274. H-Index

Description

Given an array of integers citations where citations[i] is the number of citations a researcher received for their i^th 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

Solutions

Solution 1: Sorting

We can sort the array citations in descending order. Then we enumerate the value $h$ from large to small, if there is an $h$ value satisfying $citations[h-1] \geq h$, it means that there are at least $h$ papers that have been cited at least $h$ times, just return $h$ directly. If we cannot find such an $h$ value, it means that all the papers have not been cited, return $0$.

Time complexity $O(n \times \log n)$, space complexity $O(\log n)$. Here $n$ is the length of the array citations.

Solution 2: Counting + Sum

We can use an array $cnt$ of length $n+1$, where $cnt[i]$ represents the number of papers with the reference count of $i$. We traverse the array citations and treat the papers with the reference count greater than $n$ as papers with a reference count of $n$. Then we use the reference count as the index and add $1$ to the corresponding element of $cnt$ for each paper. In this way, we have counted the number of papers for each reference count.

Then we enumerate the value $h$ from large to small, and add the element value of $cnt$ with the index of $h$ to the variable $s$, where $s$ represents the number of papers with a reference count greater than or equal to $h$. If $s \geq h$, it means that at least $h$ papers have been cited at least $h$ times, just return $h$ directly.

Time complexity $O(n)$, space complexity $O(n)$. Here $n$ is the length of the array citations.

Solution 3: Binary Search

We notice that if there is a $h$ value that satisfies at least $h$ papers are cited at least $h$ times, then for any $h’<h$, at least $h’$ papers are cited at least $h’$ times. Therefore, we can use the binary search method to find the largest $h$ such that at least $h$ papers are cited at least $h$ times.

We define the left boundary of binary search $l=0$ and the right boundary $r=n$. Each time we take $mid = \lfloor \frac{l + r + 1}{2} \rfloor$, where $\lfloor x \rfloor$ represents floor $x$. Then we count the number of elements in array citations that are greater than or equal to $mid$, and denote it as $s$. If $s \geq mid$, it means that at least $mid$ papers are cited at least $mid$ times. In this case, we change the left boundary $l$ to $mid$. Otherwise, we change the right boundary $r$ to $mid-1$. When the left boundary $l$ is equal to the right boundary $r$, we find the largest $h$ value, which is $l$ or $r$.

Time complexity $O(n \times \log n)$, where $n$ is the length of array citations. Space complexity $O(1)$.

class Solution {
    public int hIndex(int[] citations) {
        int n = citations.length;
        int[] cnt = new int[n + 1];
        for (int x : citations) {
            ++cnt[Math.min(x, n)];
        }
        for (int h = n, s = 0;; --h) {
            s += cnt[h];
            if (s >= h) {
                return h;
            }
        }
    }
}

class Solution {
public:
    int hIndex(vector<int>& citations) {
        int n = citations.size();
        int cnt[n + 1];
        memset(cnt, 0, sizeof(cnt));
        for (int x : citations) {
            ++cnt[min(x, n)];
        }
        for (int h = n, s = 0;; --h) {
            s += cnt[h];
            if (s >= h) {
                return h;
            }
        }
    }
};

class Solution:
    def hIndex(self, citations: List[int]) -> int:
        n = len(citations)
        cnt = [0] * (n + 1)
        for x in citations:
            cnt[min(x, n)] += 1
        s = 0
        for h in range(n, -1, -1):
            s += cnt[h]
            if s >= h:
                return h

func hIndex(citations []int) int {
	n := len(citations)
	cnt := make([]int, n+1)
	for _, x := range citations {
		cnt[min(x, n)]++
	}
	for h, s := n, 0; ; h-- {
		s += cnt[h]
		if s >= h {
			return h
		}
	}
}

function hIndex(citations: number[]): number {
    const n: number = citations.length;
    const cnt: number[] = new Array(n + 1).fill(0);
    for (const x of citations) {
        ++cnt[Math.min(x, n)];
    }
    for (let h = n, s = 0; ; --h) {
        s += cnt[h];
        if (s >= h) {
            return h;
        }
    }
}

impl Solution {
    #[allow(dead_code)]
    pub fn h_index(citations: Vec<i32>) -> i32 {
        let mut citations = citations;
        citations.sort_by(|&lhs, &rhs| { rhs.cmp(&lhs) });

        let n = citations.len();

        for i in (1..=n).rev() {
            if citations[i - 1] >= (i as i32) {
                return i as i32;
            }
        }

        0
    }
}

274 - H-Index

274. H-Index

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274. H-Index

Description

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