2418 - Sort the People (Easy)

Input: names = ["Mary","John","Emma"], heights = [180,165,170]
Output: ["Mary","Emma","John"]
Explanation: Mary is the tallest, followed by Emma and John.

Input: names = ["Alice","Bob","Bob"], heights = [155,185,150]
Output: ["Bob","Alice","Bob"]
Explanation: The first Bob is the tallest, followed by Alice and the second Bob.

class Solution {
public:
    vector<string> sortPeople(vector<string>& names, vector<int>& heights) {
        vector<string> ans;
        // {height, name}
        vector<pair<int, string>> v;
        // by putting height at first, we can skip writing a comparator as it'll sort by first element of the pair
        for (int i = 0; i < names.size(); i++) v.push_back({heights[i], names[i]});
        // sort in decreasing order
        sort(v.rbegin(), v.rend());
        // build the final ans. x.second is the name.
        for (auto & x : v) ans.push_back(x.second);
        return ans;
    }
};

class Solution {
    public String[] sortPeople(String[] names, int[] heights) {
        int[][] people = new int[names.length][2];
        for (int i = 0; i < names.length; i++) {
            people[i] = new int[] { heights[i], i };
        }

        Arrays.sort(people, (a, b) -> b[0] - a[0]);

        String[] result = new String[names.length];
        for (int i = 0; i < names.length; i++) {
            result[i] = names[people[i][1]];
        }

        return result;
    }
}

public class Solution {
    public string[] SortPeople(string[] names, int[] heights) {
        PriorityQueue<string, int> pq = new PriorityQueue<string, int>();
        List<string> list = new List<string>();

        for(int i = 0; i < names.Length; i++) {
            pq.Enqueue(names[i], -heights[i]);
        }

        while(pq.Count != 0) {
            list.Add(pq.Dequeue());
        }

        return list.ToArray();
    }
}