How do you solve Queries on a Permutation With Key on LeetCode?

Queries on a Permutation With Key (LeetCode #1409) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Using a HashMap can significantly reduce the time complexity of finding indices.

What is the brute force solution for Queries on a Permutation With Key?

The brute force approach for Queries on a Permutation With Key is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves scanning the permutation to find the position of each query and then moving that element to the front. This is straightforward but inefficient as it requires multiple scans of the permutation. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Queries on a Permutation With Key?

Queries on a Permutation With Key uses the following concepts: Array, Binary Indexed Tree, Simulation. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Queries on a Permutation With Key?

Always think about the time complexity of your approach. Consider using data structures like HashMaps for efficient lookups. Explain your thought process clearly during the interview.

What are common mistakes when solving Queries on a Permutation With Key?

Not updating the indices in the HashMap after moving elements. Forgetting to handle the case when the queried element is already at the front.

#1409

Queries on a Permutation With Key

Medium

ArrayBinary Indexed TreeSimulationHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

The optimal approach uses a HashMap to keep track of the indices of elements in the permutation, allowing us to find and move elements in constant time. This significantly reduces the number of operations needed.

⚙️

Algorithm

3 steps

1Step 1: Initialize the permutation P as a list from 1 to m and create a HashMap to store the current indices of each element.
2Step 2: For each query, use the HashMap to find the index of the query in constant time.
3Step 3: Move the queried element to the front of the permutation and update the indices in the HashMap accordingly.

solution.py12 lines

1def processQueries(queries, m):
2    P = list(range(1, m + 1))
3    index_map = {val: idx for idx, val in enumerate(P)}
4    result = []
5    for query in queries:
6        index = index_map[query]
7        result.append(index)
8        # Move the queried element to the front
9        P.insert(0, P.pop(index))
10        # Update the index map
11        for i in range(len(P)): index_map[P[i]] = i
12    return result

ℹ

Complexity note: The time complexity is O(n) for each query due to the efficient use of a HashMap to track indices, and the space complexity is O(n) for storing the permutation and the index map.

1Using a HashMap can significantly reduce the time complexity of finding indices.
2Moving elements in an array can be optimized by maintaining a mapping of indices.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.