Inplace rotate square matrix by 90 degrees | Set 1
Given a square matrix, turn it by 90 degrees in an anti-clockwise direction without using any extra space
Examples:
Input:
Matrix: 1 2 3
4 5 6
7 8 9Output: 3 6 9
2 5 8
1 4 7Input:
Matrix: 1 2 3 4
5 6 7 8
9 10 11 12
13 14 15 16Output: 4 8 12 16
3 7 11 15
2 6 10 14
1 5 9 13
Note: An approach that requires extra space is already discussed here.
Inplace rotate square matrix by 90 degrees by forming cycles:
To solve the problem follow the below idea:
To solve the question without any extra space, rotate the array in form of squares, dividing the matrix into squares or cycles. For example,
A 4 X 4 matrix will have 2 cycles. The first cycle is formed by its 1st row, last column, last row, and 1st column. The second cycle is formed by the 2nd row, second-last column, second-last row, and 2nd column. The idea is for each square cycle, to swap the elements involved with the corresponding cell in the matrix in an anti-clockwise direction i.e. from top to left, left to bottom, bottom to right, and from right to top one at a time using nothing but a temporary variable to achieve this
Dry run of the above approach:
First Cycle:
1 2 3 4
5 6 7 8
9 10 11 12
13 14 15 16Moving first group of four elements (elements
of 1st row, last row, 1st column and last column) of first cycle
in counter clockwise.4 2 3 16
5 6 7 8
9 10 11 12
1 14 15 13Moving next group of four elements of
first cycle in counter clockwise4 8 3 16
5 6 7 15
2 10 11 12
1 14 9 13Moving final group of four elements of
first cycle in counter clockwise4 8 12 16
3 6 7 15
2 10 11 14
1 5 9 13Second Cycle:
4 8 12 16
3 6 7 15
2 10 11 14
1 5 9 13Fixing second cycle
4 8 12 16
3 7 11 15
2 6 10 14
1 5 9 13
Follow the given steps to solve the problem:
- There are N/2 squares or cycles in a matrix of side N. Process a square one at a time. Run a loop to traverse the matrix a cycle at a time, i.e loop from 0 to N/2 – 1, loop counter is i
- Consider elements in group of 4 in current square, rotate the 4 elements at a time. So the number of such groups in a cycle is N – 2*i.
- So run a loop in each cycle from x to N – x – 1, loop counter is y
- The elements in the current group is (x, y), (y, N-1-x), (N-1-x, N-1-y), (N-1-y, x), now rotate the these 4 elements, i.e (x, y) <- (y, N-1-x), (y, N-1-x)<- (N-1-x, N-1-y), (N-1-x, N-1-y)<- (N-1-y, x), (N-1-y, x)<- (x, y)
- Print the matrix.
Below is the implementation of the above approach:
C++
// C++ program to rotate a matrix // by 90 degrees #include <bits/stdc++.h> #define N 4 using namespace std; // An Inplace function to // rotate a N x N matrix // by 90 degrees in // anti-clockwise direction void rotateMatrix(int mat[][N]) { // Consider all squares one by one for (int x = 0; x < N / 2; x++) { // Consider elements in group // of 4 in current square for (int y = x; y < N - x - 1; y++) { // Store current cell in // temp variable int temp = mat[x][y]; // Move values from right to top mat[x][y] = mat[y][N - 1 - x]; // Move values from bottom to right mat[y][N - 1 - x] = mat[N - 1 - x][N - 1 - y]; // Move values from left to bottom mat[N - 1 - x][N - 1 - y] = mat[N - 1 - y][x]; // Assign temp to left mat[N - 1 - y][x] = temp; } } } // Function to print the matrix void displayMatrix(int mat[N][N]) { for (int i = 0; i < N; i++) { for (int j = 0; j < N; j++) { cout << mat[i][j] << " "; } cout << endl; } cout << endl; } /* Driver code */int main() { // Test Case 1 int mat[N][N] = { { 1, 2, 3, 4 }, { 5, 6, 7, 8 }, { 9, 10, 11, 12 }, { 13, 14, 15, 16 } }; // Function call rotateMatrix(mat); // Print rotated matrix displayMatrix(mat); return 0; } |
Java
