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406. Queue Reconstruction by Height 👍

  • Time: $O(n^2)$
  • Space: $O(n)$
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class Solution {
 public:
  vector<vector<int>> reconstructQueue(vector<vector<int>>& people) {
    vector<vector<int>> ans;

    ranges::sort(people, [](const vector<int>& a, const vector<int>& b) {
      return a[0] == b[0] ? a[1] < b[1] : a[0] > b[0];
    });

    for (const vector<int>& p : people)
      ans.insert(ans.begin() + p[1], p);

    return ans;
  }
};

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class Solution {
  public int[][] reconstructQueue(int[][] people) {
    List<int[]> ans = new ArrayList<>();

    Arrays.sort(people,
                (a, b) -> a[0] == b[0] ? Integer.compare(a[1], b[1]) : Integer.compare(b[0], a[0]));

    for (final int[] p : people)
      ans.add(p[1], p);

    return ans.stream().toArray(int[][] ::new);
  }
}

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class Solution:
  def reconstructQueue(self, people: list[list[int]]) -> list[list[int]]:
    ans = []

    people.sort(key=lambda x: (-x[0], x[1]))

    for p in people:
      ans.insert(p[1], p)

    return ans