Rotate Image (matrix rotation in place).

Problem. Rotate an n × n matrix 90° clockwise, in place (O(1) extra space).

The two-step decomposition — transpose then reverse rows

A clockwise 90° rotation equals transpose (swap across the main diagonal) followed by reversing each row. This is the cleanest, least error-prone approach and the one to present first.

void rotate(int[][] m) {
    int n = m.length;

    // 1. Transpose: swap m[i][j] with m[j][i] for j > i (upper triangle only)
    for (int i = 0; i < n; i++)
        for (int j = i + 1; j < n; j++) {
            int tmp = m[i][j];
            m[i][j] = m[j][i];
            m[j][i] = tmp;
        }

    // 2. Reverse each row
    for (int[] row : m) {
        int lo = 0, hi = n - 1;
        while (lo < hi) {
            int tmp = row[lo]; row[lo] = row[hi]; row[hi] = tmp;
            lo++; hi--;
        }
    }
}

Why j = i + 1: the transpose loop must touch each off-diagonal pair once. Iterating the full square would swap every pair twice, undoing the transpose. Restricting to the upper triangle (j > i) fixes that.

For counterclockwise 90°, transpose then reverse each column (or reverse rows first, then transpose).

Alternative — four-way cycle swap (layer by layer)

Rotate the matrix in concentric rings, moving four elements per step with one temporary. Same O(n²) time, O(1) space, but far more index-juggling and easier to get wrong under pressure:

void rotate(int[][] m) {
    int n = m.length;
    for (int layer = 0; layer < n / 2; layer++) {
        int first = layer, last = n - 1 - layer;
        for (int i = first; i < last; i++) {
            int offset = i - first;
            int top = m[first][i];                          // save top
            m[first][i]            = m[last - offset][first];   // left  -> top
            m[last - offset][first] = m[last][last - offset];   // bottom-> left
            m[last][last - offset] = m[i][last];                // right -> bottom
            m[i][last]             = top;                        // top   -> right
        }
    }
}