2418. Sort the People

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Description

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.

 

Constraints:

• 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.

Solutions

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 $heights$.

Python3

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]

Java

class Solution {
    public String[] sortPeople(String[] names, int[] heights) {
        int n = names.length;
        Integer[] idx = new Integer[n];
        Arrays.setAll(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;
    }
}

C++

class Solution {
public:
    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) {
            ans.push_back(names[i]);
        }
        return ans;
    }
};

Go

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])
	}
	return
}

TypeScript

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) {
        ans.push(names[i]);
    }
    return ans;
}

Rust

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));
        combine.iter().map(|s| s.0.clone()).collect()
    }
}

Solution 2

Python3

class Solution:
    def sortPeople(self, names: List[str], heights: List[int]) -> List[str]:
        return [name for _, name in sorted(zip(heights, names), reverse=True)]

Java

class Solution {
    public String[] sortPeople(String[] names, int[] heights) {
        int n = names.length;
        int[][] arr = new int[n][2];
        for (int i = 0; i < n; ++i) {
            arr[i] = new int[] {heights[i], i};
        }
        Arrays.sort(arr, (a, b) -> b[0] - a[0]);
        String[] ans = new String[n];
        for (int i = 0; i < n; ++i) {
            ans[i] = names[arr[i][1]];
        }
        return ans;
    }
}

C++

class Solution {
public:
    vector<string> sortPeople(vector<string>& names, vector<int>& heights) {
        int n = names.size();
        vector<pair<int, int>> arr;
        for (int i = 0; i < n; ++i) {
            arr.emplace_back(-heights[i], i);
        }
        sort(arr.begin(), arr.end());
        vector<string> ans;
        for (int i = 0; i < n; ++i) {
            ans.emplace_back(names[arr[i].second]);
        }
        return ans;
    }
};

Go

func sortPeople(names []string, heights []int) []string {
	n := len(names)
	arr := make([][2]int, n)
	for i, h := range heights {
		arr[i] = [2]int{h, i}
	}
	sort.Slice(arr, func(i, j int) bool { return arr[i][0] > arr[j][0] })
	ans := make([]string, n)
	for i, x := range arr {
		ans[i] = names[x[1]]
	}
	return ans
}

TypeScript

function sortPeople(names: string[], heights: number[]): string[] {
    return names
        .map<[string, number]>((s, i) => [s, heights[i]])
        .sort((a, b) => b[1] - a[1])
        .map(([v]) => v);
}