Given a square matrix, turn it by 90 degrees in anti-clockwise direction without using any extra space.
Examples :
Input: Matrix: 1 2 3 4 5 6 7 8 9 Output: 3 6 9 2 5 8 1 4 7 The given matrix is rotated by 90 degree in anti-clockwise direction. Input: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 Output: 4 8 12 16 3 7 11 15 2 6 10 14 1 5 9 13 The given matrix is rotated by 90 degree in anti-clockwise direction.
An approach that requires extra space is already discussed here.
Approach: To solve the question without any extra space, rotate the array in form of squares, dividing the matrix into squares or cycles. For example,
A 4 X 4 matrix will have 2 cycles. The first cycle is formed by its 1st row, last column, last row and 1st column. The second cycle is formed by 2nd row, second-last column, second-last row and 2nd column. The idea is for each square cycle, swap the elements involved with the corresponding cell in the matrix in anti-clockwise direction i.e. from top to left, left to bottom, bottom to right and from right to top one at a time using nothing but a temporary variable to achieve this.
Demonstration:
First Cycle (Involves Red Elements) 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 Moving first group of four elements (First elements of 1st row, last row, 1st column and last column) of first cycle in counter clockwise. 4 2 3 16 5 6 7 8 9 10 11 12 1 14 15 13 Moving next group of four elements of first cycle in counter clockwise 4 8 3 16 5 6 7 15 2 10 11 12 1 14 9 13 Moving final group of four elements of first cycle in counter clockwise 4 8 12 16 3 6 7 15 2 10 11 14 1 5 9 13 Second Cycle (Involves Blue Elements) 4 8 12 16 3 6 7 15 2 10 11 14 1 5 9 13 Fixing second cycle 4 8 12 16 3 7 11 15 2 6 10 14 1 5 9 13
Algorithm:
- There is N/2 squares or cycles in a matrix of side N. Process a square one at a time. Run a loop to traverse the matrix a cycle at a time, i.e loop from 0 to N/2 – 1, loop counter is i
- Consider elements in group of 4 in current square, rotate the 4 elements at a time. So the number of such groups in a cycle is N – 2*i.
- So run a loop in each cycle from x to N – x – 1, loop counter is y
- The elements in the current group is (x, y), (y, N-1-x), (N-1-x, N-1-y), (N-1-y, x), now rotate the these 4 elements, i.e (x, y) <- (y, N-1-x), (y, N-1-x)<- (N-1-x, N-1-y), (N-1-x, N-1-y)<- (N-1-y, x), (N-1-y, x)<- (x, y)
- Print the matrix.
Javascript
<script>// Javascript program to rotate a// matrix by 90 degrees // An Inplace function to // rotate a N x N matrix // by 90 degrees in // anti-clockwise direction function rotateMatrix(N,mat) { // Consider all squares one by one for (let x = 0; x < N / 2; x++) { // Consider elements in group // of 4 in current square for (let y = x; y < N - x - 1; y++) { // Store current cell in // temp variable let temp = mat[x][y]; // Move values from right to top mat[x][y] = mat[y][N - 1 - x]; // Move values from bottom to right mat[y][N - 1 - x] = mat[N - 1 - x][N - 1 - y]; // Move values from left to bottom mat[N - 1 - x][N - 1 - y] = mat[N - 1 - y][x]; // Assign temp to left mat[N - 1 - y][x] = temp; } } } // Function to print the matrix function displayMatrix(N,mat) { for (let i = 0; i < N; i++) { for (let j = 0; j < N; j++) document.write( " " + mat[i][j]); document.write("<br>"); } document.write("<br>"); } /* Driver program to test above functions */ let N = 4; let mat=[[1, 2, 3, 4],[ 5, 6, 7, 8 ],[9, 10, 11, 12 ],[13, 14, 15, 16]]; // displayMatrix(mat); rotateMatrix(N, mat); // Print rotated matrix displayMatrix(N, mat); // This code is contributed by rag2127. </script> |
Output :
4 8 12 16 3 7 11 15 2 6 10 14 1 5 9 13
Complexity Analysis:
- Time Complexity: O(n*n), where n is size of array.
A single traversal of the matrix is needed. - Space Complexity: O(1).
As a constant space is needed
Please refer complete article on Inplace rotate square matrix by 90 degrees | Set 1 for more details!
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