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)
- Print the matrix.
// C++ program to rotate a matrix // by 90 degrees #include <bits/stdc++.h> #define N 4 using namespace std;
void displayMatrix(
int mat[N][N]);
// An Inplace function to // rotate a N x N matrix // by 90 degrees in // anti-clockwise direction void rotateMatrix( int mat[][N])
{ // Consider all squares one by one
for ( int x = 0; x < N / 2; x++) {
// Consider elements in group
// of 4 in current square
for ( int y = x; y < N - x - 1; y++) {
// Store current cell in
// temp variable
int 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 void displayMatrix( int mat[N][N])
{ for ( int i = 0; i < N; i++) {
for ( int j = 0; j < N; j++)
printf ( "%2d " , mat[i][j]);
printf ("
"); }
printf ("
"); } /* Driver program to test above functions */ int main()
{ // Test Case 1
int mat[N][N] = {
{ 1, 2, 3, 4 },
{ 5, 6, 7, 8 },
{ 9, 10, 11, 12 },
{ 13, 14, 15, 16 }
};
// Tese Case 2
/* int mat[N][N] = {
{1, 2, 3},
{4, 5, 6},
{7, 8, 9}
};
*/
// Tese Case 3
/*int mat[N][N] = {
{1, 2},
{4, 5}
};*/
// displayMatrix(mat);
rotateMatrix(mat);
// Print rotated matrix
displayMatrix(mat);
return 0;
} |
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 side 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!