# Rotate a matrix by 90 degree in clockwise direction without using any extra space

• Difficulty Level : Medium
• Last Updated : 11 Oct, 2021

Given a square matrix, turn it by 90 degrees in a clockwise direction without using any extra space.

Examples:

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```Input:
1 2 3
4 5 6
7 8 9
Output:
7 4 1
8 5 2
9 6 3

Input:
1 2
3 4
Output:
3 1
4 2 ```

Method 1

Approach: The approach is similar to Inplace rotate square matrix by 90 degrees | Set 1. The only thing that is different is to print the elements of the cycle in a clockwise direction i.e. An N x N matrix will have floor(N/2) square cycles.
For example, a 3 X 3 matrix will have 1 cycle. The cycle is formed by its 1st row, last column, last row, and 1st column.
For each square cycle, we swap the elements involved with the corresponding cell in the matrix in the clockwise direction. We just need a temporary variable for this.

Explanation:

Let size of row and column be 3.
During first iteration –
a[i][j] = Element at first index (leftmost corner top)= 1.
a[j][n-1-i]= Rightmost corner top Element = 3.
a[n-1-i][n-1-j] = Rightmost corner bottom element = 9.
a[n-1-j][i] = Leftmost corner bottom element = 7.
Move these elements in the clockwise direction.
During second iteration –
a[i][j] = 2.
a[j][n-1-i] = 6.
a[n-1-i][n-1-j] = 8.
a[n-1-j][i] = 4.
Similarly, move these elements in the clockwise direction.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of above approach``#include ``using` `namespace` `std;` `#define N 4` `// Function to rotate the matrix 90 degree clockwise``void` `rotate90Clockwise(``int` `a[N][N])``{` `    ``// Traverse each cycle``    ``for` `(``int` `i = 0; i < N / 2; i++) {``        ``for` `(``int` `j = i; j < N - i - 1; j++) {` `            ``// Swap elements of each cycle``            ``// in clockwise direction``            ``int` `temp = a[i][j];``            ``a[i][j] = a[N - 1 - j][i];``            ``a[N - 1 - j][i] = a[N - 1 - i][N - 1 - j];``            ``a[N - 1 - i][N - 1 - j] = a[j][N - 1 - i];``            ``a[j][N - 1 - i] = temp;``        ``}``    ``}``}` `// Function for print matrix``void` `printMatrix(``int` `arr[N][N])``{``    ``for` `(``int` `i = 0; i < N; i++) {``        ``for` `(``int` `j = 0; j < N; j++)``            ``cout << arr[i][j] << ``" "``;``        ``cout << ``'\n'``;``    ``}``}` `// Driver code``int` `main()``{``    ``int` `arr[N][N] = { { 1, 2, 3, 4 },``                      ``{ 5, 6, 7, 8 },``                      ``{ 9, 10, 11, 12 },``                      ``{ 13, 14, 15, 16 } };``    ``rotate90Clockwise(arr);``    ``printMatrix(arr);``    ``return` `0;``}`

## Java

 `// Java implementation of above approach``import` `java.io.*;` `class` `GFG``{``    ` `static` `int` `N = ``4``;` `// Function to rotate the matrix 90 degree clockwise``static` `void` `rotate90Clockwise(``int` `a[][])``{` `    ``// Traverse each cycle``    ``for` `(``int` `i = ``0``; i < N / ``2``; i++)``    ``{``        ``for` `(``int` `j = i; j < N - i - ``1``; j++)``        ``{` `            ``// Swap elements of each cycle``            ``// in clockwise direction``            ``int` `temp = a[i][j];``            ``a[i][j] = a[N - ``1` `- j][i];``            ``a[N - ``1` `- j][i] = a[N - ``1` `- i][N - ``1` `- j];``            ``a[N - ``1` `- i][N - ``1` `- j] = a[j][N - ``1` `- i];``            ``a[j][N - ``1` `- i] = temp;``        ``}``    ``}``}` `// Function for print matrix``static` `void` `printMatrix(``int` `arr[][])``{``    ``for` `(``int` `i = ``0``; i < N; i++)``    ``{``        ``for` `(``int` `j = ``0``; j < N; j++)``        ``System.out.print( arr[i][j] + ``" "``);``        ``System.out.println();``    ``}``}` `// Driver code` `    ``public` `static` `void` `main (String[] args)``    ``{``            ``int` `arr[][] = { { ``1``, ``2``, ``3``, ``4` `},``                      ``{ ``5``, ``6``, ``7``, ``8` `},``                      ``{ ``9``, ``10``, ``11``, ``12` `},``                      ``{ ``13``, ``14``, ``15``, ``16` `} };``    ``rotate90Clockwise(arr);``    ``printMatrix(arr);``    ``}``}` `// This code has been contributed by inder_verma.`

