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406. Queue Reconstruction by Height



Suppose you have a random list of people standing in a queue. Each person is described by a pair of integers (h, k), where h is the height of the person and k is the number of people in front of this person who have a height greater than or equal to h. Write an algorithm to reconstruct the queue.
The number of people is less than 1,100.
[[7,0], [4,4], [7,1], [5,0], [6,1], [5,2]]
[[5,0], [7,0], [5,2], [6,1], [4,4], [7,1]]


Approach #1 Greedy

  1. 1.
    Pick out tallest group of people and sort them in a subarray (S). Since there's no other groups of people taller than them, therefore each guy's index will be just as same as his k value.
  2. 2.
    For 2nd tallest group (and the rest), insert each one of them into (S) by k value. So on and so forth.
class Solution {
public int[][] reconstructQueue(int[][] people) {
Arrays.sort(people, (a, b) ->
a[0] == b[0]? a[1] - b[1] : b[0] - a[0]
List<int[]> output = new LinkedList<>();
for (int[] p: people) {
output.add(p[1], p);
int n = people.length;
return output.toArray(new int[n][2]);