Rotate a matrix by 90 degree without using any extra space | Set 2 - GeeksforGeeks

12 May, 2022
Given a square matrix, turn it by 90 degrees in anti-clockwise direction without using any extra space.
Examples: 
Input:
 1  2  3
 4  5  6
 7  8  9
Output:
 3  6  9 
 2  5  8 
 1  4  7 
Rotated the input matrix by
90 degrees 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
Rotated the input matrix by
90 degrees in anti-clockwise direction.
An approach that requires extra space is already discussed in a different article: Inplace rotate square matrix by 90 degrees | Set 1
This post discusses the same problem with a different approach which is space-optimized.Approach: The idea is to find the transpose of the matrix and then reverse the columns of the transposed matrix. Here is an example to show how this works. 
Algorithm:  
To solve the given problem there are two tasks. 1st is finding the transpose and second is reversing the columns without using extra spaceA transpose of a matrix is when the matrix is flipped over its diagonal, i.e the row index of an element becomes the column index and vice versa. So to find the transpose interchange the elements at position (i, j) with (j, i). Run two loops, the outer loop from 0 to row count and inner loop from 0 to index of the outer loop.To reverse the column of the transposed matrix, run two nested loops, the outer loop from 0 to column count and inner loop from 0 to row count/2, interchange elements at (i, j) with (i, row[count-1-j]), where i and j are indices of inner and outer loop respectively.
To solve the given problem there are two tasks. 1st is finding the transpose and second is reversing the columns without using extra space
A transpose of a matrix is when the matrix is flipped over its diagonal, i.e the row index of an element becomes the column index and vice versa. So to find the transpose interchange the elements at position (i, j) with (j, i). Run two loops, the outer loop from 0 to row count and inner loop from 0 to index of the outer loop.
To reverse the column of the transposed matrix, run two nested loops, the outer loop from 0 to column count and inner loop from 0 to row count/2, interchange elements at (i, j) with (i, row[count-1-j]), where i and j are indices of inner and outer loop respectively.
C++
Java
C#
Python3
PHP
Javascript
// C++ program for left// rotation of matrix by 90 degree// without using extra space#include <bits/stdc++.h>using namespace std;#define R 4#define C 4 // After transpose we swap// elements of column// one by one for finding left// rotation of matrix// by 90 degreevoid reverseColumns(int arr[R][C]){    for (int i = 0; i < C; i++)        for (int j = 0, k = C - 1; j < k; j++, k--)            swap(arr[j][i], arr[k][i]);} // Function for do transpose of matrixvoid transpose(int arr[R][C]){    for (int i = 0; i < R; i++)        for (int j = i; j < C; j++)            swap(arr[i][j], arr[j][i]);} // Function for print matrixvoid printMatrix(int arr[R][C]){    for (int i = 0; i < R; i++) {        for (int j = 0; j < C; j++)            cout << arr[i][j] << " ";        cout << '\n';    }} // Function to anticlockwise// rotate matrix by 90 degreevoid rotate90(int arr[R][C]){    transpose(arr);    reverseColumns(arr);} // Driven codeint main(){    int arr[R][C] = { { 1, 2, 3, 4 },                      { 5, 6, 7, 8 },                      { 9, 10, 11, 12 },                      { 13, 14, 15, 16 } };    rotate90(arr);    printMatrix(arr);    return 0;}
// JAVA Code for left Rotation of a// matrix by 90 degree without using// any extra spaceimport java.util.*; class GFG {     // After transpose we swap elements of    // column one by one for finding left    // rotation of matrix by 90 degree    static void reverseColumns(int arr[][])    {        for (int i = 0; i < arr[0].length; i++)            for (int j = 0, k = arr[0].length - 1; j < k;                 j++, k--) {                int temp = arr[j][i];                arr[j][i] = arr[k][i];                arr[k][i] = temp;            }    }     // Function for do transpose of matrix    static void transpose(int arr[][])    {        for (int i = 0; i < arr.length; i++)            for (int j = i; j < arr[0].length; j++) {                int temp = arr[j][i];                arr[j][i] = arr[i][j];                arr[i][j] = temp;            }    }     // Function for print matrix    static void printMatrix(int arr[][])    {        for (int i = 0; i < arr.length; i++) {            for (int j = 0; j < arr[0].length; j++)                System.out.print(arr[i][j] + " ");            System.out.println("");        }    }     // Function to anticlockwise rotate    // matrix by 90 degree    static void rotate90(int arr[][])    {        transpose(arr);        reverseColumns(arr);    }     /* Driver program to test above function */    public static void main(String[] args)    {        int arr[][] = { { 1, 2, 3, 4 },                        { 5, 6, 7, 8 },                        { 9, 10, 11, 12 },                        { 13, 14, 15, 16 } };         rotate90(arr);        printMatrix(arr);    }} // This code is contributed by Arnav Kr. Mandal.
