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

11 Jul, 2022
Given a square matrix, turn it by 90 degrees in a clockwise direction without using any extra space.
Examples: 
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++
Java
Python
C#
PHP
Javascript
// C++ implementation of above approach#include <bits/stdc++.h>using namespace std; #define N 4 // Function to rotate the matrix 90 degree clockwisevoid 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 matrixvoid 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 codeint 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 implementation of above approachimport java.io.*; class GFG{     static int N = 4; // Function to rotate the matrix 90 degree clockwisestatic 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 matrixstatic 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.
# Function to rotate the matrix# 90 degree clockwisedef 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 matrixdef printMatrix(A):    N = len(A[0])    for i in range(N):        print(A[i]) # Driver codeA = [[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# implementation of above approachusing System; class GFG{static int N = 4; // Function to rotate the matrix// 90 degree clockwisestatic 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 matrixstatic 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 codepublic 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// PHP implementation of above approach$N = 4; // Function to rotate the matrix// 90 degree clockwisefunction rotate90Clockwise(&$a){    global $N;         // Traverse each cycle    for ($i = 0; $i < $N / 2; $i++)    {        for ($j = $i;             $j < $N - $i - 1; $j++)        {             // Swap elements of each cycle            // in clockwise direction            $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] = $temp;        }    }} // Function for print matrixfunction printMatrix(&$arr){    global $N;    for ($i = 0; $i < $N; $i++)    {        for ($j = 0; $j < $N; $j++)            echo $arr[$i][$j] . " ";        echo "\n";    }} // Driver code$arr = array(array(1, 2, 3, 4),             array(5, 6, 7, 8),             array(9, 10, 11, 12),             array(13, 14, 15, 16));rotate90Clockwise($arr);printMatrix($arr); // This code is contributed// by ChitraNayal?>
<script> // Javascript implementation of above approach     var N = 4;     // Function to rotate the matrix 90 degree clockwise    function rotate90Clockwise(a) {         // Traverse each cycle        for (i = 0; i < parseInt(N / 2); i++) {            for (j = i; j < N - i - 1; j++) {                 // Swap elements of each cycle                // in clockwise direction                var 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    function printMatrix(arr) {        for (i = 0; i < N; i++) {            for (j = 0; j < N; j++)                document.write(arr[i][j] + " ");            document.write("<br/>");        }    }     // Driver code              var arr = [ [ 1, 2, 3, 4 ],                    [ 5, 6, 7, 8 ],                    [ 9, 10, 11, 12 ],                    [ 13, 14, 15, 16 ] ];        rotate90Clockwise(arr);        printMatrix(arr); // This code contributed by Rajput-Ji </script>
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)
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 into20 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++
Java
Python3
C#
Javascript
// C++ implementation of above approach#include <bits/stdc++.h>using namespace std; #define N 4 // Function to rotate the matrix 90 degree clockwisevoid 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 codeint 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 implementation of above approachimport 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 implementation of above approachN = 4  # Function to rotate the matrix 90 degree clockwisedef 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# implementation of above approachusing 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.
<script>// javascript implementation of above approach    var N = 4;     // Function to rotate the matrix 90 degree clockwise    function rotate90Clockwise(arr) {        // printing the matrix on the basis of        // observations made on indices.        for (j = 0; j < N; j++) {            for (i = N - 1; i >= 0; i--)                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 ] ];        rotate90Clockwise(arr); // This code contributed by Rajput-Ji</script>
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)
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++
Java
Python3
C#
Javascript
#include <bits/stdc++.h>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 codeint 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
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<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;            }        }     }     // 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
# Python3 implementation of above approachN = 4 # Function to rotate the matrix 90 degree clockwisedef rotate(arr):    global N     # First rotation# with respect to main diagonal    for i in range(N):        for j in range(i):            temp = arr[i][j]            arr[i][j] = arr[j][i]            arr[j][i] = temp     # Second rotation# with respect to middle column    for i in range(N):        for j in range(int(N/2)):            temp = arr[i][j]            arr[i][j] = arr[i][N-j-1]            arr[i][N-j-1] = temp  # Driver codearr = [[1, 2, 3, 4],       [5, 6, 7, 8],       [9, 10, 11, 12],       [13, 14, 15, 16]] rotate(arr) for i in range(N):    for j in range(N):        print(arr[i][j], end=" ")    print() # This code is contributed by Aarti_Rathi
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<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;            }        }     }     // 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 contributed by Rajput-Ji
<script> let N = 4 function print(arr){    for(let i = 0; i < N; ++i)    {        for(let j = 0; j < N; ++j)            document.write(arr[i][j] + " ");                     document.write("<br>");    }} function rotate(arr){         // First rotation    // with respect to main diagonal    for(let i = 0; i < N; ++i)    {        for(let j = 0; j < i; ++j)        {            let temp = arr[i][j];            arr[i][j] = arr[j][i];            arr[j][i] = temp;        }    }         // Second rotation    // with respect to middle column    for(let i = 0; i < N; ++i)    {        for(let j = 0; j < N / 2; ++j)        {            let temp = arr[i][j];            arr[i][j] = arr[i][N - j - 1];            arr[i][N - j - 1] = temp;        }    }} // Driver code     let arr = [ [ 1, 2, 3, 4 ],                    [ 5, 6, 7, 8 ],                    [ 9, 10, 11, 12 ],                    [ 13, 14, 15, 16 ] ];    rotate(arr);    print(arr);     //This code is contributed by Mayank Tyagi</script>