## Python

 `# Function to rotate the matrix``# 90 degree clockwise``def` `rotate90Clockwise(A):``    ``N ``=` `len``(A[``0``])``    ``for` `i ``in` `range``(N ``/``/` `2``):``        ``for` `j ``in` `range``(i, N ``-` `i ``-` `1``):``            ``temp ``=` `A[i][j]``            ``A[i][j] ``=` `A[N ``-` `1` `-` `j][i]``            ``A[N ``-` `1` `-` `j][i] ``=` `A[N ``-` `1` `-` `i][N ``-` `1` `-` `j]``            ``A[N ``-` `1` `-` `i][N ``-` `1` `-` `j] ``=` `A[j][N ``-` `1` `-` `i]``            ``A[j][N ``-` `1` `-` `i] ``=` `temp` `# Function to print the matrix``def` `printMatrix(A):``    ``N ``=` `len``(A[``0``])``    ``for` `i ``in` `range``(N):``        ``print``(A[i])` `# Driver code``A ``=` `[[``1``, ``2``, ``3``, ``4``],``     ``[``5``, ``6``, ``7``, ``8``],``     ``[``9``, ``10``, ``11``, ``12``],``     ``[``13``, ``14``, ``15``, ``16``]]``rotate90Clockwise(A)``printMatrix(A)` `# This code was contributed``# by pk_tautolo`

## C#

 `// C# implementation of above approach``using` `System;` `class` `GFG``{``static` `int` `N = 4;` `// Function to rotate the matrix``// 90 degree clockwise``static` `void` `rotate90Clockwise(``int``[,] a)``{` `    ``// Traverse each cycle``    ``for` `(``int` `i = 0; i < N / 2; i++)``    ``{``        ``for` `(``int` `j = i; j < N - i - 1; j++)``        ``{` `            ``// Swap elements of each cycle``            ``// in clockwise direction``            ``int` `temp = a[i, j];``            ``a[i, j] = a[N - 1 - j, i];``            ``a[N - 1 - j, i] = a[N - 1 - i, N - 1 - j];``            ``a[N - 1 - i, N - 1 - j] = a[j, N - 1 - i];``            ``a[j, N - 1 - i] = temp;``        ``}``    ``}``}` `// Function for print matrix``static` `void` `printMatrix(``int``[,] arr)``{``    ``for` `(``int` `i = 0; i < N; i++)``    ``{``        ``for` `(``int` `j = 0; j < N; j++)``        ``Console.Write( arr[i, j] + ``" "``);``        ``Console.Write(``"\n"``);``    ``}``}` `// Driver code``public` `static` `void` `Main ()``    ``{``    ``int` `[,]arr = {{1, 2, 3, 4},``                  ``{5, 6, 7, 8},``                  ``{9, 10, 11, 12},``                  ``{13, 14, 15, 16}};``    ``rotate90Clockwise(arr);``    ``printMatrix(arr);``}``}` `// This code is contributed``// by ChitraNayal`