// C# program for left rotation// of matrix by 90 degree// without using extra spaceusing System; class GFG {    static int R = 4;    static int C = 4;     // After transpose we swap    // elements of column one    // by one for finding left    // rotation of matrix by    // 90 degree    static void reverseColumns(int[, ] arr)    {        for (int i = 0; i < C; i++)            for (int j = 0, k = C - 1; j < k; j++, k--) {                int temp = arr[j, i];                arr[j, i] = arr[k, i];                arr[k, i] = temp;            }    }     // Function for do    // transpose of matrix    static void transpose(int[, ] arr)    {        for (int i = 0; i < R; i++)            for (int j = i; j < C; j++) {                int temp = arr[j, i];                arr[j, i] = arr[i, j];                arr[i, j] = temp;            }    }     // Function for print matrix    static void printMatrix(int[, ] arr)    {         for (int i = 0; i < R; i++) {            for (int j = 0; j < C; j++)                Console.Write(arr[i, j] + " ");            Console.WriteLine("");        }    }     // Function to anticlockwise    // rotate matrix by 90 degree    static void rotate90(int[, ] arr)    {        transpose(arr);        reverseColumns(arr);    }     // Driver code    static void Main()    {        int[, ] arr = { { 1, 2, 3, 4 },                        { 5, 6, 7, 8 },                        { 9, 10, 11, 12 },                        { 13, 14, 15, 16 } };         rotate90(arr);        printMatrix(arr);    }     // This code is contributed    // by Sam007}
# Python 3 program for left rotation of matrix by 90# degree without using extra space R = 4C = 4 # After transpose we swap elements of column# one by one for finding left rotation of matrix# by 90 degree  def reverseColumns(arr):    for i in range(C):        j = 0        k = C-1        while j < k:            t = arr[j][i]            arr[j][i] = arr[k][i]            arr[k][i] = t            j += 1            k -= 1 # Function for do transpose of matrix  def transpose(arr):    for i in range(R):        for j in range(i, C):            t = arr[i][j]            arr[i][j] = arr[j][i]            arr[j][i] = t # Function for print matrix  def printMatrix(arr):    for i in range(R):        for j in range(C):            print(str(arr[i][j]), end=" ")        print() # Function to anticlockwise rotate matrix# by 90 degree  def rotate90(arr):    transpose(arr)    reverseColumns(arr)  # Driven codearr = [[1, 2, 3, 4],       [5, 6, 7, 8],       [9, 10, 11, 12],       [13, 14, 15, 16]       ]rotate90(arr)printMatrix(arr)
<?php// PHP program for left rotation of matrix by 90 $R = 4;$C = 4;// Function to rotate the matrix by 90 degreefunction reverseColumns(&$arr){    global $C;    for ($i = 0; $i < $C; $i++)    {        for ($j = 0, $k = $C - 1; $j < $k; $j++, $k--)        {            $t = $arr[$j][$i];            $arr[$j][$i] = $arr[$k][$i];            $arr[$k][$i] = $t;        }    }      }  // Function for transpose of matrixfunction transpose(&$arr){    global $R, $C;    for ($i = 0; $i < $R; $i++)    {        for ($j = $i; $j < $C; $j++)        {            $t = $arr[$i][$j];            $arr[$i][$j] = $arr[$j][$i];            $arr[$j][$i] = $t;        }    }}  // Function for display the matrixfunction printMatrix(&$arr){    global $R, $C;    for ($i = 0; $i < $R; $i++) {        for ($j = 0; $j < $C; $j++)            echo $arr[$i][$j]." ";        echo "\n";    }}  // Function to anticlockwise rotate matrix// by 90 degreefunction rotate90(&$arr){    transpose($arr);    reverseColumns($arr);}  // Driven code $arr = array( array( 1, 2, 3, 4 ),                 array( 5, 6, 7, 8 ),                 array( 9, 10, 11, 12 ),                 array( 13, 14, 15, 16 ) );rotate90($arr);printMatrix($arr);return 0;?>
