Day011 - 274. H-Index

274. H-Index

Given an array of integers citations where citations[i] is the number of citations a researcher received for their ith paper, return the researcher’s h-index.

According to the definition of h-index on Wikipedia: The h-index is defined as the maximum value of h such that the given researcher has published at least h papers that have each been cited at least h times.

Example 1:

Input: citations = [3,0,6,1,5]
Output: 3
Explanation: [3,0,6,1,5] means the researcher has 5 papers in total and each of them had received 3, 0, 6, 1, 5 citations respectively.
Since the researcher has 3 papers with at least 3 citations each and the remaining two with no more than 3 citations each, their h-index is 3.

Example 2:

Input: citations = [1,3,1]
Output: 1

Constraints:

• n == citations.length
• 1 <= n <= 5000
• 0 <= citations[i] <= 1000

내 풀이

from collections import Counter

class Solution:
    def hIndex(self, citations: List[int]) -> int:
        citations = sorted(Counter(citations).items(), key=lambda x:x[0], reverse=True)

        if len(citations) == 1:
            return min(citations[0][0], citations[0][1])

        print(citations)

        before_score = citations[0][0]
        all_paper_cnt = citations[0][1]

        for i in citations[1:]:
            print(f"for loop {i}")
            score, paper_cnt = i
            if score == 0:
                continue
            all_paper_cnt += paper_cnt

            print(f"score:{score} / paper_cnt:{all_paper_cnt}", '\n')

            if score == all_paper_cnt:
                return score
            
            elif score < all_paper_cnt:
                h_index_1 = min(score, all_paper_cnt)
                h_index_2 = min(before_score, all_paper_cnt - paper_cnt)
                h_index = max(h_index_1, h_index_2)
                return h_index

            before_score = score

        print("end for loop")
        print(f"before_score:{before_score} / paper_cnt:{paper_cnt}")
        return min(before_score, all_paper_cnt)

이

# Time Complexity : \(O(N)\)