Stealth Interview
  • Features
  • Pricing
  • Blog
  • Login
  • Sign up

Leetcode #1409: Queries on a Permutation With Key

In this guide, we solve Leetcode #1409 Queries on a Permutation With Key in Python and focus on the core idea that makes the solution efficient.

You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.

Leetcode

Problem Statement

Given the array queries of positive integers between 1 and m, you have to process all queries[i] (from i=0 to i=queries.length-1) according to the following rules: In the beginning, you have the permutation P=[1,2,3,...,m]. For the current i, find the position of queries[i] in the permutation P (indexing from 0) and then move this at the beginning of the permutation P.

Quick Facts

  • Difficulty: Medium
  • Premium: No
  • Tags: Binary Indexed Tree, Array, Simulation

Intuition

The rules are explicit, so simulating the process step by step is safest.

Careful state updates prevent subtle bugs.

Approach

Translate the rules into state updates and apply them in order.

Track the final state or aggregate as required.

Steps:

  • Translate rules into state updates.
  • Iterate for each step.
  • Return the final state.

Example

Input: queries = [3,1,2,1], m = 5 Output: [2,1,2,1] Explanation: The queries are processed as follow: For i=0: queries[i]=3, P=[1,2,3,4,5], position of 3 in P is 2, then we move 3 to the beginning of P resulting in P=[3,1,2,4,5]. For i=1: queries[i]=1, P=[3,1,2,4,5], position of 1 in P is 1, then we move 1 to the beginning of P resulting in P=[1,3,2,4,5]. For i=2: queries[i]=2, P=[1,3,2,4,5], position of 2 in P is 2, then we move 2 to the beginning of P resulting in P=[2,1,3,4,5]. For i=3: queries[i]=1, P=[2,1,3,4,5], position of 1 in P is 1, then we move 1 to the beginning of P resulting in P=[1,2,3,4,5]. Therefore, the array containing the result is [2,1,2,1].

Python Solution

class Solution: def processQueries(self, queries: List[int], m: int) -> List[int]: p = list(range(1, m + 1)) ans = [] for v in queries: j = p.index(v) ans.append(j) p.pop(j) p.insert(0, v) return ans

Complexity

The time complexity is O(n). The space complexity is O(1) to O(n).

Edge Cases and Pitfalls

Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.

Summary

This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.


Ace your next coding interview

We're here to help you ace your next coding interview.

Subscribe
Stealth Interview
© 2026 Stealth Interview®Stealth Interview is a registered trademark. All rights reserved.
Product
  • Blog
  • Pricing
Company
  • Terms of Service
  • Privacy Policy