506. Relative Ranks

Easy

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

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

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

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

Example 1:

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

Example 2:

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

Constraints:

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

 class Solution:
    def findRelativeRanks(self, scores: List[int]) -> List[str]:

        heap = []
        for i, score in enumerate(scores):
             heapq.heappush(heap, (-score, i))


        medels = ["Gold Medal","Silver Medal","Bronze Medal"]
        i = 0
        ans = [None]*len(scores)

        while i < len(scores):
            val, pos = heapq.heappop(heap)
            if i < 3:                
                ans[pos] = medels[i]
            else:
                ans[pos] = str(i+1)
            i+=1

        return ans

506. Relative Ranks

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