Last updated on March 7th, 2025 at 09:39 pm

Here, we see a Rotate Image LeetCode Solution. This Leetcode problem is solved using different approaches in many programming languages, such as C++, Java, JavaScript, Python, etc.

List of all LeetCode Solution

Topics

Array, Math, Matrix

Companies

Amazon, Apple, Microsoft

Level of Question

Medium

Rotate Image LeetCode Solution

1. Problem Statement

You are given an n x n 2D matrix representing an image, rotate the image by 90 degrees (clockwise).

Example 1: (fig-1)
Input: matrix = [[1,2,3],[4,5,6],[7,8,9]]
Output: [[7,4,1],[8,5,2],[9,6,3]]

Example 2: (fig-2)
Input: matrix = [[5,1,9,11],[2,4,8,10],[13,3,6,7],[15,14,12,16]]
Output: [[15,13,2,5],[14,3,4,1],[12,6,8,9],[16,7,10,11]]

2. Coding Pattern Used in Solution

The coding pattern used in all the provided implementations is “Matrix Manipulation”. This pattern involves performing operations directly on a 2D matrix, such as transposing, rotating, flipping, or other transformations.

3. Code Implementation in Different Languages

3.1 Rotate Image C++

class Solution {
public:
    void rotate(vector<vector<int>>& matrix) {
    int n = matrix.size();
    
    for(int i=0; i<n; i++)
    {
        for(int j=i; j<n; j++)
        {
            int temp = matrix[i][j];
            matrix[i][j] = matrix[j][i];
            matrix[j][i] = temp;
        }
    } 
    for(int i=0; i<n; i++)
    {
        reverse(matrix[i].begin(), matrix[i].end());
    }        
    }
};

3.2 Rotate Image Java

class Solution {
    public void rotate(int[][] matrix) {
        for(int i = 0; i<matrix.length; i++){
            for(int j = i; j<matrix[0].length; j++){
                int temp = 0;
                temp = matrix[i][j];
                matrix[i][j] = matrix[j][i];
                matrix[j][i] = temp;
            }
        }
        for(int i =0 ; i<matrix.length; i++){
            for(int j = 0; j<matrix.length/2; j++){
                int temp = 0;
                temp = matrix[i][j];
                matrix[i][j] = matrix[i][matrix.length-1-j];
                matrix[i][matrix.length-1-j] = temp;
            }
        }        
    }
}

3.3 Rotate Image JavaScript

var rotate = function(matrix) {
    let n = matrix.length, depth = ~~(n / 2)
    for (let i = 0; i < depth; i++) {
        let len = n - 2 * i - 1, opp = n - 1 - i
        for (let j = 0; j < len; j++) {
            let temp = matrix[i][i+j]
            matrix[i][i+j] = matrix[opp-j][i]
            matrix[opp-j][i] = matrix[opp][opp-j]
            matrix[opp][opp-j] = matrix[i+j][opp]
            matrix[i+j][opp] = temp
        }
    }    
};

3.4 Rotate Image Python

class Solution(object):
    def rotate(self, matrix):
        l = 0
        r = len(matrix) -1
        while l < r:
	        matrix[l], matrix[r] = matrix[r], matrix[l]
	        l += 1
	        r -= 1
        for i in range(len(matrix)):
	        for j in range(i):
		        matrix[i][j], matrix[j][i] = matrix[j][i], matrix[i][j]

4. Time and Space Complexity

	Time Complexity	Space Complexity
C++	O(n²)	O(1)
Java	O(n²)	O(1)
JavaScript	O(n²)	O(1)
Python	O(n²)	O(1)

The code uses the Matrix Manipulation pattern to rotate the matrix in-place.
The time complexity for all implementations is O(n²) because every element in the matrix is processed once.
The space complexity is O(1) because the rotation is performed in-place without using additional data structures.