Program to convert the diagonal elements of the matrix to 0
Last Updated :
15 Sep, 2022
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 <iostream>
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
<?php
// PHP program to change value of
// diagonal elements of a matrix to 0.
// to print the resultant matrix
function printsd(&$mat, $n, $m)
{
for ($i = 0; $i < $n; $i++)
{
for ($j = 0; $j < $m; $j++)
{
echo ($mat[$i][$j]);
echo (" ");
}
echo ("\n");
}
}
// function to change the values
// of diagonal elements to 0
function makediagonalzero(&$mat, $n, $m)
{
for ($i = 0; $i < $n; $i++)
{
for ($j = 0; $j < $m; $j++)
{
// right and left diagonal condition
if ($i == $j || ($i + $j + 1) == $n)
$mat[$i][$j] = 0;
}
}
// print resultant matrix
printsd($mat, $n, $m);
}
// Driver code
$n = 3; $m = 3;
$mat = array(array(2, 1, 7),
array(3, 7, 2),
array(5, 4, 9));
makediagonalzero($mat, $n, $m);
// This code is contributed by Shivi_Aggarwal
?>
<script>
// Javascript program to change value of
// diagonal elements of a matrix to 0.
const MAX = 100;
// to print the resultant matrix
function print(mat, n, m)
{
for (let i = 0; i < n; i++) {
for (let j = 0; j < m; j++)
document.write(mat[i][j] + " ");
document.write("<br>");
}
}
// function to change the values
// of diagonal elements to 0
function makediagonalzero(mat, n, m)
{
for (let i = 0; i < n; i++)
{
for (let 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
let n = 3, m = 3;
let mat = [ [ 2, 1, 7 ],
[ 3, 7, 2 ],
[ 5, 4, 9 ] ];
makediagonalzero(mat, n, m);
// This code is contributed by souravmahato348.
</script>
// C program to change value of
// diagonal elements of a matrix to 0.
#include <stdio.h>
#define 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++)
printf("%d ",mat[i][j]);
printf("\n");
}
}
// 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;
}
Time Complexity: O(n*m)
Auxiliary Space: O(1)
Another Approach:
C++
// C++ program to change value of
// diagonal elements of a matrix to 0.
#include <bits/stdc++.h>
#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[4] 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--;
} // otherwise the diagonal will be equivalent 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
// C program to change value of
// diagonal elements of a matrix to 0.
#include <stdio.h>
#define COL 100
// 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[4] 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--;
} //otherwise the diagonal will be equivalent 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)
{
printf(" %d",m[i][j]);
j++;
}
i++;
printf("\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 kothavvsaakash
// 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[4] 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--;
}
//otherwise the diagonal will be equivalent 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){
System.out.print(" "+ m[i][j]);
j++;
}
i++;
System.out.println();
}
}
}
# 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[4] 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;
# otherwise the diagonal will be equivalent 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[4] 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--;
}
//otherwise the diagonal will be equivalent 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
<script>
// Javascript program to change value of
// diagonal elements of a matrix to 0.
const COL = 100;
// Method to replace the diagonal matrix with zeros
function diagonalMat(m, row, col)
{
// l is the left iterator which is
// iterationg from 0 to col-1[4] here
//k is the right iterator which is
// iterating from col-1 to 0
let i = 0, l = 0, k = col - 1;
// i used to iterate over rows of the matrix
while (i < row)
{
let j = 0;
// condition to check if it is
// the centre of the matrix
if (l == k)
{
m[l][k] = 0;
l++;
k--;
} //otherwise the diagonal will be equivalent 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)
{
document.write(" " + m[i][j]);
j++;
}
i++;
document.write("<br>");
}
}
// Driver code
let m =[[ 2, 1, 7],
[ 3, 7, 2],
[ 5, 4, 9]];
let row = 3, col = 3;
diagonalMat( m, row, col);
</script>
Time Complexity: O(n*m)
Auxiliary Space: O(1)
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