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Kotlin-based LeetCode algorithm problem solutions, regularly updated
406\. Queue Reconstruction by Height
Medium
You are given an array of people, `people`, which are the attributes of some people in a queue (not necessarily in order). Each people[i] = [hi, ki] represents the ith person of height hi with **exactly** ki other people in front who have a height greater than or equal to hi.
Reconstruct and return _the queue that is represented by the input array_ `people`. The returned queue should be formatted as an array `queue`, where queue[j] = [hj, kj] is the attributes of the jth person in the queue (`queue[0]` is the person at the front of the queue).
**Example 1:**
**Input:** people = [[7,0],[4,4],[7,1],[5,0],[6,1],[5,2]]
**Output:** [[5,0],[7,0],[5,2],[6,1],[4,4],[7,1]]
**Explanation:**
Person 0 has height 5 with no other people taller or the same height in front.
Person 1 has height 7 with no other people taller or the same height in front.
Person 2 has height 5 with two persons taller or the same height in front, which is person 0 and 1.
Person 3 has height 6 with one person taller or the same height in front, which is person 1.
Person 4 has height 4 with four people taller or the same height in front, which are people 0, 1, 2, and 3.
Person 5 has height 7 with one person taller or the same height in front, which is person 1.
Hence [[5,0],[7,0],[5,2],[6,1],[4,4],[7,1]] is the reconstructed queue.
**Example 2:**
**Input:** people = [[6,0],[5,0],[4,0],[3,2],[2,2],[1,4]]
**Output:** [[4,0],[5,0],[2,2],[3,2],[1,4],[6,0]]
**Constraints:**
* `1 <= people.length <= 2000`
* 0 <= hi <= 106
* 0 <= ki < people.length
* It is guaranteed that the queue can be reconstructed.