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

Given a matrix mat[][] of dimensions M * N, the task is to print the matrix obtained after rotating every i^th 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 0^th row is rotated 0 times. Therefore, the 0^th row remains the same as {1, 2, 3}.
The 1^st row is rotated 1 times. Therefore, the 1^st row modifies to {6, 4, 5}.
The 2^nd row is rotated 2 times. Therefore, the 2^nd 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:

• Traverse the given matrix in row – wise manner and for every i^th row, perform the following steps:
  • Reverse the current row of the matrix.
  • Reverse the first i elements of the current row.
  • Reverse the last (N – i) elements of the current row, where N is the current size of the row.
• After completing the above steps, print the matrix mat[][].

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 << "
";
    }
}

// Driver Code
int main()
{
    vector<vector<int> > mat
        = { { 1, 2, 3 }, { 4, 5, 6 }, { 7, 8, 9 } };
    rotateMatrix(mat);

    return 0;
}

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.

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