Given a square matrix, turn it by 90 degrees in 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 cycle in 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] = Righmost 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-j] = 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 above approach:
C++
// C++ implementation of above approach #include <bits/stdc++.h> 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
<?php // PHP implementation of above approach $N = 4; // Function to rotate the matrix // 90 degree clockwise function 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 matrix function 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 ?> |
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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 <bits/stdc++.h> 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. |
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