Approach: Binary search on index

def hIndex(nums):
    n = len(nums)
    lo, hi = 0, n - 1
    while lo <= hi:
        mid = (lo + hi) // 2
        if nums[mid] >= n - mid: hi = mid - 1
        else: lo = mid + 1
    return n - lo

Complexity: O(log n) time, O(1) space.

Full Solution — all languages

class Solution:
    def hIndex(self, nums):
        n = len(nums)
        lo, hi = 0, n - 1
        while lo <= hi:
            mid = (lo + hi) // 2
            if nums[mid] >= n - mid:
                hi = mid - 1
            else:
                lo = mid + 1
        return n - lo

import java.util.*;
class Solution {
    public int hIndex(int[] nums) {
        int n = nums.length;
        int lo = 0, hi = n - 1;
        while (lo <= hi) {
            int mid = (lo + hi) / 2;
            if (nums[mid] >= n - mid) hi = mid - 1;
            else lo = mid + 1;
        }
        return n - lo;
    }
}

#include <bits/stdc++.h> using namespace std;
class Solution {
    public: int hIndex(vector<int>& nums) {
        int n = (int)nums.size();
        int lo = 0, hi = n - 1;
        while (lo <= hi) {
            int mid = (lo + hi) / 2;
            if (nums[mid] >= n - mid) hi = mid - 1;
            else lo = mid + 1;
        }
        return n - lo;
    }
};

/**
 * @param {number[]} nums
 * @return {number}
 */ var hIndex = function(nums) {
    const n = nums.length;
    let lo = 0, hi = n - 1;
    while (lo <= hi) {
        const mid = (lo + hi) >> 1;
        if (nums[mid] >= n - mid) hi = mid - 1;
        else lo = mid + 1;
    }
    return n - lo;
};

H-Index II

Example 1

Example 2

Example 3

Constraints