LeetCode 0275. H-Index II Solution in Java, Python, C++, JavaScript, Go & Rust | Explanation + Code

Description

Given an array of integers citations where citations[i] is the number of citations a researcher received for their ith paper and citations is sorted in non-descending order, 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.

You must write an algorithm that runs in logarithmic time.

 

Example 1:

Input: citations = [0,1,3,5,6]
Output: 3
Explanation: [0,1,3,5,6] means the researcher has 5 papers in total and each of them had received 0, 1, 3, 5, 6 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 = [1,2,100]
Output: 2

 

Constraints:

• n == citations.length
• 1 <= n <= 105
• 0 <= citations[i] <= 1000
• citations is sorted in ascending order.

Solutions

Solution 1: Binary Search

We notice that if there are at least x papers with citation counts greater than or equal to x, then for any y \lt x, its citation count must also be greater than or equal to y. This exhibits monotonicity.

Therefore, we use binary search to enumerate h and obtain the maximum h that satisfies the condition. Since we need to satisfy that h papers are cited at least h times, we have citations[n - mid] \ge mid.

The time complexity is O(log n), where n is the length of the array citations. The space complexity is O(1).

PythonJavaC++GoTypeScriptRustC#

class Solution: def hIndex(self, citations: List[int]) -> int: n = len(citations) left, right = 0, n while left < right: mid = (left + right + 1) >> 1 if citations[n - mid] >= mid: left = mid else: right = mid - 1 return left(code-box)

class Solution { public int hIndex(int[] citations) { int n = citations.length; int left = 0, right = n; while (left < right) { int mid = (left + right) >>> 1; if (citations[mid] >= n - mid) { right = mid; } else { left = mid + 1; } } return n - left; } }(code-box)

class Solution { public: int hIndex(vector<int>& citations) { int n = citations.size(); int left = 0, right = n; while (left < right) { int mid = (left + right + 1) >> 1; if (citations[n - mid] >= mid) left = mid; else right = mid - 1; } return left; } };(code-box)

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

function hIndex(citations: number[]): number { const n = citations.length; let left = 0, right = n; while (left < right) { const mid = (left + right + 1) >> 1; if (citations[n - mid] >= mid) { left = mid; } else { right = mid - 1; } } return left; }(code-box)

impl Solution { pub fn h_index(citations: Vec<i32>) -> i32 { let n = citations.len(); let (mut left, mut right) = (0, n); while left < right { let mid = ((left + right + 1) >> 1) as usize; if citations[n - mid] >= (mid as i32) { left = mid; } else { right = mid - 1; } } left as i32 } }(code-box)

public class Solution { public int HIndex(int[] citations) { int n = citations.Length; int left = 0, right = n; while (left < right) { int mid = (left + right + 1) >> 1; if (citations[n - mid] >= mid) { left = mid; } else { right = mid - 1; } } return left; } }(code-box)