<script> // JAVASCRIPT Code for left Rotation of a// matrix by 90 degree without using// any extra space         // After transpose we swap elements of    // column one by one for finding left    // rotation of matrix by 90 degree    function reverseColumns(arr)    {        for (let i = 0; i < arr[0].length; i++)            for (let j = 0, k = arr[0].length - 1;                 j < k; j++, k--) {                let temp = arr[j][i];                arr[j][i] = arr[k][i];                arr[k][i] = temp;            }    }         // Function for do transpose of matrix    function transpose(arr)    {        for (let i = 0; i < arr.length; i++)            for (let j = i; j < arr[0].length;                 j++) {                let temp = arr[j][i];                arr[j][i] = arr[i][j];                arr[i][j] = temp;            }    }         // Function for print matrix    function printMatrix(arr)    {        for (let i = 0; i < arr.length; i++) {            for (let j = 0; j < arr[0].length;                 j++)                document.write(arr[i][j] + " ");            document.write("<br>");        }    }         // Function to anticlockwise rotate    // matrix by 90 degree    function rotate90(arr)    {        transpose(arr);        reverseColumns(arr);    }         /* Driver program to test above function */    let arr = [[1, 2, 3, 4],        [5, 6, 7, 8],        [9, 10, 11, 12],        [13, 14, 15, 16]    ];    rotate90(arr)    printMatrix(arr)         // This code is contributed by rag2127 </script>
  
4 8 12 16 
3 7 11 15 
2 6 10 14 
1 5 9 13 
 
Complexity Analysis: 
 
Time complexity :O(R*C). The matrix is traversed twice, so the complexity is O(R*C).
Space complexity :O(1). The space complexity is constant as no extra space is required.
 
Another Approach using a single traversal of the matrix :
 
The idea is to traverse along the boundaries of the matrix and shift the positions of the elements in 900 anticlockwise directions in each boundary. There are such (n/2-1) boundaries in the matrix.
 
Algorithm: 
 
Iterate over all the boundaries in the matrix. There are total (n/2-1) boundariesFor each boundary take the 4 corner elements and swap them such that the 4 corner elements get rotated in anticlockwise directions. Then take the next 4 elements along the edges(left, right, top, bottom) and swap them in an anticlockwise direction. Continue as long as all the elements in that particular boundary get rotated in 900 anticlockwise directions.Then move on to the next inner boundary and continue the process as long the whole matrix is rotated in 900 anticlockwise direction.
Iterate over all the boundaries in the matrix. There are total (n/2-1) boundaries
For each boundary take the 4 corner elements and swap them such that the 4 corner elements get rotated in anticlockwise directions. Then take the next 4 elements along the edges(left, right, top, bottom) and swap them in an anticlockwise direction. Continue as long as all the elements in that particular boundary get rotated in 900 anticlockwise directions.