## PHP

 ``

## Javascript

 ``
Output
```13 9 5 1
14 10 6 2
15 11 7 3
16 12 8 4 ```

Method 2:

Approach: The approach is based on the pattern made by indices after rotating the matrix. Consider the following illustration to have a clear insight into it.

Consider a 3 x 3 matrix having indices (i, j) as follows.

00 01 02
10 11 12
20 21 22

After rotating the matrix by 90 degrees in clockwise direction, indices transform into
20 10 00  current_row_index = 0, i = 2, 1, 0
21 11 01 current_row_index = 1, i = 2, 1, 0
22 12 02  current_row_index = 2, i = 2, 1, 0

Observation: In any row, for every decreasing row index i, there exists a constant column index j, such that j = current_row_index

This pattern can be printed using 2 nested loops.
(This pattern of writing indices is achieved by writing the exact indices of the desired elements of
where they actually existed in the original array.)

Below is the implementation of the above approach:

## C++

 `// C++ implementation of above approach``#include ``using` `namespace` `std;` `#define N 4` `// Function to rotate the matrix 90 degree clockwise``void` `rotate90Clockwise(``int` `arr[N][N])``{``    ``// printing the matrix on the basis of``    ``// observations made on indices.``    ``for` `(``int` `j = 0; j < N; j++)``    ``{``        ``for` `(``int` `i = N - 1; i >= 0; i--)``            ``cout << arr[i][j] << ``" "``;``        ``cout << ``'\n'``;``    ``}``}` `// Driver code``int` `main()``{``    ``int` `arr[N][N] = { { 1, 2, 3, 4 },``                      ``{ 5, 6, 7, 8 },``                      ``{ 9, 10, 11, 12 },``                      ``{ 13, 14, 15, 16 } };``    ``rotate90Clockwise(arr);``    ``return` `0;``}` `// This code is contributed by yashbeersingh42`

## Java

 `// Java implementation of above approach``import` `java.io.*;` `class` `GFG {``    ``static` `int` `N = ``4``;` `    ``// Function to rotate the matrix 90 degree clockwise``    ``static` `void` `rotate90Clockwise(``int` `arr[][])``    ``{``        ``// printing the matrix on the basis of``        ``// observations made on indices.``        ``for` `(``int` `j = ``0``; j < N; j++)``        ``{``            ``for` `(``int` `i = N - ``1``; i >= ``0``; i--)``                ``System.out.print(arr[i][j] + ``" "``);``            ``System.out.println();``        ``}``    ``}``    ``public` `static` `void` `main(String[] args)``    ``{``        ``int` `arr[][] = { { ``1``, ``2``, ``3``, ``4` `},``                        ``{ ``5``, ``6``, ``7``, ``8` `},``                        ``{ ``9``, ``10``, ``11``, ``12` `},``                        ``{ ``13``, ``14``, ``15``, ``16` `} };``        ``rotate90Clockwise(arr);``    ``}``}``// This code is contributed by yashbeersingh42`

## Python3

 `# Python3 implementation of above approach``N ``=` `4` `# Function to rotate the matrix 90 degree clockwise``def` `rotate90Clockwise(arr) :``    ``global` `N``    ` `    ``# printing the matrix on the basis of``    ``# observations made on indices.``    ``for` `j ``in` `range``(N) :``        ``for` `i ``in` `range``(N ``-` `1``, ``-``1``, ``-``1``) :``            ``print``(arr[i][j], end ``=` `" "``)``        ``print``()``        ` `# Driver code       ``arr ``=` `[ [ ``1``, ``2``, ``3``, ``4` `],``          ``[ ``5``, ``6``, ``7``, ``8` `],``          ``[ ``9``, ``10``, ``11``, ``12` `],``          ``[ ``13``, ``14``, ``15``, ``16` `] ]``rotate90Clockwise(arr);` `# This code is contributed by divyesh072019.`

## C#

 `// C# implementation of above approach``using` `System;``class` `GFG {``    ` `    ``static` `int` `N = 4;``    ` `    ``// Function to rotate the matrix 90 degree clockwise``    ``static` `void` `rotate90Clockwise(``int``[,] arr)``    ``{``      ` `        ``// printing the matrix on the basis of``        ``// observations made on indices.``        ``for` `(``int` `j = 0; j < N; j++)``        ``{``            ``for` `(``int` `i = N - 1; i >= 0; i--)``                ``Console.Write(arr[i, j] + ``" "``);``            ``Console.WriteLine();``        ``}``    ``}``    ` `  ``// Driver code``  ``static` `void` `Main() {``    ``int``[,] arr = { { 1, 2, 3, 4 },``                  ``{ 5, 6, 7, 8 },``                  ``{ 9, 10, 11, 12 },``                  ``{ 13, 14, 15, 16 } };``    ``rotate90Clockwise(arr);``  ``}``}` `// This code is contributed by divyeshrabadiya07.`

## Javascript

 ``
Output
```13 9 5 1
14 10 6 2
15 11 7 3
16 12 8 4 ```

Method 3:

Approach: The Approach is to rotate the given matrix two times, first time with respect to the Main diagonal, next time rotate the resultant matrix with respect to the middle column, Consider the following illustration to have a clear insight into it. Rotate square matrix 90 degrees in a clockwise direction

Below is the implementation of the above approach:

