2418 - Sort the People (Easy)

Problem Link

https://leetcode.com/problems/sort-the-people/

Problem Statement

You are given an array of strings names, and an array heights that consists of distinct positive integers. Both arrays are of length n.

For each index i, names[i] and heights[i] denote the name and height of the ith person.

Return namessorted in descending order by the people's heights.

Example 1:

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

Example 2:

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.

Constraints:

• n == names.length == heights.length
• 1 <= n <= 103
• 1 <= names[i].length <= 20
• 1 <= heights[i] <= 10^5
• names[i] consists of lower and upper case English letters.
• All the values of heights are distinct.

Approach 1: Sorting

C++

Written by @wkw

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;
    }
};

Java

Written by @ganajayant

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;
    }
}

Approach 2: Heap (Priority Queue)

C#

Written by @lonyehan

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();
    }
}