Then move on to the next inner boundary and continue the process as long the whole matrix is rotated in 900 anticlockwise direction.
Original array
total n/2-1 boundaries
for outermost boundary
for next inner boundary
 
C++14
Java
Python3
C#
Javascript
// C++ program for left rotation of matrix// by 90 degree without using extra space#include <bits/stdc++.h>using namespace std;#define R 4#define C 4 // Function to rotate matrix anticlockwise by 90 degrees.void rotateby90(int arr[R][C]){    int n = R; // n=size of the square matrix    int a = 0, b = 0, c = 0, d = 0;     // iterate over all the boundaries of the matrix    for (int i = 0; i <= n / 2 - 1; i++) {         // for each boundary, keep on taking 4 elements (one        // each along the 4 edges) and swap them in        // anticlockwise manner        for (int j = 0; j <= n - 2 * i - 2; j++) {            a = arr[i + j][i];            b = arr[n - 1 - i][i + j];            c = arr[n - 1 - i - j][n - 1 - i];            d = arr[i][n - 1 - i - j];             arr[i + j][i] = d;            arr[n - 1 - i][i + j] = a;            arr[n - 1 - i - j][n - 1 - i] = b;            arr[i][n - 1 - i - j] = c;        }    }} // Function for print matrixvoid printMatrix(int arr[R][C]){    for (int i = 0; i < R; i++) {        for (int j = 0; j < C; j++)            cout << arr[i][j] << " ";        cout << '\n';    }} // Driven codeint main(){    int arr[R][C] = { { 1, 2, 3, 4 },                      { 5, 6, 7, 8 },                      { 9, 10, 11, 12 },                      { 13, 14, 15, 16 } };    rotateby90(arr);    printMatrix(arr);    return 0;} // This code is contributed by Md. Enjamum// Hossain(enja_2001)
// JAVA Code for left Rotation of a matrix// by 90 degree without using any extra spaceimport java.util.*; class GFG {     // Function to rotate matrix anticlockwise by 90    // degrees.    static void rotateby90(int arr[][])    {        int n = arr.length;        int a = 0, b = 0, c = 0, d = 0;         // iterate over all the boundaries of the matrix        for (int i = 0; i <= n / 2 - 1; i++) {             // for each boundary, keep on taking 4 elements            // (one each along the 4 edges) and swap them in            // anticlockwise manner            for (int j = 0; j <= n - 2 * i - 2; j++) {                a = arr[i + j][i];                b = arr[n - 1 - i][i + j];                c = arr[n - 1 - i - j][n - 1 - i];                d = arr[i][n - 1 - i - j];                 arr[i + j][i] = d;                arr[n - 1 - i][i + j] = a;                arr[n - 1 - i - j][n - 1 - i] = b;                arr[i][n - 1 - i - j] = c;            }        }    }    // Function for print matrix    static void printMatrix(int arr[][])    {        for (int i = 0; i < arr.length; i++) {            for (int j = 0; j < arr[0].length; j++)                System.out.print(arr[i][j] + " ");            System.out.println("");        }    }     /* Driver program to test above function */    public static void main(String[] args)    {        int arr[][] = { { 1, 2, 3, 4 },                        { 5, 6, 7, 8 },                        { 9, 10, 11, 12 },                        { 13, 14, 15, 16 } };         rotateby90(arr);        printMatrix(arr);    }} // This code is contributed by Md. Enjamum// Hossain(enja_2001)
