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

• Last Updated : 30 May, 2022

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 ` `using` `namespace` `std;`   `// Function to rotate every i-th` `// row of the matrix i times` `void` `rotateMatrix(vector >& 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 > 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.

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

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