// Java program to rotate a // matrix by 90 degrees import java.io.*; class GFG { // An Inplace function to // rotate a N x N matrix // by 90 degrees in // anti-clockwise direction static void rotateMatrix(int N, int mat[][]) { // Consider all squares one by one for (int x = 0; x < N / 2; x++) { // Consider elements in group // of 4 in current square for (int y = x; y < N - x - 1; y++) { // Store current cell in // temp variable int temp = mat[x][y]; // Move values from right to top mat[x][y] = mat[y][N - 1 - x]; // Move values from bottom to right mat[y][N - 1 - x] = mat[N - 1 - x][N - 1 - y]; // Move values from left to bottom mat[N - 1 - x][N - 1 - y] = mat[N - 1 - y][x]; // Assign temp to left mat[N - 1 - y][x] = temp; } } } // Function to print the matrix static void displayMatrix(int N, int mat[][]) { for (int i = 0; i < N; i++) { for (int j = 0; j < N; j++) System.out.print(" " + mat[i][j]); System.out.print("\n"); } System.out.print("\n"); } /* Driver code*/ public static void main(String[] args) { int N = 4; int mat[][] = { { 1, 2, 3, 4 }, { 5, 6, 7, 8 }, { 9, 10, 11, 12 }, { 13, 14, 15, 16 } }; // Function call rotateMatrix(N, mat); // Print rotated matrix displayMatrix(N, mat); } } // This code is contributed by Prakriti Gupta |
Python3
# Python3 program to rotate a matrix by 90 degrees N = 4 # An Inplace function to rotate # N x N matrix by 90 degrees in # anti-clockwise direction def rotateMatrix(mat): # Consider all squares one by one for x in range(0, int(N / 2)): # Consider elements in group # of 4 in current square for y in range(x, N-x-1): # store current cell in temp variable temp = mat[x][y] # move values from right to top mat[x][y] = mat[y][N-1-x] # move values from bottom to right mat[y][N-1-x] = mat[N-1-x][N-1-y] # move values from left to bottom mat[N-1-x][N-1-y] = mat[N-1-y][x] # assign temp to left mat[N-1-y][x] = temp # Function to print the matrix def displayMatrix(mat): for i in range(0, N): for j in range(0, N): print(mat[i][j], end=' ') print("") # Driver Code if __name__ == "__main__": mat = [[0 for x in range(N)] for y in range(N)] mat = [[1, 2, 3, 4], [5, 6, 7, 8], [9, 10, 11, 12], [13, 14, 15, 16]] # Function call rotateMatrix(mat) # Print rotated matrix displayMatrix(mat) # This code is contributed by saloni1297 |
C#
// C# program to rotate a // matrix by 90 degrees using System; class GFG { // An Inplace function to // rotate a N x N matrix // by 90 degrees in anti- // clockwise direction static void rotateMatrix(int N, int[, ] mat) { // Consider all // squares one by one for (int x = 0; x < N / 2; x++) { // Consider elements // in group of 4 in // current square for (int y = x; y < N - x - 1; y++) { // store current cell // in temp variable int temp = mat[x, y]; // move values from // right to top mat[x, y] = mat[y, N - 1 - x]; // move values from // bottom to right mat[y, N - 1 - x] = mat[N - 1 - x, N - 1 - y]; // move values from // left to bottom mat[N - 1 - x, N - 1 - y] = mat[N - 1 - y, x]; // assign temp to left mat[N - 1 - y, x] = temp; } } } // Function to print the matrix static void displayMatrix(int N, int[, ] mat) { for (int i = 0; i < N; i++) { for (int j = 0; j < N; j++) Console.Write(" " + mat[i, j]); Console.WriteLine(); } Console.WriteLine(); } // Driver Code static public void Main() { int N = 4; int[, ] mat = { { 1, 2, 3, 4 }, { 5, 6, 7, 8 }, { 9, 10, 11, 12 }, { 13, 14, 15, 16 } }; // Function call rotateMatrix(N, mat); // Print rotated matrix displayMatrix(N, mat); } } // This code is contributed by ajit |
PHP
