Leetcode 275 - H-index II

Note:

  • Use binary search to reach log runtime.
  • When citations[mid] >= n - mid, mid might be the answer, but our h can be evern bigger, so we should move left with right = mid (Not right = mid - 1 in case we miss our ans.)
  • When citations[mid] < n - mid, either citation[mid] is too small or mid is too small, so we should move right by left = mid + 1.
  • When the loop stops, the exception is array like [0], we need to check if cs[left] >= n - left.
    If not, return 0.

Question:

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 an ascending order, 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.

You must write an algorithm that runs in logarithmic time.

Example:

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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.

Code:

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/**
* @param {number[]} citations
* @return {number}
*/
var hIndex = function(citations) {
let left = 0, right = citations.length - 1;
const n = citations.length;
while (left < right) {
const mid = ~~((left + right) / 2);
if (citations[mid] >= n - mid) {
right = mid;
} else {
left = mid + 1;
}
}
return citations[left] >= n - left ? n - left : 0;
};