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48. Rotate Image

mediumAsked at Cisco

Rotate Image at Cisco tests whether you can manipulate a 2D matrix in place without falling back to an auxiliary matrix. The interviewer wants the transpose-then-reverse trick because it generalizes to any 90-degree rotation and uses zero extra space.

By Alex Chen, Founder, InterviewChamp.AI · Last verified

Source citations

Public interview reports confirming this problem appears in Cisco loops.

  • Glassdoor (2026-Q1)— Cisco SDE-II interview reports cite Rotate Image as a 20-30 minute matrix round.
  • Blind (2025-09)— Cisco interview thread lists Rotate Image as a recurring problem.

Problem

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.

Constraints

  • n == matrix.length == matrix[i].length
  • 1 <= n <= 20
  • -1000 <= matrix[i][j] <= 1000

Examples

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]]

Approaches

1. Brute-force: allocate a fresh matrix

Read every cell from the input and write into result[j][n-1-i].

Time
O(n^2)
Space
O(n^2)
function rotateBrute(matrix) {
  const n = matrix.length;
  const result = Array.from({ length: n }, () => new Array(n));
  for (let i = 0; i < n; i++) {
    for (let j = 0; j < n; j++) {
      result[j][n - 1 - i] = matrix[i][j];
    }
  }
  for (let i = 0; i < n; i++) {
    for (let j = 0; j < n; j++) {
      matrix[i][j] = result[i][j];
    }
  }
}

Tradeoff: Violates the in-place constraint — Cisco interviewers reject this immediately. Useful only to anchor the index math before optimizing.

2. Transpose, then reverse each row (optimal)

Two passes: (1) transpose the matrix (swap matrix[i][j] with matrix[j][i] for i < j), (2) reverse each row in place. Composition of those two operations is a 90-degree clockwise rotation.

Time
O(n^2)
Space
O(1)
function rotate(matrix) {
  const n = matrix.length;
  // Transpose
  for (let i = 0; i < n; i++) {
    for (let j = i + 1; j < n; j++) {
      [matrix[i][j], matrix[j][i]] = [matrix[j][i], matrix[i][j]];
    }
  }
  // Reverse each row
  for (let i = 0; i < n; i++) {
    matrix[i].reverse();
  }
}

Tradeoff: Zero extra space, idiomatic. The key insight: a 90-degree rotation = reflect across main diagonal + reflect across vertical center. The 'j = i + 1' bound on the transpose loop is what prevents double-swapping (which would undo the transpose).

3. Four-way cyclic swap (also optimal)

Rotate four cells at a time — for each (i, j) in the top-left quadrant, swap matrix[i][j], matrix[j][n-1-i], matrix[n-1-i][n-1-j], matrix[n-1-j][i].

Time
O(n^2)
Space
O(1)
function rotateCyclic(matrix) {
  const n = matrix.length;
  for (let i = 0; i < Math.floor(n / 2); i++) {
    for (let j = i; j < n - 1 - i; j++) {
      const tmp = matrix[i][j];
      matrix[i][j] = matrix[n - 1 - j][i];
      matrix[n - 1 - j][i] = matrix[n - 1 - i][n - 1 - j];
      matrix[n - 1 - i][n - 1 - j] = matrix[j][n - 1 - i];
      matrix[j][n - 1 - i] = tmp;
    }
  }
}

Tradeoff: Single pass, identical Big-O. The index math is harder to get right under interview pressure. Cisco interviewers respect the cleaner transpose-then-reverse version unless you specifically claim 'single pass.'

Cisco-specific tips

Cisco interviewers grade this on the index math AND the verbal explanation. Say the geometric framing FIRST: '90-degree rotation = transpose + horizontal flip.' Then write the two loops. The `j = i + 1` bound is the gotcha — write it and explain 'I skip the diagonal because swapping it with itself is a no-op AND skip the lower triangle because those swaps were already done.'

Common mistakes

  • Writing the transpose loop as `for j = 0 to n` — that double-swaps every cell and ends with the original matrix.
  • Forgetting to reverse the rows — half the algorithm; produces a transposed matrix, not a rotated one.
  • Mixing up clockwise and counterclockwise — clockwise = transpose + reverse-rows; counterclockwise = transpose + reverse-cols (or reverse-rows then transpose).

Follow-up questions

An interviewer at Cisco may pivot to one of these next:

  • Spiral Matrix (LC 54) — traverse in spiral order.
  • Set Matrix Zeroes (LC 73) — in-place using first row/col as flags.
  • Rotate by 180 or 270 degrees — what changes?
  • Rotate a rectangular (m x n) matrix by 90 — requires allocation, not in-place possible.

Solve it now

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Output

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FAQ

Why does transpose + reverse equal 90-degree clockwise?

Transpose maps (i, j) -> (j, i) — it's a reflection across the main diagonal. Reversing each row then maps (j, i) -> (j, n-1-i). Composition is (i, j) -> (j, n-1-i), which is exactly the 90-degree clockwise rule. Geometrically: 'reflect across diagonal, then mirror horizontally.'

Should I default to transpose+reverse or cyclic swap?

Transpose+reverse for the whiteboard — it's two simple loops with obvious correctness. Use cyclic swap only if the interviewer asks for 'single pass.' Cisco interviewers care more about clarity than performance on n=20.

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