Modify a matrix by rotating ith row exactly i times in clockwise direction
Last Updated :
23 Jul, 2025
Given a matrix mat[][] of dimensions M * N, the task is to print the matrix obtained after rotating every ith row of the matrix i times in a clockwise direction.
Examples:
Input: mat[][] = {{1, 2, 3}, {4, 5, 6}, {7, 8, 9}}
Output:
1 2 3
6 4 5
8 9 7
Explanation:
The 0th row is rotated 0 times. Therefore, the 0th row remains the same as {1, 2, 3}.
The 1st row is rotated 1 times. Therefore, the 1st row modifies to {6, 4, 5}.
The 2nd row is rotated 2 times. Therefore, the 2nd row modifies to {8, 9, 7}.
After completing the above operations, the given matrix modifies to {{1, 2, 3}, {6, 4, 5}, {8, 9, 7}}.
Input: mat[][] = {{1, 2, 3, 4}, {4, 5, 6, 7}, {7, 8, 9, 8}, {7, 8, 9, 8}}
Output:
1 2 3 4
7 4 5 6
9 8 7 8
8 9 8 7
Approach: Follow the steps below to solve the problem:
Below is the implementation of the above approach:
C++
// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to rotate every i-th
// row of the matrix i times
void rotateMatrix(vector<vector<int> >& mat)
{
int i = 0;
// Traverse the matrix row-wise
for (auto& it : mat) {
// Reverse the current row
reverse(it.begin(), it.end());
// Reverse the first i elements
reverse(it.begin(), it.begin() + i);
// Reverse the last (N - i) elements
reverse(it.begin() + i, it.end());
// Increment count
i++;
}
// Print final matrix
for (auto rows : mat) {
for (auto cols : rows) {
cout << cols << " ";
}
cout << "\n";
}
}
// Driver Code
int main()
{
vector<vector<int> > mat
= { { 1, 2, 3 }, { 4, 5, 6 }, { 7, 8, 9 } };
rotateMatrix(mat);
return 0;
}
// java program for the above approach
import java.io.*;
import java.lang.*;
import java.util.*;
class GFG {
// Function to reverse arr[] from start to end
static void reverse(int arr[], int start, int end)
{
while (start < end) {
int temp = arr[start];
arr[start] = arr[end];
arr[end] = temp;
start++;
end--;
}
}
// Function to rotate every i-th
// row of the matrix i times
static void rotateMatrix(int mat[][])
{
int i = 0;
// Traverse the matrix row-wise
for (int rows[] : mat) {
// Reverse the current row
reverse(rows, 0, rows.length - 1);
// Reverse the first i elements
reverse(rows, 0, i - 1);
// Reverse the last (N - i) elements
reverse(rows, i, rows.length - 1);
// Increment count
i++;
}
// Print final matrix
for (int rows[] : mat) {
for (int cols : rows) {
System.out.print(cols + " ");
}
System.out.println();
}
}
// Driver Code
public static void main(String[] args)
{
int mat[][] = { { 1, 2, 3 },
{ 4, 5, 6 },
{ 7, 8, 9 } };
rotateMatrix(mat);
}
}
// This code is contributed by Kingash.
# Python3 program for the above approach
# Function to rotate every i-th
# row of the matrix i times
def rotateMatrix(mat):
i = 0
mat1 = []
# Traverse the matrix row-wise
for it in mat:
# Reverse the current row
it.reverse()
# Reverse the first i elements
it1 = it[:i]
it1.reverse()
# Reverse the last (N - i) elements
it2 = it[i:]
it2.reverse()
# Increment count
i += 1
mat1.append(it1 + it2)
# Print final matrix
for rows in mat1:
for cols in rows:
print(cols, end = " ")
print()
# Driver Code
if __name__ == "__main__":
mat = [ [ 1, 2, 3 ], [ 4, 5, 6 ], [ 7, 8, 9 ] ]
rotateMatrix(mat)
# This code is contributed by ukasp
// C# program to Modify a matrix
// by rotating ith row exactly
// i times in clockwise direction
using System;
class GFG
{
// Reverse each row of matrix
static void reverse(int N, int[, ] mat, int start, int end)
{
// Till start < end, swap the element
// at start and end index
while (start < end) {
// Swap the element
int temp = mat[N,start];
mat[N, start] = mat[N, end];
mat[N, end] = temp;
// Increment start and decrement
// end for next pair of swapping
start++;
end--;
}
}
// An Inplace function to
// rotate a N x N matrix
// by 90 degrees in anti-
// clockwise direction
static void rotateMatrix(int N, int[, ] mat)
{
int i = 0;
// Performing Transpose
for(int j = 0; j < N; j++)
{
// Reverse the current row
reverse(j, mat, 0, N - 1);
// Reverse the first i elements
reverse(j, mat, 0, i - 1);
// Reverse the last (N - i) elements
reverse(j, mat, i, N - 1);
// Increment count
i++;
}
for (i = 0; i < N; i++) {
for (int j = 0; j < N; j++)
Console.Write(mat[i, j] + " ");
Console.Write("\n");
}
}
// Driver Code
static public void Main()
{
int N = 3;
// Test Case 1
int[, ] mat = { { 1, 2, 3 },
{ 4, 5, 6 },
{ 7, 8, 9 }};
rotateMatrix(N, mat);
}
}
// This code is contributed by Aarti_Rathi
<script>
// javascript program for the above approach
// Function to reverse arr[] from start to end
function reverse(arr,start,end)
{
while (start < end) {
let temp = arr[start];
arr[start] = arr[end];
arr[end] = temp;
start++;
end--;
}
}
// Function to rotate every i-th
// row of the matrix i times
function rotateMatrix(mat)
{
let i = 0;
// Traverse the matrix row-wise
for (let rows=0;rows<mat.length;rows++) {
// Reverse the current row
reverse(mat[rows], 0, mat[rows].length - 1);
// Reverse the first i elements
reverse(mat[rows], 0, i - 1);
// Reverse the last (N - i) elements
reverse(mat[rows], i, mat[rows].length - 1);
// Increment count
i++;
}
// Print final matrix
for (let rows=0;rows< mat.length;rows++) {
for (let cols=0;cols< mat[rows].length;cols++) {
document.write(mat[rows][cols] + " ");
}
document.write("<br>");
}
}
// Driver Code
let mat=[[ 1, 2, 3 ],
[ 4, 5, 6 ],
[ 7, 8, 9 ]];
rotateMatrix(mat);
// This code is contributed by avanitrachhadiya2155
</script>
Output: 1 2 3
6 4 5
8 9 7
Time Complexity: O(M * N)
Auxiliary Space: O(1)
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