506. Relative Ranks

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Description

You are given an integer array score of size n, where score[i] is the score of the ith athlete in a competition. All the scores are guaranteed to be unique.

The athletes are placed based on their scores, where the 1st place athlete has the highest score, the 2nd place athlete has the 2nd highest score, and so on. The placement of each athlete determines their rank:

• The 1st place athlete's rank is "Gold Medal".
• The 2nd place athlete's rank is "Silver Medal".
• The 3rd place athlete's rank is "Bronze Medal".
• For the 4th place to the nth place athlete, their rank is their placement number (i.e., the xth place athlete's rank is "x").

Return an array answer of size n where answer[i] is the rank of the ith athlete.

 

Example 1:

Input: score = [5,4,3,2,1]
Output: ["Gold Medal","Silver Medal","Bronze Medal","4","5"]
Explanation: The placements are [1st, 2nd, 3rd, 4th, 5th].

Example 2:

Input: score = [10,3,8,9,4]
Output: ["Gold Medal","5","Bronze Medal","Silver Medal","4"]
Explanation: The placements are [1st, 5th, 3rd, 2nd, 4th].

 

Constraints:

• n == score.length
• 1 <= n <= 104
• 0 <= score[i] <= 106
• All the values in score are unique.

Solutions

Solution 1: Sorting

We use an array $\textit{idx}$ to store the indices from $0$ to $n-1$, then sort $\textit{idx}$ based on the values in $\textit{score}$ in descending order.

Next, we define an array $\textit{top3} = [\text{Gold Medal}, \text{Silver Medal}, \text{Bronze Medal}]$. We traverse $\textit{idx}$, and for each index $j$, if $j$ is less than $3$, then $\textit{ans}[j]$ is $\textit{top3}[j]$; otherwise, it is $j+1$.

The time complexity is $O(n \times \log n)$, and the space complexity is $O(n)$. Here, $n$ is the length of the array $\textit{score}$.

Python3

class Solution:
    def findRelativeRanks(self, score: List[int]) -> List[str]:
        n = len(score)
        idx = list(range(n))
        idx.sort(key=lambda x: -score[x])
        top3 = ["Gold Medal", "Silver Medal", "Bronze Medal"]
        ans = [None] * n
        for i, j in enumerate(idx):
            ans[j] = top3[i] if i < 3 else str(i + 1)
        return ans

Java

class Solution {
    public String[] findRelativeRanks(int[] score) {
        int n = score.length;
        Integer[] idx = new Integer[n];
        for (int i = 0; i < n; ++i) {
            idx[i] = i;
        }
        Arrays.sort(idx, (i1, i2) -> score[i2] - score[i1]);
        String[] ans = new String[n];
        String[] top3 = new String[] {"Gold Medal", "Silver Medal", "Bronze Medal"};
        for (int i = 0; i < n; ++i) {
            ans[idx[i]] = i < 3 ? top3[i] : String.valueOf(i + 1);
        }
        return ans;
    }
}

C++

class Solution {
public:
    vector<string> findRelativeRanks(vector<int>& score) {
        int n = score.size();
        vector<int> idx(n);
        iota(idx.begin(), idx.end(), 0);
        sort(idx.begin(), idx.end(), [&score](int a, int b) {
            return score[a] > score[b];
        });
        vector<string> ans(n);
        vector<string> top3 = {"Gold Medal", "Silver Medal", "Bronze Medal"};
        for (int i = 0; i < n; ++i) {
            ans[idx[i]] = i < 3 ? top3[i] : to_string(i + 1);
        }
        return ans;
    }
};

Go

func findRelativeRanks(score []int) []string {
	n := len(score)
	idx := make([][]int, n)
	for i := 0; i < n; i++ {
		idx[i] = []int{score[i], i}
	}
	sort.Slice(idx, func(i1, i2 int) bool {
		return idx[i1][0] > idx[i2][0]
	})
	ans := make([]string, n)
	top3 := []string{"Gold Medal", "Silver Medal", "Bronze Medal"}
	for i := 0; i < n; i++ {
		if i < 3 {
			ans[idx[i][1]] = top3[i]
		} else {
			ans[idx[i][1]] = strconv.Itoa(i + 1)
		}
	}
	return ans
}

TypeScript

function findRelativeRanks(score: number[]): string[] {
    const n = score.length;
    const idx = Array.from({ length: n }, (_, i) => i);
    idx.sort((a, b) => score[b] - score[a]);
    const top3 = ['Gold Medal', 'Silver Medal', 'Bronze Medal'];
    const ans: string[] = Array(n);
    for (let i = 0; i < n; i++) {
        if (i < 3) {
            ans[idx[i]] = top3[i];
        } else {
            ans[idx[i]] = (i + 1).toString();
        }
    }
    return ans;
}