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++
#include <bits/stdc++.h>
using namespace std;
#define N 4
void rotate90Clockwise(int a[N][N])
{
for (int i = 0; i < N / 2; i++) {
for (int j = i; j < N - i - 1; j++) {
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;
}
}
}
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';
}
}
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
import java.io.*;
class GFG
{
static int N = 4;
static void rotate90Clockwise(int a[][])
{
for (int i = 0; i < N / 2; i++)
{
for (int j = i; j < N - i - 1; j++)
{
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;
}
}
}
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();
}
}
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);
}
}
|
Python
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
def printMatrix(A):
N = len(A[0])
for i in range(N):
print(A[i])
A = [[1, 2, 3, 4],
[5, 6, 7, 8],
[9, 10, 11, 12],
[13, 14, 15, 16]]
rotate90Clockwise(A)
printMatrix(A)
|
C#
using System;
class GFG
{
static int N = 4;
static void rotate90Clockwise(int[,] a)
{
for (int i = 0; i < N / 2; i++)
{
for (int j = i; j < N - i - 1; j++)
{
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;
}
}
}
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");
}
}
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);
}
}
|
PHP
<?php
$N = 4;
function rotate90Clockwise(&$a)
{
global $N;
for ($i = 0; $i < $N / 2; $i++)
{
for ($j = $i;
$j < $N - $i - 1; $j++)
{
$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] = $temp;
}
}
}
function printMatrix(&$arr)
{
global $N;
for ($i = 0; $i < $N; $i++)
{
for ($j = 0; $j < $N; $j++)
echo $arr[$i][$j] . " ";
echo "\n";
}
}
$arr = array(array(1, 2, 3, 4),
array(5, 6, 7, 8),
array(9, 10, 11, 12),
array(13, 14, 15, 16));
rotate90Clockwise($arr);
printMatrix($arr);
?>
|
Javascript
<script>
var N = 4;
function rotate90Clockwise(a) {
for (i = 0; i < parseInt(N / 2); i++) {
for (j = i; j < N - i - 1; j++) {
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 printMatrix(arr) {
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 ] ];
rotate90Clockwise(arr);
printMatrix(arr);
</script>
|
Output13 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 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++
#include <bits/stdc++.h>
using namespace std;
#define N 4
void rotate90Clockwise(int arr[N][N])
{
for (int j = 0; j < N; j++)
{
for (int i = N - 1; i >= 0; i--)
cout << arr[i][j] << " ";
cout << '\n';
}
}
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;
}
|
Java
import java.io.*;
class GFG {
static int N = 4;
static void rotate90Clockwise(int arr[][])
{
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);
}
}
|
Python3
N = 4
def rotate90Clockwise(arr) :
global N
for j in range(N) :
for i in range(N - 1, -1, -1) :
print(arr[i][j], end = " ")
print()
arr = [ [ 1, 2, 3, 4 ],
[ 5, 6, 7, 8 ],
[ 9, 10, 11, 12 ],
[ 13, 14, 15, 16 ] ]
rotate90Clockwise(arr);
|
C#
using System;
class GFG {
static int N = 4;
static void rotate90Clockwise(int[,] arr)
{
for (int j = 0; j < N; j++)
{
for (int i = N - 1; i >= 0; i--)
Console.Write(arr[i, j] + " ");
Console.WriteLine();
}
}
static void Main() {
int[,] arr = { { 1, 2, 3, 4 },
{ 5, 6, 7, 8 },
{ 9, 10, 11, 12 },
{ 13, 14, 15, 16 } };
rotate90Clockwise(arr);
}
}
|
Javascript
<script>
var N = 4;
function rotate90Clockwise(arr) {
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);
</script>
|
Output13 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++
#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])
{
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;
}
}
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;
}
}
}
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;
}
|
Java
import java.io.*;
class GFG {
static void rotate(int[][] arr) {
int n=arr.length;
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;
}
}
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;
}
}
}
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 } };
rotate(arr);
printMatrix(arr);
}
}
|
Python3
N = 4
def rotate(arr):
global N
for i in range(N):
for j in range(i):
temp = arr[i][j]
arr[i][j] = arr[j][i]
arr[j][i] = temp
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
arr = [[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()
|
C#
using System;
using System.Collections.Generic;
public class GFG {
static void rotate(int[,] arr) {
int n=arr.GetLength(0);
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;
}
}
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;
}
}
}
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 } };
rotate(arr);
printMatrix(arr);
}
}
|
Javascript
<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)
{
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;
}
}
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;
}
}
}
let arr = [ [ 1, 2, 3, 4 ],
[ 5, 6, 7, 8 ],
[ 9, 10, 11, 12 ],
[ 13, 14, 15, 16 ] ];
rotate(arr);
print(arr);
</script>
|
Output13 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++
#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])
{
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;
}
}
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;
}
}
}
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;
}
|
Java
import java.io.*;
class GFG {
static void rotate(int[][] arr)
{
int n = arr.length;
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;
}
}
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;
}
}
}
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 } };
rotate(arr);
printMatrix(arr);
}
}
|
Python3
N = 4
def display(arr):
for i in range(N) :
for j in range(N) :
print(arr[i][j],end=" ")
print()
def rotate90Clockwise(arr) :
global N
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]
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]
arr = [ [ 1, 2, 3, 4 ],
[ 5, 6, 7, 8 ],
[ 9, 10, 11, 12 ],
[ 13, 14, 15, 16 ] ]
rotate90Clockwise(arr)
display(arr)
|
C#
using System;
using System.Collections.Generic;
class GFG{
static void rotate(int[,] arr)
{
int n = arr.GetLength(0);
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;
}
}
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;
}
}
}
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 } };
rotate(arr);
printMatrix(arr);
}
}
|
Javascript
<script>
function rotate(arr) {
var n = arr.length;
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;
}
}
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;
}
}
}
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 ] ];
rotate(arr);
printMatrix(arr);
</script>
|
Output13 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++
#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])
{
for (int i = 0; i < n; i++)
for (int j = i + 1; j < n; j++)
swap(mat[i][j], mat[j][i]);
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);
}
|
Java
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[][])
{
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;
}
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--;
}
}
}
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);
}
}
|
Python3
N = 4
def display(arr):
for i in range(N) :
for j in range(N) :
print(arr[i][j],end=" ")
print()
def rotate90Clockwise(arr) :
global N
for i in range(N) :
for j in range(i+1,N) :
arr[i][j],arr[j][i]=arr[j][i],arr[i][j]
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
arr = [ [ 1, 2, 3, 4 ],
[ 5, 6, 7, 8 ],
[ 9, 10, 11, 12 ],
[ 13, 14, 15, 16 ] ]
rotate90Clockwise(arr)
display(arr)
|
C#
using System;
using System.Collections.Generic;
class GFG{
static void rotate90clockwise(int[,] arr)
{
int n = arr.GetLength(0);
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--;
}
}
}
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 } };
rotate90clockwise(arr);
printMatrix(arr);
}
}
|
Javascript
<script>
function rotate(arr) {
var n = arr.length;
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;
}
}
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--;
}
}
}
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 ] ];
rotate(arr);
printMatrix(arr);
</script>
|
Output13 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)