LeetCode #1409 — MEDIUM

Queries on a Permutation With Key

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

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. Notice that the position of queries[i] in P is the result for queries[i].

Return an array containing the result for the given queries.

Example 1:

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].

Example 2:

Input: queries = [4,1,2,2], m = 4
Output: [3,1,2,0]

Example 3:

Input: queries = [7,5,5,8,3], m = 8
Output: [6,5,0,7,5]

Constraints:

1 <= m <= 10^3
1 <= queries.length <= m
1 <= queries[i] <= m

Step 01

Brute Force Baseline

Problem summary: 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. Notice that the position of queries[i] in P is the result for queries[i]. Return an array containing the result for the given queries.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Segment Tree

Example 1

[3,1,2,1]
5

Example 2

[4,1,2,2]
4

Example 3

[7,5,5,8,3]
8

Step 02

Core Insight

What unlocks the optimal approach

Create the permutation P=[1,2,...,m], it could be a list for example.
For each i, find the position of queries[i] with a simple scan over P and then move this to the beginning.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #1409: Queries on a Permutation With Key
class Solution {
    public int[] processQueries(int[] queries, int m) {
        List<Integer> p = new LinkedList<>();
        for (int i = 1; i <= m; ++i) {
            p.add(i);
        }
        int[] ans = new int[queries.length];
        int i = 0;
        for (int v : queries) {
            int j = p.indexOf(v);
            ans[i++] = j;
            p.remove(j);
            p.add(0, v);
        }
        return ans;
    }
}

// Accepted solution for LeetCode #1409: Queries on a Permutation With Key
func processQueries(queries []int, m int) []int {
	p := make([]int, m)
	for i := range p {
		p[i] = i + 1
	}
	ans := []int{}
	for _, v := range queries {
		j := 0
		for i := range p {
			if p[i] == v {
				j = i
				break
			}
		}
		ans = append(ans, j)
		p = append(p[:j], p[j+1:]...)
		p = append([]int{v}, p...)
	}
	return ans
}

# Accepted solution for LeetCode #1409: Queries on a Permutation With Key
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

// Accepted solution for LeetCode #1409: Queries on a Permutation With Key
struct Solution;

impl Solution {
    fn process_queries(queries: Vec<i32>, m: i32) -> Vec<i32> {
        let mut v: Vec<i32> = (1..=m).collect();
        let mut res = vec![];
        for q in queries {
            let p = v.iter().position(|&x| x == q).unwrap();
            v.remove(p);
            v.insert(0, q);
            res.push(p as i32);
        }
        res
    }
}

#[test]
fn test() {
    let queries = vec![3, 1, 2, 1];
    let m = 5;
    let res = vec![2, 1, 2, 1];
    assert_eq!(Solution::process_queries(queries, m), res);
    let queries = vec![4, 1, 2, 2];
    let m = 4;
    let res = vec![3, 1, 2, 0];
    assert_eq!(Solution::process_queries(queries, m), res);
    let queries = vec![7, 5, 5, 8, 3];
    let m = 8;
    let res = vec![6, 5, 0, 7, 5];
    assert_eq!(Solution::process_queries(queries, m), res);
}

// Accepted solution for LeetCode #1409: Queries on a Permutation With Key
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #1409: Queries on a Permutation With Key
// class Solution {
//     public int[] processQueries(int[] queries, int m) {
//         List<Integer> p = new LinkedList<>();
//         for (int i = 1; i <= m; ++i) {
//             p.add(i);
//         }
//         int[] ans = new int[queries.length];
//         int i = 0;
//         for (int v : queries) {
//             int j = p.indexOf(v);
//             ans[i++] = j;
//             p.remove(j);
//             p.add(0, v);
//         }
//         return ans;
//     }
// }

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n + q log n)

Space

O(n)

Approach Breakdown

BRUTE FORCE

O(n × q) time

O(1) space

For each of q queries, scan the entire range to compute the aggregate (sum, min, max). Each range scan takes up to O(n) for a full-array query. With q queries: O(n × q) total. Point updates are O(1) but queries dominate.

SEGMENT TREE

O(n + q log n) time

O(n) space

Building the tree is O(n). Each query or update traverses O(log n) nodes (tree height). For q queries: O(n + q log n) total. Space is O(4n) ≈ O(n) for the tree array. Lazy propagation adds O(1) per node for deferred updates.

Shortcut: Build O(n), query/update O(log n) each. When you need both range queries AND point updates.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.