<?php // PHP program to rotate a // matrix by 90 degrees $N = 4; // An Inplace function to // rotate a N x N matrix // by 90 degrees in // anti-clockwise direction function rotateMatrix(&$mat) { global $N; // Consider all // squares one by one for ($x = 0; $x < $N / 2; $x++) { // Consider elements // in group of 4 in // current square for ($y = $x; $y < $N - $x - 1; $y++) { // store current cell // in temp variable $temp = $mat[$x][$y]; // move values from // right to top $mat[$x][$y] = $mat[$y][$N - 1 - $x]; // move values from // bottom to right $mat[$y][$N - 1 - $x] = $mat[$N - 1 - $x][$N - 1 - $y]; // move values from // left to bottom $mat[$N - 1 - $x][$N - 1 - $y] = $mat[$N - 1 - $y][$x]; // assign temp to left $mat[$N - 1 - $y][$x] = $temp; } } } // Function to // print the matrix function displayMatrix(&$mat) { global $N; for ($i = 0; $i < $N; $i++) { for ($j = 0; $j < $N; $j++) echo $mat[$i][$j] . " "; echo "\n"; } echo "\n"; } // Driver code $mat = array(array(1, 2, 3, 4), array(5, 6, 7, 8), array(9, 10, 11, 12), array(13, 14, 15, 16)); // Function call rotateMatrix($mat); // Print rotated matrix displayMatrix($mat); // This code is contributed // by ChitraNayal ?> |
Javascript
<script> // Javascript program to rotate a // matrix by 90 degrees // An Inplace function to // rotate a N x N matrix // by 90 degrees in // anti-clockwise direction function rotateMatrix(N,mat) { // Consider all squares one by one for (let x = 0; x < N / 2; x++) { // Consider elements in group // of 4 in current square for (let y = x; y < N - x - 1; y++) { // Store current cell in // temp variable let temp = mat[x][y]; // Move values from right to top mat[x][y] = mat[y][N - 1 - x]; // Move values from bottom to right mat[y][N - 1 - x] = mat[N - 1 - x][N - 1 - y]; // Move values from left to bottom mat[N - 1 - x][N - 1 - y] = mat[N - 1 - y][x]; // Assign temp to left mat[N - 1 - y][x] = temp; } } } // Function to print the matrix function displayMatrix(N,mat) { for (let i = 0; i < N; i++) { for (let j = 0; j < N; j++) document.write( " " + mat[i][j]); document.write("<br>"); } document.write("<br>"); } /* Driver program to test above functions */ let N = 4; let mat=[[1, 2, 3, 4],[ 5, 6, 7, 8 ],[9, 10, 11, 12 ],[13, 14, 15, 16]]; // displayMatrix(mat); rotateMatrix(N, mat); // Print rotated matrix displayMatrix(N, mat); // This code is contributed by rag2127. </script> |
Output
4 8 12 16 3 7 11 15 2 6 10 14 1 5 9 13
Time Complexity: O(N2), where n is the side of the array. A single traversal of the matrix is needed.
Auxiliary Space: O(1). As a constant space is needed
Inplace rotate square matrix by 90 degrees by transposing and reversing the matrix:
Follow the given steps to solve the problem:
- Reverse every individual row of the matrix
- Perform Transpose of the matrix
Below is the implementation of the above approach:
C++
// C++ program to rotate a matrix // by 90 degrees #include <bits/stdc++.h> using namespace std; #define N 4 // An Inplace function to // rotate a N x N matrix // by 90 degrees in // anti-clockwise direction void rotateMatrix(int mat[][N]) { // REVERSE every row for (int i = 0; i < N; i++) reverse(mat[i], mat[i] + N); // Performing Transpose for (int i = 0; i < N; i++) { for (int j = i; j < N; j++) swap(mat[i][j], mat[j][i]); } } // Function to print the matrix void displayMatrix(int mat[N][N]) { for (int i = 0; i < N; i++) { for (int j = 0; j < N; j++) { cout << mat[i][j] << " "; } cout << endl; } cout << endl; } /* Driver code */int main() { int mat[N][N] = { { 1, 2, 3, 4 }, { 5, 6, 7, 8 }, { 9, 10, 11, 12 }, { 13, 14, 15, 16 } }; // Function call rotateMatrix(mat); // Print rotated matrix displayMatrix(mat); return 0; } |
Java
// Java program to rotate a // matrix by 90 degrees import java.io.*; class GFG { // Function to reverse // the given 2D arr[][] static void Reverse(int i, int mat[][], int N) { // Initialise start and end index int start = 0; int end = N - 1; // Till start < end, swap the element // at start and end index while (start < end) { // Swap the element int temp = mat[i][start]; mat[i][start] = mat[i][end]; mat[i][end] = temp; // Increment start and decrement // end for next pair of swapping start++; end--; } } // An Inplace function to // rotate a N x N matrix // by 90 degrees in // anti-clockwise direction static void rotateMatrix(int N, int mat[][]) { // REVERSE every row for (int i = 0; i < N; i++) Reverse(i, mat, N); // Performing