h-index-ii.js

/**
 * H-Index II
 *
 * Given an array of citations sorted in ascending order (each citation is a non-negative integer)
 * of a researcher, write a function to compute the researcher's h-index.
 *
 * According to the definition of h-index on Wikipedia: "A scientist has index h if h of his/her N papers
 * have at least h citations each, and the other N − h papers have no more than h citations each."
 *
 * Example:
 *
 * 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, her h-index is 3.
 * Note:
 *
 * If there are several possible values for h, the maximum one is taken as the h-index.
 *
 * Follow up:
 *
 * This is a follow up problem to H-Index, where citations is now guaranteed to be sorted in ascending order.
 * Could you solve it in logarithmic time complexity?
 *
 * Solution
 *
 * case 1: citations[mid] == len-mid, then it means there are citations[mid] papers
 * that have at least citations[mid] citations.
 *
 * case 2: citations[mid] > len-mid, then it means there are citations[mid] papers that
 * have moret than citations[mid] citations, so we should continue searching in the left half
 *
 * case 3: citations[mid] < len-mid, we should continue searching in the right side
 * After iteration, it is guaranteed that right+1 is the one we need to find (i.e. len-(right+1)
 * papars have at least len-(righ+1) citations)
 */

/**
 * @param {number[]} citations
 * @return {number}
 */
const hIndex = citations => {
  const n = citations.length;
  let lo = 0;
  let hi = n - 1;

  while (lo <= hi) {
    const mid = lo + Math.floor((hi - lo) / 2);

    if (citations[mid] >= n - mid) {
      hi = mid - 1;
    } else {
      lo = mid + 1;
    }
  }

  return n - lo;
};

export { hIndex };