Given a matrix, clockwise rotate elements in it.
Examples:
Input 1 2 3 4 5 6 7 8 9 Output: 4 1 2 7 5 3 8 9 6 For 4*4 matrix Input: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 Output: 5 1 2 3 9 10 6 4 13 11 7 8 14 15 16 12
The idea is to use loops similar to the program for printing a matrix in spiral form. One by one rotate all rings of elements, starting from the outermost. To rotate a ring, we need to do following.
1) Move elements of top row.
2) Move elements of last column.
3) Move elements of bottom row.
4) Move elements of first column.
Repeat above steps for inner ring while there is an inner ring.
Below is the implementation of above idea. Thanks to Gaurav Ahirwar for suggesting below solution.
Javascript
<script>// Javascript program to rotate a matrix let R = 4;let C = 4;// A function to rotate a matrix // mat[][] of size R x C.// Initially, m = R and n = Cfunction rotatematrix(m, n, mat){ let row = 0, col = 0; let prev, curr; /* row - Starting row index m - ending row index col - starting column index n - ending column index i - iterator */ while (row < m && col < n) { if (row + 1 == m || col + 1 == n) break; // Store the first element of next // row, this element will replace // first element of current row prev = mat[row + 1][col]; // Move elements of first row // from the remaining rows for(let i = col; i < n; i++) { curr = mat[row][i]; mat[row][i] = prev; prev = curr; } row++; // Move elements of last column // from the remaining columns for(let i = row; i < m; i++) { curr = mat[i][n - 1]; mat[i][n - 1] = prev; prev = curr; } n--; // Move elements of last row // from the remaining rows if (row < m) { for(let i = n - 1; i >= col; i--) { curr = mat[m - 1][i]; mat[m - 1][i] = prev; prev = curr; } } m--; // Move elements of first column // from the remaining rows if (col < n) { for(let i = m - 1; i >= row; i--) { curr = mat[i][col]; mat[i][col] = prev; prev = curr; } } col++; } // Print rotated matrix for(let i = 0; i < R; i++) { for(let j = 0; j < C; j++) document.write( mat[i][j] + " "); document.write("<br>"); }}// Driver code// Test Case 1let a = [ [ 1, 2, 3, 4 ], [ 5, 6, 7, 8 ], [ 9, 10, 11, 12 ], [ 13, 14, 15, 16 ] ]; rotatematrix(R, C, a);// This code is contributed by avanitrachhadiya2155</script> |
Output:
5 1 2 3 9 10 6 4 13 11 7 8 14 15 16 12
Time Complexity: O(max(m,n) * max(m,n))
Auxiliary Space: O(m*n)
Please refer complete article on Rotate Matrix Elements for more details!
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