README_EN.md

274. H-Index

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Description

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

According to the definition of h-index on Wikipedia: A scientist has an index h if h of their n papers have at least h citations each, and the other n − h papers have no more than h citations each.

If there are several possible values for h, the maximum one is taken as the h-index.

 

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

Solutions

The simplest solution is to judge after sort, but because H cannot be greater than the total number of papers, it can be optimized by counting sort.

Python3

class Solution:
    def hIndex(self, citations: List[int]) -> int:
        n = len(citations)
        cnt = [0 for i in range(n + 1)]
        for c in citations:
            if c <= n:
                cnt[c] += 1
            else:
                cnt[n] += 1
        sum = 0
        for i in range(n, -1, -1):
            sum += cnt[i]
            if sum >= i:
                return i
        return 0

Java

class Solution {
    public int hIndex(int[] citations) {
        int n = citations.length;
        int[] cnt = new int[n + 1];
        for (int c : citations) {
            if (c <= n) {
                ++cnt[c];
            } else {
                ++cnt[n];
            }
        }
        int sum = 0;
        for (int i = n; i >= 0; --i) {
            sum += cnt[i];
            if (sum >= i) {
                return i;
            }
        }
        return 0;
    }
}

TypeScript

function hIndex(citations: number[]): number {
    let n = citations.length;
    let cnt = new Array(n + 1).fill(0);
    for (let c of citations) {
        if (c <= n) {
            ++cnt[c];
        } else {
            ++cnt[n];
        }
    }
    let sum = 0;
    for (let i = n; i > -1; --i) {
        sum += cnt[i];
        if (sum >= i) {
            return i;
        }
    }
    return 0;
}

Go

Use binary search to locate the maximum value that meets the conditions

func hIndex(citations []int) int {
	n := len(citations)
	left, right := 0, n
	for left+1 < right {
		mid := int(uint(left+right) >> 1)
		if check(citations, mid) {
			left = mid
		} else {
			right = mid
		}
	}
	if check(citations, right) {
		return right
	}
	return left
}

func check(citations []int, mid int) bool {
	cnt := 0
	for _, citation := range citations {
		if citation >= mid {
			cnt++
		}
	}
	return cnt >= mid
}

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