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

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}}
```

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

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++ 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 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 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# 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 `

 ` `

Output:
```0 1 0
3 0 2
0 4 0
```
Another approach :
 `// 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 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 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# 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 `

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

Time complexity : O(n*m)

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