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)
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++
Java
Python3
C#
Javascript
#include <bits/stdc++.h>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 codeint 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
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
# Python3 implementation of above approachN = 4 def display(arr):    for i in range(N) :        for j in range(N) :            print(arr[i][j],end=" ")        print()             # Function to rotate the matrix 90 degree clockwisedef rotate90Clockwise(arr) :    global N          # First rotation    # with respect to Secondary diagonal    for i in range(N) :        for j in range(N-i) :            arr[i][j],arr[N - 1 - j][N - 1 - i]=arr[N - 1 - j][N - 1 - i],arr[i][j]         # Second rotation    # with respect to middle row    for i in range(N//2) :        for j in range(N) :            arr[i][j],arr[N - 1 - i][j]=arr[N - 1 - i][j],arr[i][j]          # Driver code      arr = [ [ 1, 2, 3, 4 ],          [ 5, 6, 7, 8 ],          [ 9, 10, 11, 12 ],          [ 13, 14, 15, 16 ] ]rotate90Clockwise(arr)display(arr)  # This code is contributed by Aarti_Rathi
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 matrixstatic 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 codepublic 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
<script>    function rotate(arr) {         var n = arr.length;         // first rotation        // with respect to Secondary diagonal        for (i = 0; i < n; i++) {            for (j = 0; j < n - i; j++) {                var 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 (i = 0; i < n / 2; i++) {            for (j = 0; j < n; j++) {                var temp = arr[i][j];                arr[i][j] = arr[n - 1 - i][j];                arr[n - 1 - i][j] = temp;            }        }    }     // to print 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/>");        }    }     // Driver code        var 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 gauravrajput1</script>
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)
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++
Java
Python3
C#
Javascript
#include <iostream>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);}
import java.util.*; class GFG {    static int n = 4;     static void print(int mat[][]) {        for (int i = 0; i < n; i++) {            for (int j = 0; j < n; j++)                System.out.print(mat[i][j] + " ");            System.out.println();        }    }     static void rotate90clockwise(int mat[][])    {               // Transpose of matrix        for (int i = 0; i < n; i++)            for (int j = i + 1; j < n; j++) {                int t = mat[i][j];                mat[i][j] = mat[j][i];                mat[j][i] = t;            }         // Reverse individual rows        for (int i = 0; i < n; i++) {            int low = 0, high = n - 1;            while (low < high) {                int t = mat[i][low];                mat[i][low] = mat[i][high];                mat[i][high] = t;                low++;                high--;            }        }    }   // Driver code    public static void main(String[] args) {        int mat[][] = { { 1, 2, 3, 4 },                       { 5, 6, 7, 8 },                       { 9, 10, 11, 12 },                       { 13, 14, 15, 16 } };        rotate90clockwise(mat);        print(mat);    }}  // This code is contributed by umadevi9616
# Python3 implementation of above approachN = 4 def display(arr):    for i in range(N) :        for j in range(N) :            print(arr[i][j],end=" ")        print()             # Function to rotate the matrix 90 degree clockwisedef rotate90Clockwise(arr) :    global N          # Transpose of matrix    for i in range(N) :        for j in range(i+1,N) :            arr[i][j],arr[j][i]=arr[j][i],arr[i][j]         # Reverse individual rows    for i in range(N//2) :        low = 0        high = N-1        while (low<high) :            arr[i][low],arr[i][high]=arr[i][high],arr[i][low]            low = low + 1            high = high - 1          # Driver code      arr = [ [ 1, 2, 3, 4 ],          [ 5, 6, 7, 8 ],          [ 9, 10, 11, 12 ],          [ 13, 14, 15, 16 ] ]rotate90Clockwise(arr)display(arr)  # This code is contributed by Aarti_Rathi
using System;using System.Collections.Generic;  class GFG{  static void rotate90clockwise(int[,] arr){    int n = arr.GetLength(0);      // Transpose of matrix    for(int i = 0; i < n; i++)    {        for(int j = i+1; j < n; j++)        {            int temp = arr[i, j];            arr[i, j] = arr[j, i];            arr[j, i] = temp;        }    }          /// Reverse individual rows    for (int i = 0; i < n; i++) {        int low = 0, high = n - 1;        while (low < high) {            int temp = arr[i, low];            arr[i, low] = arr[i, high];            arr[i, high] = temp;            low++;            high--;        }    }}  // To print matrixstatic 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 codepublic 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 is contributed by Aarti_Rathi
<script>    function rotate(arr) {          var n = arr.length;          // Transpose of matrix        for (i = 0; i < n; i++) {            for (j = i+1; j < n; j++) {                var temp = arr[i][j];                arr[i][j] = arr[j][i];                arr[j][i] = temp;            }        }                  // Reverse individual rows        for (i = 0; i < n; i++) {            var low = 0, high = n-1;            while (low < high) {                var temp = arr[i][low];                arr[i][low] = arr[i][high];                arr[i][high] = temp;                low++;                high--;            }        }    }      // to print 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/>");        }    }      // Driver code    var 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 Aarti_Rathi</script>
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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