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# C++ Program to Inplace rotate square matrix by 90 degrees | Set 1

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:

1. 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
2. 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.
3. So run a loop in each cycle from x to N – x – 1, loop counter is y
4. 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)
5. Print the matrix.

## C++

 `// C++ program to rotate a matrix``// by 90 degrees``#include ``#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

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