How do you solve Queue Reconstruction by Height on LeetCode?

Queue Reconstruction by Height (LeetCode #406) can be solved using Brute Force. The optimal approach is Brute Force with undefined time complexity and undefined space complexity. Key insight: The brute force approach involves trying to place each person in the queue one by one, checking how many taller people are in front of them. This is straightforward but inefficient.

What is the time complexity of Queue Reconstruction by Height?

The optimal solution for Queue Reconstruction by Height runs in undefined time and uses undefined space. undefined

What algorithm or data structure is used to solve Queue Reconstruction by Height?

Queue Reconstruction by Height uses the following concepts: Array, Binary Indexed Tree, Segment Tree, Sorting. The recommended patterns to study are: .

#406

Queue Reconstruction by Height

Medium

ArrayBinary Indexed TreeSegment TreeSorting

LeetCode ↗

Approaches

💡

Intuition

Time Space

The brute force approach involves trying to place each person in the queue one by one, checking how many taller people are in front of them. This is straightforward but inefficient.

⚙️

Algorithm

3 steps

1Step 1: Sort the people array by height in descending order and by k in ascending order for those with the same height.
2Step 2: Initialize an empty queue.
3Step 3: For each person in the sorted list, insert them into the queue at the index equal to their k value.

solution.py8 lines

1# Full working Python code
2people = [[7,0],[4,4],[7,1],[5,0],[6,1],[5,2]]
3
4people.sort(key=lambda x: (-x[0], x[1]))
5queue = []
6for person in people:
7    queue.insert(person[1], person)
8print(queue)

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