O(n) Easy Understand Solution

• ``````class Solution {
public:
int hIndex(vector<int>& citations) {
int n = citations.size();
vector<int> tbl = vector<int>(n+1, 0);
for (int i = 0; i < n; i++) {
int tmp = citations[i];
if (tmp >= n) tbl[n]++;
else tbl[tmp]++;
}
int sum = 0;
for (int i = n; i >= 0; i--) {
sum += tbl[i];
if (sum >= i) return i;
}
return -1;
}
};``````

• Hi,

I came across a situation: given citations = [3, 1, 7, 8, 9]. What will be the answer be?

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."

When h=3, the definition will be "A scientist has index 3 if 3 of his/her 5 papers have at least 3 citations each, and the other 2 papers have no more than 3 citations each." Obviously wrong....

When h=4, the definition will be "A scientist has index 4 if 4 of his/her 5 papers have at least 4 citations each, and the other 1 papers have no more than 4 citations each." Obviously wrong....

When h=2, the definition will be "A scientist has index 2 if 2 of his/her 5 papers have at least 2 citations each, and the other 3 papers have no more than 3 citations each." Obviously wrong....

• Hello,

h = 3 is correct. At first, I was as confused as you were.

In fact, that is trick. 7, 8, 9 are at least 3, whereas 1, 3 are no more than 3.

Looks like your connection to LeetCode Discuss was lost, please wait while we try to reconnect.