Complete the function to find the count of the most frequent item of an array. You can assume that input is an array of integers. For an empty array return `0`

## Example
```python
input array: [3, -1, -1, -1, 2, 3, -1, 3, -1, 2, 4, 9, 3]
ouptut: 5 
```
The most frequent number in the array is `-1` and it occurs `5` times.

Source: codewars | Difficulty: EASY
Tags: Fundamentals

Data structures

Starter code: def most_frequent_item_count(collection):
	

Solutions:
def most_frequent_item_count(collection):
	if collection:
		return max([collection.count(item) for item in collection])
	return 0


from collections import Counter

def most_frequent_item_count(collection):
	return Counter(collection).most_common(1)[0][1] if collection else 0


def most_frequent_item_count(collection):
	return max([collection.count(i) for i in collection] + [0])


from collections import Counter

def most_frequent_item_count(collection):
	return max(Counter(collection).values(), default=0)


def most_frequent_item_count(collection):
	return bool(collection) and max((collection.count(item) for item in set(collection)))


def most_frequent_item_count(collection):
	return max((collection.count(i) for i in collection)) if collection else 0


def getNumAndFreq(myList):
	curNum = myList.pop(0)
	freqCurNum = 1
	while myList and myList[0] == curNum:
		myList.pop(0)
		freqCurNum += 1
	return (curNum, freqCurNum)

def most_frequent_item_count(myList):
	myList.sort()
	maxNum = 0
	maxFreq = 0
	while myList:
		(curNum, curFreq) = getNumAndFreq(myList)
		if maxFreq < curFreq:
			maxNum = curNum
			maxFreq = curFreq
	return maxFreq


from collections import Counter
most_frequent_item_count = lambda c: Counter(c).most_common(1)[0][1] if c else 0