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2418. Sort the People


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 names sorted 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.



  • n == names.length == heights.length
  • 1 <= n <= 103
  • 1 <= names[i].length <= 20
  • 1 <= heights[i] <= 105
  • names[i] consists of lower and upper case English letters.
  • All the values of heights are distinct.


Solution 1: Sorting

According to the problem description, we can create an index array $idx$ of length $n$, where $idx[i]=i$. Then we sort each index in $idx$ in descending order according to the corresponding height in $heights$. Finally, we traverse each index $i$ in the sorted $idx$ and add $names[i]$ to the answer array.

We can also create an array $arr$ of length $n$, where each element is a tuple $(heights[i], i)$. Then we sort $arr$ in descending order by height. Finally, we traverse each element $(heights[i], i)$ in the sorted $arr$ and add $names[i]$ to the answer array.

The time complexity is $O(n \times \log n)$, and the space complexity is $O(n)$. Here, $n$ is the length of the arrays $names$ and $heights$.

  • class Solution {
        public String[] sortPeople(String[] names, int[] heights) {
            int n = names.length;
            Integer[] idx = new Integer[n];
            for (int i = 0; i < n; ++i) {
                idx[i] = i;
            Arrays.sort(idx, (i, j) -> heights[j] - heights[i]);
            String[] ans = new String[n];
            for (int i = 0; i < n; ++i) {
                ans[i] = names[idx[i]];
            return ans;
  • class Solution {
        vector<string> sortPeople(vector<string>& names, vector<int>& heights) {
            int n = names.size();
            vector<int> idx(n);
            iota(idx.begin(), idx.end(), 0);
            sort(idx.begin(), idx.end(), [&](int i, int j) { return heights[j] < heights[i]; });
            vector<string> ans;
            for (int i : idx) {
            return ans;
  • class Solution:
        def sortPeople(self, names: List[str], heights: List[int]) -> List[str]:
            idx = list(range(len(heights)))
            idx.sort(key=lambda i: -heights[i])
            return [names[i] for i in idx]
  • func sortPeople(names []string, heights []int) (ans []string) {
    	n := len(names)
    	idx := make([]int, n)
    	for i := range idx {
    		idx[i] = i
    	sort.Slice(idx, func(i, j int) bool { return heights[idx[j]] < heights[idx[i]] })
    	for _, i := range idx {
    		ans = append(ans, names[i])
  • function sortPeople(names: string[], heights: number[]): string[] {
        const n = names.length;
        const idx = new Array(n);
        for (let i = 0; i < n; ++i) {
            idx[i] = i;
        idx.sort((i, j) => heights[j] - heights[i]);
        const ans: string[] = [];
        for (const i of idx) {
        return ans;
  • impl Solution {
        pub fn sort_people(names: Vec<String>, heights: Vec<i32>) -> Vec<String> {
            let mut combine: Vec<(String, i32)> = names.into_iter().zip(heights.into_iter()).collect();
            combine.sort_by(|a, b| b.1.cmp(&a.1));
                .map(|s| s.0.clone())

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