# Function to rotate matrix anticlockwise by 90 degrees.def rotateby90(arr):     n = len(arr)    a,b,c,d = 0,0,0,0     # iterate over all the boundaries of the matrix    for i in range(n // 2):         # for each boundary, keep on taking 4 elements        # (one each along the 4 edges) and swap them in        # anticlockwise manner        for j in range(n - 2 * i - 1):            a = arr[i + j][i]            b = arr[n - 1 - i][i + j]            c = arr[n - 1 - i - j][n - 1 - i]            d = arr[i][n - 1 - i - j]             arr[i + j][i] = d            arr[n - 1 - i][i + j] = a            arr[n - 1 - i - j][n - 1 - i] = b            arr[i][n - 1 - i - j] = c              # Function for print matrixdef printMatrix(arr):     for i in range(len(arr)):            for j in range(len(arr[0])):                print(arr[i][j] ,end = " ")            print()      # Driver program to test above functionarr=[[ 1, 2, 3, 4 ],     [ 5, 6, 7, 8 ],     [ 9, 10, 11, 12 ],     [ 13, 14, 15, 16 ]]rotateby90(arr)printMatrix(arr) # this code is contributed by shinjanpatra
// C# Code for left Rotation of a matrix// by 90 degree without using any extra spaceusing System;using System.Collections.Generic; public class GFG {     // Function to rotate matrix anticlockwise by 90    // degrees.    static void rotateby90(int [,]arr)    {        int n = arr.GetLength(0);        int a = 0, b = 0, c = 0, d = 0;         // iterate over all the boundaries of the matrix        for (int i = 0; i <= n / 2 - 1; i++) {             // for each boundary, keep on taking 4 elements            // (one each along the 4 edges) and swap them in            // anticlockwise manner            for (int j = 0; j <= n - 2 * i - 2; j++) {                a = arr[i + j,i];                b = arr[n - 1 - i,i + j];                c = arr[n - 1 - i - j,n - 1 - i];                d = arr[i,n - 1 - i - j];                 arr[i + j,i] = d;                arr[n - 1 - i,i + j] = a;                arr[n - 1 - i - j,n - 1 - i] = b;                arr[i,n - 1 - i - j] = c;            }        }    }    // Function for print matrix    static void printMatrix(int [,]arr)    {        for (int i = 0; i < arr.GetLength(0); i++) {            for (int j = 0; j < arr.GetLength(1); j++)                Console.Write(arr[i,j] + " ");            Console.WriteLine("");        }    }     /* Driver program to test above function */    public static void Main(String[] args)    {        int [,]arr = { { 1, 2, 3, 4 },                        { 5, 6, 7, 8 },                        { 9, 10, 11, 12 },                        { 13, 14, 15, 16 } };         rotateby90(arr);        printMatrix(arr);    }} // This code is contributed by umadevi9616
<script> // JavaScript Code for left Rotation of a matrix// by 90 degree without using any extra space // Function to rotate matrix anticlockwise by 90    // degrees.function  rotateby90(arr){    let n = arr.length;        let a = 0, b = 0, c = 0, d = 0;          // iterate over all the boundaries of the matrix        for (let i = 0; i <= n / 2 - 1; i++) {              // for each boundary, keep on taking 4 elements            // (one each along the 4 edges) and swap them in            // anticlockwise manner            for (let j = 0; j <= n - 2 * i - 2; j++) {                a = arr[i + j][i];                b = arr[n - 1 - i][i + j];                c = arr[n - 1 - i - j][n - 1 - i];                d = arr[i][n - 1 - i - j];                  arr[i + j][i] = d;                arr[n - 1 - i][i + j] = a;                arr[n - 1 - i - j][n - 1 - i] = b;                arr[i][n - 1 - i - j] = c;            }        }}  // Function for print matrixfunction printMatrix(arr){    for (let i = 0; i < arr.length; i++) {            for (let j = 0; j < arr[0].length; j++)                document.write(arr[i][j] + " ");            document.write("<br>");        }}     /* Driver program to test above function */let arr=[[ 1, 2, 3, 4 ],                        [ 5, 6, 7, 8 ],                        [ 9, 10, 11, 12 ],                        [ 13, 14, 15, 16 ]];rotateby90(arr);printMatrix(arr); // This code is contributed by avanitrachhadiya2155 </script>
4 8 12 16 
3 7 11 15 
2 6 10 14 
1 5 9 13 
Complexity Analysis: 
Time complexity :O(R*C). The matrix is traversed only once, so the time complexity is O(R*C).
