Rotate Image

mediumAmazon-tagged: Arrays & Strings39 stepsJava · Python · C++

Rotate an n×n matrix 90° clockwise in place. Transpose it (swap across the main diagonal), then reverse every row. The two reflections together are a rotation.

▶ Watch the animated trace

Traced input: matrix = [[1, 2, 3], [4, 5, 6], [7, 8, 9]] → [[7, 4, 1], [8, 5, 2], [9, 6, 3]]. Original problem: neetcode.io/problems/rotate-matrix

Complexity

BestO(n²)

AverageO(n²)

WorstO(n²)

SpaceO(1)

Each cell is swapped a constant number of times.

Key idea

Clockwise rotation = transpose, then mirror left-right.
Transpose only swaps cells above the diagonal (j > i), or it undoes itself.
Counter-clockwise would reverse columns instead of rows.

Common mistakes

Transposing the whole matrix (j from 0) swaps everything twice.
Using a second matrix breaks the in-place requirement.

Solution

Java

void rotate(int[][] matrix) {
    int n = matrix.length;
    for (int i = 0; i < n; i++)
        for (int j = i + 1; j < n; j++) {
            int t = matrix[i][j];
            matrix[i][j] = matrix[j][i];
            matrix[j][i] = t;
        }
    for (int[] row : matrix) {
        for (int l = 0, r = n - 1; l < r; l++, r--) {
            int t = row[l];
            row[l] = row[r];
            row[r] = t;
        }
    }
}

Python

def rotate(matrix: List[List[int]]) -> None:
    n = len(matrix)
    for i in range(n):
        for j in range(i + 1, n):
            matrix[i][j], matrix[j][i] = matrix[j][i], matrix[i][j]
    for row in matrix:
        row.reverse()

C++

void rotate(vector<vector<int>>& matrix) {
    int n = matrix.size();
    for (int i = 0; i < n; i++)
        for (int j = i + 1; j < n; j++)
            swap(matrix[i][j], matrix[j][i]);
    for (auto& row : matrix)
        reverse(row.begin(), row.end());
}

Edge cases

Each one has its own trace in the interactive view.

Case	Input	Answer
1×1 Unchanged.	`matrix = [[1]]`	`[[1]]`
2×2 One swap, then two row reversals.	`matrix = [[1, 2], [3, 4]]`	`[[3, 1], [4, 2]]`