winningPlayerCount: easy O(m+n) solution. BEAT 100%. ✅
IntuitionThe goal is to determine the number of players who have won based on their color picks. Each player can pick a color multiple times. A player is consid
# Intuition
The goal is to determine the number of players who have won based on their color picks. Each player can pick a color multiple times. A player is considered a winner if they have picked any color more times than their own ID.

# Approach
1. Initialize a counter `winning` to zero to keep track of the number of winners.
2. Create a hashmap to store the count of each color picked by each player. The keys are player IDs, and the values are arrays of size 11 initialized to zero.
3. Iterate through each pick and update the corresponding color count for each player in the hashmap.
4. Iterate through each player and check if they have picked any color more times than their own ID. If so, increment the `winning` counter.
5. Return the `winning` counter.

# Complexity
- Time complexity:
O(m+n)

- Space complexity:
O(n)

# Code
```python3 []
class Solution:
    def winningPlayerCount(self, n: int, pick: List[List[int]]) -> int:
        winning=0
        hashmap={Id:[0]*11 for Id in range(n)}
        for Id,color in pick:
            hashmap[Id][color]+=1
        for Id in range(n):
            if(max(hashmap[Id])>Id): winning+=1
        return winning
```
['Array', 'Hash Table', 'Counting', 'Python3']