Space complexity :O(1). The space complexity is O(1) as no extra space is required.
Implementation: Let’s see a method of Python numpy that can be used to arrive at the particular solution. 
Python3
# Alternative implementation using numpyimport numpy # Driven codearr = [[1, 2, 3, 4],       [5, 6, 7, 8],       [9, 10, 11, 12],       [13, 14, 15, 16]       ] # Define flip algorithm (Note numpy.flip is a builtin f# function for versions > v.1.12.0)  def flip(m, axis):    if not hasattr(m, 'ndim'):        m = asarray(m)    indexer = [slice(None)] * m.ndim    try:        indexer[axis] = slice(None, None, -1)    except IndexError:        raise ValueError("axis =% i is invalid for the % i-dimensional input array"                         % (axis, m.ndim))    return m[tuple(indexer)]  # Transpose the matrixtrans = numpy.transpose(arr) # Flip the matrix anti-clockwise (1 for clockwise)flipmat = flip(trans, 0) print("\nnumpy implementation\n")print(flipmat)
Output:  
numpy implementation

[[ 4  8 12 16]
 [ 3  7 11 15]
 [ 2  6 10 14]
 [ 1  5  9 13]]
Note: The above steps/programs do left (or anticlockwise) rotation. Let’s see how to do the right rotation or clockwise rotation. The approach would be similar. Find the transpose of the matrix and then reverse the rows of the transposed matrix. This is how it is done. 
This article is contributed by DANISH_RAZA . If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above. 
Rotating Along the Boundaries
We can start at the first 4 corners of the given matrix and then keep incrementing the row and column indices to moves around.
At any given moment we will have four corners lu (left-up),ld(left-down),ru(right-up),rd(right-down).
To left rotate we will first swap the ru and ld,  then lu and ld and lastly ru and rd.
C++
Java
C#
Javascript
#include <bits/stdc++.h>using namespace std;  //Function to rotate matrix anticlockwise by 90 degrees.void rotateby90(vector<vector<int> >& matrix){      int n=matrix.size();    int mid;     if(n%2==0)         mid=n/2-1;     else         mid=n/2;     for(int i=0,j=n-1;i<=mid;i++,j--)     {         for(int k=0;k<j-i;k++)         {             swap(matrix[i][j-k],matrix[j][i+k]);     //ru  ld             swap(matrix[i+k][i],matrix[j][i+k]);     //lu ld             swap(matrix[i][j-k],matrix[j-k][j]);     //ru rd          }     }     } void printMatrix(vector<vector<int>>& arr){   int n=arr.size();  for(int i=0;i<n;i++)  {      for(int j=0;j<n;j++)    {        cout<<arr[i][j]<<" ";    }    cout<<endl;  }}int main() {        vector<vector<int>> arr = { { 1, 2, 3, 4 },                      { 5, 6, 7, 8 },                      { 9, 10, 11, 12 },                      { 13, 14, 15, 16 } };    rotateby90(arr);    printMatrix(arr);    return 0;}
import java.util.*; class GFG{    private static int[][] swap(int[][] matrix, int x1,int x2,int y1, int y2) {          int temp = matrix[x1][x2] ;          matrix[x1][x2] = matrix[y1][y2];          matrix[y1][y2] = temp;          return matrix;        }  //Function to rotate matrix anticlockwise by 90 degrees.static int[][] rotateby90(int[][] matrix){      int n=matrix.length;    int mid;     if(n % 2 == 0)         mid = n / 2 - 1;     else         mid = n / 2;     for(int i = 0, j = n - 1; i <= mid; i++, j--)     {         for(int k = 0; k < j - i; k++)         {             matrix= swap(matrix, i, j - k, j, i + k);     //ru  ld             matrix= swap(matrix, i + k, i, j, i + k);     //lu ld             matrix= swap(matrix, i, j - k, j - k, j);     //ru rd          }     }     return 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();  }}public static void main(String[] args) {        int[][] arr = { { 1, 2, 3, 4 },                      { 5, 6, 7, 8 },                      { 9, 10, 11, 12 },                      { 13, 14, 15, 16 } };    arr = rotateby90(arr);    printMatrix(arr);}} // This code is contributed by umadevi9616
