
Output:
``` 0 1 0
3 0 2
0 4 0```

Time complexity : O(n*m)

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