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:
- Traverse the given matrix in row – wise manner and for every ith 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:
Javascript
<script>// javascript program for the above approach// Function to reverse arr[] from start to endfunction 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 timesfunction 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 Codelet mat=[[ 1, 2, 3 ], [ 4, 5, 6 ], [ 7, 8, 9 ]];rotateMatrix(mat);// This code is contributed by avanitrachhadiya2155</script> |
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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