## C++

 `#include ``using` `namespace` `std;` `#define N 4` `void` `print(``int` `arr[N][N])``{``    ``for``(``int` `i = 0; i < N; ++i)``    ``{``        ``for``(``int` `j = 0; j < N; ++j)``            ``cout << arr[i][j] << ``" "``;``            ` `        ``cout << ``'\n'``;``    ``}``}` `void` `rotate(``int` `arr[N][N])``{``    ` `    ``// First rotation``    ``// with respect to main diagonal``    ``for``(``int` `i = 0; i < N; ++i)``    ``{``        ``for``(``int` `j = 0; j < i; ++j)``        ``{``            ``int` `temp = arr[i][j];``            ``arr[i][j] = arr[j][i];``            ``arr[j][i] = temp;``        ``}``    ``}``    ` `    ``// Second rotation``    ``// with respect to middle column``    ``for``(``int` `i = 0; i < N; ++i)``    ``{``        ``for``(``int` `j = 0; j < N / 2; ++j)``        ``{``            ``int` `temp = arr[i][j];``            ``arr[i][j] = arr[i][N - j - 1];``            ``arr[i][N - j - 1] = temp;``        ``}``    ``}``}` `// Driver code``int` `main()``{``    ``int` `arr[N][N] = { { 1, 2, 3, 4 },``                      ``{ 5, 6, 7, 8 },``                      ``{ 9, 10, 11, 12 },``                      ``{ 13, 14, 15, 16 } };``    ``rotate(arr);``    ``print(arr);``    ``return` `0;``}` `// This code is contributed  by Rahul Verma`

## Java

 `import` `java.io.*;` `class` `GFG {``  ` `  ``static` `void` `rotate(``int``[][] arr) {` `        ``int` `n=arr.length;``    ` `    ``// first rotation``    ``// with respect to main diagonal``        ``for``(``int` `i=``0``;i

## C#

 `using` `System;``using` `System.Collections.Generic;``public` `class` `GFG {``  ` `  ``static` `void` `rotate(``int``[,] arr) {` `        ``int` `n=arr.GetLength(0);``    ` `    ``// first rotation``    ``// with respect to main diagonal``        ``for``(``int` `i=0;i

## Javascript

 ``
Output

```13 9 5 1
14 10 6 2
15 11 7 3
16 12 8 4 ```

Method 4:

Approach: This approach is similar to method 3 the only difference is that in first rotation we rotate about the Secondary Diagonal and after that about the Middle row. Rotate square matrix 90 degrees in a clockwise direction

Below is the implementation of the above approach:

## C++

 `#include ``using` `namespace` `std;` `#define N 4` `void` `print(``int` `arr[N][N])``{``    ``for``(``int` `i = 0; i < N; ++i)``    ``{``        ``for``(``int` `j = 0; j < N; ++j)``            ``cout << arr[i][j] << ``" "``;``            ` `        ``cout << ``'\n'``;``    ``}``}` `void` `rotate(``int` `arr[N][N])``{``    ` `    ``// First rotation``    ``// with respect to Secondary diagonal``    ``for``(``int` `i = 0; i < N; i++)``    ``{``        ``for``(``int` `j = 0; j < N - i; j++)``        ``{``            ``int` `temp = arr[i][j];``            ``arr[i][j] = arr[N - 1 - j][N - 1 - i];``            ``arr[N - 1 - j][N - 1 - i] = temp;``        ``}``    ``}``    ` `    ``// Second rotation``    ``// with respect to middle row``    ``for``(``int` `i = 0; i < N / 2; i++)``    ``{``        ``for``(``int` `j = 0; j < N; j++)``        ``{``            ``int` `temp = arr[i][j];``            ``arr[i][j] = arr[N - 1 - i][j];``            ``arr[N - 1 - i][j] = temp;``        ``}``    ``}``}` `// Driver code``int` `main()``{``    ``int` `arr[N][N] = { { 1, 2, 3, 4 },``                      ``{ 5, 6, 7, 8 },``                      ``{ 9, 10, 11, 12 },``                      ``{ 13, 14, 15, 16 } };``    ``rotate(arr);``    ``print(arr);``    ``return` `0;``}` `// This code is contributed  by Rahul Verma`

## Java

 `import` `java.io.*;` `class` `GFG {` `    ``static` `void` `rotate(``int``[][] arr)``    ``{` `        ``int` `n = arr.length;` `        ``// first rotation``        ``// with respect to Secondary diagonal``        ``for` `(``int` `i = ``0``; i < n; i++) {``            ``for` `(``int` `j = ``0``; j < n - i; j++) {``                ``int` `temp = arr[i][j];``                ``arr[i][j] = arr[n - ``1` `- j][n - ``1` `- i];``                ``arr[n - ``1` `- j][n - ``1` `- i] = temp;``            ``}``        ``}``        ``// Second rotation``        ``// with respect to middle row``        ``for` `(``int` `i = ``0``; i < n / ``2``; i++) {``            ``for` `(``int` `j = ``0``; j < n; j++) {``                ``int` `temp = arr[i][j];``                ``arr[i][j] = arr[n - ``1` `- i][j];``                ``arr[n - ``1` `- i][j] = temp;``            ``}``        ``}``    ``}` `    ``// to print matrix``    ``static` `void` `printMatrix(``int` `arr[][])``    ``{``        ``int` `n = arr.length;``        ``for` `(``int` `i = ``0``; i < n; i++) {``            ``for` `(``int` `j = ``0``; j < n; j++)``                ``System.out.print(arr[i][j] + ``" "``);``            ``System.out.println();``        ``}``    ``}``    ``// Driver code``    ``public` `static` `void` `main(String[] args)``    ``{``        ``int` `arr[][] = { { ``1``, ``2``, ``3``, ``4` `},``                        ``{ ``5``, ``6``, ``7``, ``8` `},``                        ``{ ``9``, ``10``, ``11``, ``12` `},``                        ``{ ``13``, ``14``, ``15``, ``16` `} };``        ``rotate(arr);``        ``printMatrix(arr);``    ``}``}``// This code is contributed  by Rahul Verma`

## C#

 `using` `System;``using` `System.Collections.Generic;` `class` `GFG{` `static` `void` `rotate(``int``[,] arr)``{``    ``int` `n = arr.GetLength(0);` `    ``// First rotation``    ``// with respect to Secondary diagonal``    ``for``(``int` `i = 0; i < n; i++)``    ``{``        ``for``(``int` `j = 0; j < n - i; j++)``        ``{``            ``int` `temp = arr[i, j];``            ``arr[i, j] = arr[n - 1 - j, n - 1 - i];``            ``arr[n - 1 - j, n - 1 - i] = temp;``        ``}``    ``}``    ` `    ``// Second rotation``    ``// with respect to middle row``    ``for``(``int` `i = 0; i < n / 2; i++)``    ``{``        ``for``(``int` `j = 0; j < n; j++)``        ``{``            ``int` `temp = arr[i, j];``            ``arr[i, j] = arr[n - 1 - i, j];``            ``arr[n - 1 - i, j] = temp;``        ``}``    ``}``}` `// To print matrix``static` `void` `printMatrix(``int` `[,]arr)``{``    ``int` `n = arr.GetLength(0);``    ``for``(``int` `i = 0; i < n; i++)``    ``{``        ``for``(``int` `j = 0; j < n; j++)``            ``Console.Write(arr[i, j] + ``" "``);``            ` `        ``Console.WriteLine();``    ``}``}` `// Driver code``public` `static` `void` `Main(String[] args)``{``    ``int` `[,]arr = { { 1, 2, 3, 4 },``                   ``{ 5, 6, 7, 8 },``                   ``{ 9, 10, 11, 12 },``                   ``{ 13, 14, 15, 16 } };``    ``rotate(arr);``    ``printMatrix(arr);``}``}` `// This code is contributed by aashish1995`
Output
```13 9 5 1
14 10 6 2
15 11 7 3
16 12 8 4 ```

