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C++ Program to Modify a matrix by rotating ith row exactly i times in clockwise direction

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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

// Java program for the above approach
 
import java.util.*;
 
class Main {
    // Function to rotate every i-th row of the matrix i
    // times
    static void rotateMatrix(List<List<Integer> > mat)
    {
        int i = 0;
 
        // Traverse the matrix row-wise
        for (List<Integer> it : mat) {
            // Reverse the current row
            Collections.reverse(it);
 
            // Reverse the first i elements
            Collections.reverse(it.subList(0, i));
 
            // Reverse the last (N - i) elements
            Collections.reverse(it.subList(i, it.size()));
 
            // Increment count
            i++;
        }
 
        // Print final matrix
        for (List<Integer> rows : mat) {
            for (int cols : rows) {
                System.out.print(cols + " ");
            }
            System.out.println();
        }
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        List<List<Integer> > mat = new ArrayList<>();
        mat.add(Arrays.asList(1, 2, 3));
        mat.add(Arrays.asList(4, 5, 6));
        mat.add(Arrays.asList(7, 8, 9));
        rotateMatrix(mat);
    }
}

                    

Output: 
1 2 3 
6 4 5 
8 9 7

 

Time Complexity: O(N*M), as we are using nested loops to traverse N*M times.

Auxiliary Space: O(1), as we are not using any extra space.

Please refer complete article on Modify a matrix by rotating ith row exactly i times in clockwise direction for more details!



Last Updated : 31 Mar, 2023
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