Unique number of occurrences

Time: O(N); Space: O(N); easy

Given an array of integers nums, write a function that returns true if and only if the number of occurrences of each value in the array is unique.

Example 1:

Input: nums = [1,2,2,1,1,3]

Output: True

Explanation:

• The value 1 has 3 occurrences, 2 has 2 and 3 has 1.

• No two values have the same number of occurrences.

Example 2:

Input: nums = [1,2]

Output: False

Example 3:

Input: nums = [-3,0,1,-3,1,1,1,-3,10,0]

Output: True

Constraints:

• 1 <= len(nums) <= 1000

• -1000 <= arr[i] <= 1000

import collections

class Solution1(object):
    """
    Time: O(N)
    Space: O(N)
    """
    def uniqueOccurrences(self, nums):
        """
        :type nums: List[int]
        :rtype: bool
        """
        count = collections.Counter(nums)

        lookup = set()
        for v in count.values():
            if v in lookup:
                return False
            lookup.add(v)
        return True

s = Solution1()

nums =  [1,2,2,1,1,3]
assert s.uniqueOccurrences(nums) == True

nums = [1,2]
assert s.uniqueOccurrences(nums) == False

nums = [-3,0,1,-3,1,1,1,-3,10,0]
assert s.uniqueOccurrences(nums) == True

import collections

class Solution2(object):
    def uniqueOccurrences(self, nums):
        """
        :type arr: List[int]
        :rtype: bool
        """
        count = collections.Counter(nums)

        return len(count) == len(set(count.values()))

s = Solution2()
nums =  [1,2,2,1,1,3]
assert s.uniqueOccurrences(nums) == True
nums = [1,2]
assert s.uniqueOccurrences(nums) == False
nums = [-3,0,1,-3,1,1,1,-3,10,0]
assert s.uniqueOccurrences(nums) == True

See also:

https://leetcode.com/problems/unique-number-of-occurrences