Rotate Image medium

Problem Statement

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

You have to rotate the image in-place, which means you have to modify the input 2D matrix directly. DO NOT allocate another 2D matrix and do the rotation.

Example 1:

Input: matrix = [[1,2,3],[4,5,6],[7,8,9]]
Output: [[7,4,1],[8,5,2],[9,6,3]]

Example 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]]

Steps and Explanation

The key to solving this problem efficiently and in-place is to understand the pattern of how elements move during a 90-degree clockwise rotation. We can achieve this rotation in three steps:

1. Transpose: First, we transpose the matrix. Transposing a matrix swaps rows and columns. This means element matrix[i][j] becomes matrix[j][i].

2. Reverse Each Row: After transposing, we reverse each row of the matrix. This completes the 90-degree clockwise rotation.

Let's illustrate with a 3x3 matrix:

Original:
[[1, 2, 3],
 [4, 5, 6],
 [7, 8, 9]]

Transpose:
[[1, 4, 7],
 [2, 5, 8],
 [3, 6, 9]]

Reverse each Row:
[[7, 4, 1],
 [8, 5, 2],
 [9, 6, 3]]

Code (C++)

#include <vector>
#include <algorithm>

using namespace std;

void rotate(vector<vector<int>>& matrix) {
    int n = matrix.size();

    // Transpose the matrix
    for (int i = 0; i < n; i++) {
        for (int j = i; j < n; j++) {
            swap(matrix[i][j], matrix[j][i]);
        }
    }

    // Reverse each row
    for (int i = 0; i < n; i++) {
        reverse(matrix[i].begin(), matrix[i].end());
    }
}

Complexity Analysis

• Time Complexity: O(n^2). We iterate through the matrix twice (once for transpose and once for reversing rows).
• Space Complexity: O(1). We perform the rotation in-place, without using any extra space proportional to the input size. We only use a constant amount of extra space for variables like n, i, and j.