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.

Difficulty: introductory

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

from collections import Counter
def most_frequent_item_count(collection):
    return Counter(collection).most_common()[0][1] 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
#---end function getNumAndFreq


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 maxNum, maxFreq     #if you want num with max freq
    return maxFreq

#-----end fucntion


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