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Queue Reconstruction by Height⚓︎

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Description⚓︎

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 i-th 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 j-th 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:
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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 <= 10^6
  • 0 <= ki < people.length
  • It is guaranteed that the queue can be reconstructed.

Solution⚓︎

Algorithm:

  • Sort people:
    • In descending order by height.
    • Among the guys of the same height, in the ascending order by k-values.
  • Take guys one by one, and place them in the output array at the indexes equal to their k-values.
  • Return output array.
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class Solution {
public:
    vector<vector<int>> reconstructQueue(vector<vector<int>>& people) {
        sort(people.begin(), people.end(), [](vector<int>& lhs, vector<int>& rhs) {
            if (lhs[0] != rhs[0]) return lhs[0] > rhs[0];
            else return lhs[1] < rhs[1];
        });

        vector<vector<int>> res;
        for (int i = 0; i < people.size(); i++) {
            res.insert(res.begin() + people[i][1], people[i]);
        }
        return res;
    }
};
  • Time complexity: \(O\left( n\log n \right) +O\left( n^2 \right) \approx O\left( n^2 \right)\);
  • Space complexity: \(O(n)\).

Using List:

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class Solution {
public:
    vector<vector<int>> reconstructQueue(vector<vector<int>>& people) {
        sort(people.begin(), people.end(), [](vector<int>& lhs, vector<int>& rhs) {
            if (lhs[0] != rhs[0]) return lhs[0] > rhs[0];
            else return lhs[1] < rhs[1];
        });

        list<vector<int>> res;
        for (int i = 0; i < people.size(); i++) {
            int pos = people[i][1];
            auto it = res.begin();
            while (pos--) ++it;
            res.insert(it, people[i]);
        }

        return vector<vector<int>>(res.begin(), res.end());
    }
};