Maximum Element After Decreasing and Rearranging LeetCode solution in Python

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class Solution:
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    def maximumElementAfterDecrementingAndRearranging(self, arr: List[int]) -> int:
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        counter = collections.Counter(arr)
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        available = sum(n > len(arr) for n in arr)
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        i = ans = len(arr)
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        while i > 0:
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            # This number is not in arr
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            if not counter[i]:
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                # Use another number to fill in its place. If we cannot, we have to decrease our max
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                if available:
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                    available -= 1
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                else:
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                    ans -= 1
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                # Other occurences can be used for future.
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            else:
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                available += counter[i] - 1
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            i -= 1
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        return ans