using System; public class GFG{    private static int[,] swap(int[,] matrix, int x1,int x2,int y1, int y2) {          int temp = matrix[x1,x2] ;          matrix[x1,x2] = matrix[y1,y2];          matrix[y1,y2] = temp;          return matrix;        }  //Function to rotate matrix anticlockwise by 90 degrees.static int[,] rotateby90(int[,] matrix){      int n=matrix.GetLength(0);    int mid;     if(n % 2 == 0)         mid = (n / 2 - 1);     else         mid = (n / 2);     for(int i = 0, j = n - 1; i <= mid; i++, j--)     {         for(int k = 0; k < j - i; k++)         {             matrix= swap(matrix, i, j - k, j, i + k);     //ru  ld             matrix= swap(matrix, i + k, i, j, i + k);     //lu ld             matrix= swap(matrix, i, j - k, j - k, j);     //ru rd          }     }     return 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();  }}public static void Main(String[] args) {        int[,] arr = { { 1, 2, 3, 4 },                      { 5, 6, 7, 8 },                      { 9, 10, 11, 12 },                      { 13, 14, 15, 16 } };    arr = rotateby90(arr);    printMatrix(arr);}} // This code is contributed by Rajput-Ji
<script>    function swap(matrix , x1 , x2 , y1 , y2) {        var temp = matrix[x1][x2];        matrix[x1][x2] = matrix[y1][y2];        matrix[y1][y2] = temp;        return matrix;    }     // Function to rotate matrix anticlockwise by 90 degrees.     function rotateby90(matrix) {        var n = matrix.length;        var mid;        if (n % 2 == 0)            mid = n / 2 - 1;        else            mid = n / 2;        for (i = 0, j = n - 1; i <= mid; i++, j--) {            for (k = 0; k < j - i; k++) {                matrix = swap(matrix, i, j - k, j, i + k); // ru ld                matrix = swap(matrix, i + k, i, j, i + k); // lu ld                matrix = swap(matrix, i, j - k, j - k, j); // ru rd            }        }        return matrix;    }     function printMatrix(arr) {         var n = arr.length;        for (i = 0; i < n; i++) {            for (j = 0; j < n; j++) {                document.write(arr[i][j] + " ");            }            document.write("<br/>");        }    }         var arr = [ [ 1, 2, 3, 4 ],        [ 5, 6, 7, 8 ],        [ 9, 10, 11, 12 ],        [ 13, 14, 15, 16 ] ];        arr = rotateby90(arr);        printMatrix(arr); // This code is contributed by Rajput-Ji</script>
Output:  
4 8 12 16 
3 7 11 15 
2 6 10 14 
1 5 9 13 
Time Complexity: O(n^2)
Space Complexity : O(1)
Sam007
ukasp
bpgreyling
andrew1234
rag2127
enja_2001
avanitrachhadiya2155
anikakapoor
piyushbisht3
umadevi9616
surindertarika1234
Rajput-Ji
shinjanpatra
rotation
Matrix
Matrix
Writing code in comment?
Please use ide.geeksforgeeks.org,
generate link and share the link here.
Matrix Chain Multiplication | DP-8
Program to find largest element in an array
Print a given matrix in spiral form
Rat in a Maze | Backtracking-2
Maximum size square sub-matrix with all 1s
Sudoku | Backtracking-7
Find the number of islands | Set 1 (Using DFS)
Divide and Conquer | Set 5 (Strassen's Matrix Multiplication)
Count all possible paths from top left to bottom right of a mXn matrix
Maximum size rectangle binary sub-matrix with all 1s