In-place rotate matrix by 180 degrees

Given a square matrix, rotate the matrix by 180 degrees in a clockwise direction. The transformation should be done in-place in quadratic time.

For example,

Input:

1   2   3   4
5   6   7   8
9   10  11  12
13  14  15  16

Output:

16  15  14  13
12  11  10  9
8   7   6   5
4   3   2   1

Practice this problem

If we swap elements of the first row with the last row elements in reverse order, elements of the second row with the elements of the second last row in reverse order, and so on… we will get our desired matrix. Note that if the matrix has odd dimensions, reverse elements of the middle row.

The algorithm can be implemented as follows in C++, Java, and Python:

C++

#include <iostream>

#include <vector>

#include <iomanip>

using namespace std;

// In-place rotate the matrix by 180 degrees in an anti-clockwise direction

void rotateMatrix(vector<vector<int>> &mat)

{

// `N × N` matrix

int N = mat.size();

// base case

if (N == 0) {

return;

}

// rotate the matrix by 180 degrees

for (int i = 0; i < N / 2; i++)

{

for (int j = 0; j < N; j++) {

swap(mat[i][j], mat[N - i - 1][N - j - 1]);

}

// handle the case when the matrix has odd dimensions

if (N & 1)

{

for (int j = 0; j < N/2; j++) {

swap(mat[N/2][j], mat[N/2][N - j - 1]);

}

// Function to print the matrix

void printMatrix(vector<vector<int>> const &mat)

{

for (auto &row: mat) {

for (auto &i: row) {

cout << setw(3) << i;

}

cout << endl;

}

cout << endl;

}

int main()

{

vector<vector<int>> mat =

{

{ 1, 2, 3, 4 },

{ 5, 6, 7, 8 },

{ 9, 10, 11, 12 },

{ 13, 14, 15, 16 }

};

rotateMatrix(mat);

printMatrix(mat);

return 0;

}

Download Run Code

Output:

16  15  14  13
12  11  10  9
8   7   6   5
4   3   2   1

Java

import java.util.Arrays;

class Main

{

// In-place rotate it by 180 degrees in an anti-clockwise direction

public static void rotateMatrix(int[][] mat)

{

// base case

if (mat == null || mat.length == 0) {

return;

}

// `N × N` matrix

int N = mat.length;

// rotate the matrix by 180 degrees

for (int i = 0; i < N /2; i++)

{

for (int j = 0; j < N; j++)

{

int temp = mat[i][j];

mat[i][j] = mat[N - i - 1][N - j - 1];

mat[N - i - 1][N - j - 1] = temp;

}

// handle the case when the matrix has odd dimensions

if (N % 2 == 1)

{

for (int j = 0; j < N/2; j++)

{

int temp = mat[N/2][j];

mat[N/2][j] = mat[N/2][N - j - 1];

mat[N/2][N - j - 1] = temp;

}

public static void main(String[] args)

{

int[][] mat =

{

{ 1, 2, 3, 4 },

{ 5, 6, 7, 8 },

{ 9, 10, 11, 12 },

{ 13, 14, 15, 16 }

};

rotateMatrix(mat);

// print the matrix

for (var r: mat) {

System.out.println(Arrays.toString(r));

}

Download Run Code

Output:

[16, 15, 14, 13]
[12, 11, 10, 9]
[8, 7, 6, 5]
[4, 3, 2, 1]

Python

# In-place rotate it by 180 degrees in an anti-clockwise direction

def rotateMatrix(mat):

# base case

if not mat or not len(mat):

return

# `N × N` matrix

N = len(mat)

# rotate the matrix by 180 degrees

for i in range(N // 2):

for j in range(N):

temp = mat[i][j]

mat[i][j] = mat[N - i - 1][N - j - 1]

mat[N - i - 1][N - j - 1] = temp

# handle the case when the matrix has odd dimensions

if N % 2 == 1:

for j in range(N // 2):

temp = mat[N // 2][j]

mat[N // 2][j] = mat[N // 2][N - j - 1]

mat[N // 2][N - j - 1] = temp

if __name__ == '__main__':

mat = [

[1, 2, 3, 4],

[5, 6, 7, 8],

[9, 10, 11, 12],

[13, 14, 15, 16]

]

rotateMatrix(mat)

# print the matrix

for r in mat:

print(r)

Download Run Code

Output:

[16, 15, 14, 13]
[12, 11, 10, 9]
[8, 7, 6, 5]
[4, 3, 2, 1]

The time complexity of the proposed solution is O(N²) for an N × N matrix and doesn’t require any extra space.

Related Posts:

In-place rotate matrix by 90 degrees in a clockwise direction