Transpose for (int i = 0; i < N; i++) { for (int j = i; j < N; j++) { int temp = mat[i][j]; mat[i][j] = mat[j][i]; mat[j][i] = temp; } } } // Function to print the matrix static void displayMatrix(int N, int mat[][]) { for (int i = 0; i < N; i++) { for (int j = 0; j < N; j++) System.out.print(" " + mat[i][j]); System.out.print("\n"); } System.out.print("\n"); } /* Driver code*/ public static void main(String[] args) { int N = 4; int mat[][] = { { 1, 2, 3, 4 }, { 5, 6, 7, 8 }, { 9, 10, 11, 12 }, { 13, 14, 15, 16 } }; // Function call rotateMatrix(N, mat); // Print rotated matrix displayMatrix(N, mat); } } // This code is contributed by Aarti_Rathi |
Python3
# Python program to rotate # a matrix by 90 degrees def rotateMatrix(mat): # reversing the matrix for i in range(len(mat)): mat[i].reverse() # make transpose of the matrix for i in range(len(mat)): for j in range(i, len(mat)): # swapping mat[i][j] and mat[j][i] mat[i][j], mat[j][i] = mat[j][i], mat[i][j] # Function to print the matrix def displayMatrix(mat): for i in range(0, len(mat)): for j in range(0, len(mat)): print(mat[i][j], end=' ') print() # Driver code if __name__ == "__main__": mat = [[1, 2, 3, 4], [5, 6, 7, 8], [9, 10, 11, 12], [13, 14, 15, 16]] # Function call rotateMatrix(mat) # Print rotated matrix displayMatrix(mat) # This code is contributed by shivambhagat02(CC). |
C#
// C# program to rotate a // matrix by 90 degrees using System; class GFG { // Reverse each row of matrix static void reverse(int N, int[, ] mat) { // Traverse each row of [,]mat for (int i = 0; i < N; i++) { // Initialise start and end index int start = 0; int end = N - 1; // Till start < end, swap the element // at start and end index while (start < end) { // Swap the element int temp = mat[i, start]; mat[i, start] = mat[i, end]; mat[i, end] = temp; // Increment start and decrement // end for next pair of swapping start++; end--; } } } // An Inplace function to // rotate a N x N matrix // by 90 degrees in anti- // clockwise direction static void rotateMatrix(int N, int[, ] mat) { reverse(N, mat); // Performing Transpose for (int i = 0; i < N; i++) { for (int j = i; j < N; j++) { int temp = mat[i, j]; mat[i, j] = mat[j, i]; mat[j, i] = temp; } } } // Function to print the matrix static void displayMatrix(int N, int[, ] mat) { for (int i = 0; i < N; i++) { for (int j = 0; j < N; j++) Console.Write(mat[i, j] + " "); Console.Write("\n"); } } // Driver Code static public void Main() { int N = 4; int[, ] mat = { { 1, 2, 3, 4 }, { 5, 6, 7, 8 }, { 9, 10, 11, 12 }, { 13, 14, 15, 16 } }; // Function call rotateMatrix(N, mat); // Print rotated matrix displayMatrix(N, mat); } } // This code is contributed by Aarti_Rathi |
Javascript
<script> // JavaScript program to rotate // a matrix by 90 degrees function rotateMatrix(mat){ // reversing the matrix for(let i = 0; i < mat.length; i++){ mat[i].reverse() } // make transpose of the matrix for(let i = 0; i < mat.length; i++){ for(let j = i; j < mat.length; j++){ // swapping mat[i][j] and mat[j][i] let temp = mat[i][j] mat[i][j] = mat[j][i] mat[j][i] = temp } } } // Function to print the matrix function displayMatrix(mat){ for(let i = 0; i < mat.length; i++){ for(let j = 0; j < mat.length; j++){ document.write(mat[i][j],' ') } document.write("</br>") } } let mat = [[1, 2, 3, 4], [5, 6, 7, 8], [9, 10, 11, 12], [13, 14, 15, 16]] rotateMatrix(mat) // Print rotated matrix displayMatrix(mat) // This code is contributed by shinjanpatra </script> |
Output
4 8 12 16 3 7 11 15 2 6 10 14 1 5 9 13
Time Complexity: O(N2) + O(N2) where N is the size of the array.
Auxiliary Space: O(1). As a constant space is needed
Exercise: Turn the 2D matrix by 90 degrees in a clockwise direction without using extra space.
Rotate a matrix by 90 degrees without using any extra space | Set 2
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