DSA With AI
Week 10 ยท Backtracking

Permutations

Solve Permutations by recognizing the Backtracking pattern and turning the prompt into a small invariant before coding.

Medium Backtracking Week 10 Practice runner

Frame the problem

  • Implement permute with the exact signature used by the interactive runner.
  • Use the visible tests to confirm the input and output shape before reading the final solution.
  • Treat challenge tests as edge-case pressure: empty inputs, repeated values, boundary shapes, or impossible states.
  • State the invariant before code, then dry-run one passing case and one failing-looking case.
1. Reveal example inputs and outputs

Example 1

Input:

permute([
  1,
  2,
  3
])

Output:

[
  [
    1,
    2,
    3
  ],
  [
    1,
    3,
    2
  ],
  [
    2,
    1,
    3
  ],
  [
    2,
    3,
    1
  ],
  [
    3,
    1,
    2
  ],
  [
    3,
    2,
    1
  ]
]
2. Brute force first

What direct brute force would be correct for a tiny input? Name the exact repeated work that the target pattern removes.

3. Reveal the insight ladder
  1. Map the prompt to the Backtracking pattern instead of starting from syntax.
  2. Choose each remaining number once for the next position.
  3. A length two input has two orderings.
  4. Only reveal the final code after you can explain why each state update is safe.
4. Dry run before code
  1. permute-three: input [[1,2,3]] should produce [[1,2,3],[1,3,2],[2,1,3],[2,3,1],[3,1,2],[3,2,1]]. Hint to check your state: Choose each remaining number once for the next position.
  2. permute-two: input [[0,1]] should produce [[0,1],[1,0]]. Hint to check your state: A length two input has two orderings.
5. Reveal final Python solution 
def permute(nums: list[int]) -> list[list[int]]:
    result: list[list[int]] = []

    def backtrack(path: list[int], remaining: list[int]) -> None:
        if not remaining:
            result.append(path[:])
            return
        for index, value in enumerate(remaining):
            backtrack(path + [value], remaining[:index] + remaining[index + 1 :])

    backtrack([], nums)
    return result

Complexity: Derive the exact bounds from permute: count how often each input item is visited and the maximum size of the main state structure.

Interview narration

  • I will first describe the invariant in plain language.
  • Then I will explain what data structure carries that invariant across the traversal, loop, recursion, or DP transition.
  • Finally I will walk one edge case before writing the optimized version.

Common traps

  • Solving only the visible example instead of the invariant.
  • Forgetting empty input, singleton input, duplicate values, or impossible-state cases.
  • Revealing the solution before doing a dry run from the starter signature.

Follow-up drills

1. How do you turn this into a timed interview answer?

Start with the invariant, give the brute force in one sentence, name the optimized state, code the core loop or recursion, and run one visible test aloud before mentioning complexity.

2. How do you scale the same pattern to a larger input?

Track which state grows with the input: hash maps and sets grow with distinct values, queues grow with frontier width, recursion grows with depth, heaps grow with active candidates, and DP tables grow with state count.

3. What should you practice from blank tomorrow?

Rewrite permute without looking at the solution, then compare only the invariant and state updates before checking syntax.

Interactive runner

Write the required Python function. Your code runs locally in this browser. Hints reveal one failing case at a time.