Find H-Index

"""
Intuition for H-Index (Step-by-Step Breakdown)

Step 1: Identify the Problem Type
---------------------------------
The H-Index measures the impact of a researcher. An H-Index of 'h'
means at least 'h' papers have 'h' or more citations. While we could
solve this by sorting (O(n log n)), we can use a 'Counting Sort'
style approach to achieve O(n) time.

Step 2: Defining the Constraints (The 'Cap' Logic)
--------------------------------------------------
The maximum possible H-Index can be the N(lenght of the array of count of research papers) even for eg. [8,8,8,8] still H-index would be 4 only, maximum ie. all the research paper have more than 4 citations for 5 it won't fit.
- Therefore, any citation count greater than 'n' can be treated
  exactly the same as 'n'.

Step 3: Creating the Frequency Map (The Bucket)
-----------------------------------------------
1. Create a 'buckets' array of size n+1.
2. Iterate through the citations:
   - If citation count >= n: Increment buckets[n].
   - Otherwise: Increment buckets[citation_count].
This bucket array now tells us: "How many papers have exactly X citations?"

Step 4: The Cumulative Counting (Reverse Thinking)
--------------------------------------------------
To find 'h' papers with at least 'h' citations, it's easier to look
from the highest possible 'h' down to 0.
- Start a 'count' variable at 0.
- Iterate through the buckets array from index 'n' down to 0.

Step 5: The Decision Moment
---------------------------
As we iterate backwards from i = n down to 0:
1. Add the number of papers in the current bucket to our 'count'.
   (count += buckets[i])
2. The 'count' now represents: "How many papers have AT LEAST 'i'
   citations?"
3. If this 'count' >= 'i', then 'i' is our H-Index!
   Since we are moving from largest to smallest, the first 'i' that
   satisfies this is the MAXIMUM H-Index.

Step 6: Final Result
--------------------
Return 'i'. If no such 'i' is found (though it always will be for
non-negative citations), return 0.

Step 7: Complexity
------------------
- Time: O(n) - Two linear passes (one to fill buckets, one to find h).
- Space: O(n) - To store the buckets array.
"""

class Solution:
    def hIndex(self, citations: List[int]) -> int:
        arr=[0]*(len(citations)+1)
        n=len(citations)
        for i in citations:
            if i>=len(citations):
                arr[n]+=1
            else:
                arr[i]+=1
        print(arr)
        n=len(citations)
        prev=0
        while n>=0:
            if prev + arr[n]>=n:
                return n
            prev+=arr[n]
            n-=1
        return 0