**Physical Address**

304 North Cardinal St.

Dorchester Center, MA 02124

Given an array of integers `citations`

where `citations[i]`

is the number of citations a researcher received for their `i`

paper, return compute the researcher’s ^{th}`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**.

**Example 1:**

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

**Constraints:**

`n == citations.length`

`1 <= n <= 5000`

`0 <= citations[i] <= 1000`

```
public int hIndex(int[] citations) {
int n = citations.length;
int[] buckets = new int[n+1];
for(int c : citations) {
if(c >= n) {
buckets[n]++;
} else {
buckets[c]++;
}
}
int count = 0;
for(int i = n; i >= 0; i--) {
count += buckets[i];
if(count >= i) {
return i;
}
}
return 0;
}
```

```
return sum(i < j for i, j in enumerate(sorted(citations, reverse=True)))
```

```
class Solution {
public:
int hIndex(vector<int>& c) {
// 3 support variables for us
int s = 0, e = c.size() - 1, avg;
// the base of every happy binary search ever: having a sorted dataset
sort(begin(c), end(c));
// some good old binary search here to find the maximum element meeting the conditions
while (s <= e) {
if (c[avg = (e + s) / 2] < c.size() - avg) s = avg + 1;
else e = avg - 1;
}
return c.size() - s;
}
};
```

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