Method 5:

Approach: We first transpose the given matrix, and then reverse the content of individual rows to get the resultant 90 degree clockwise rotated matrix.

1  2  3                                                1  4  7                                                                 7  4  1

4  5  6        ——Transpose——>    2  5  8         —-Reverse individual rows—->    8  5  2     (Resultant matrix)

7  8  9                                                3  6  9                                                                 9  6  3

Below is the implementation of the above approach:

## C++

 `#include ``using` `namespace` `std;``const` `int` `n = 4;``void` `print(``int` `mat[n][n])``{``    ``for` `(``int` `i = 0; i < n; i++) {``        ``for` `(``int` `j = 0; j < n; j++)``            ``cout << mat[i][j] << ``" "``;``        ``cout << endl;``    ``}``}``void` `rotate90clockwise(``int` `mat[n][n])``{``    ``// Transpose of matrix``    ``for` `(``int` `i = 0; i < n; i++)``        ``for` `(``int` `j = i + 1; j < n; j++)``            ``swap(mat[i][j], mat[j][i]);``    ``// Reverse individual rows``    ``for` `(``int` `i = 0; i < n; i++) {``        ``int` `low = 0, high = n - 1;``        ``while` `(low < high) {``            ``swap(mat[i][low], mat[i][high]);``            ``low++;``            ``high--;``        ``}``    ``}``}``int` `main()``{``    ``int` `mat[n][n]``        ``= { { 1, 2, 3, 4 },``                      ``{ 5, 6, 7, 8 },``                      ``{ 9, 10, 11, 12 },``                      ``{ 13, 14, 15, 16 } };``    ``rotate90clockwise(mat);``    ``print(mat);``}`
Output
```13 9 5 1
14 10 6 2
15 11 7 3
16 12 8 4 ```

Complexity Analysis:

Time Complexity – O(n*n)

Auxiliary Space – O(1)

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