Problem

Solution

Intuition

This problem requires us to return True if the number of occurrences of each item in a given list is unique. For example, if an item shows up exactly twice in the list, no other item can have exactly two occurrences as well.

To solve this, we simply can count the number of occurrences of each number and store this in memory, and then see if there are any repeated numbers in the occurrences.

Approach

1. Initialize our dictionary occurrences which we will use to store the number of occurrences of each integer in arr.
2. Iterate through arr to count the total number of occurences of each integer.
3. Return True if the total number of occurrences is equal to the unique number of occurrences found by comparing the sizes of a list and a set made using the occurrence numbers.

Complexity

• Time complexity: $O(n)$. We iterate through each number in arr once, we then iterate through the number of occurrences; however, this is dominated by the length of the array.
• Space complexity: $O(n)$. Our dictionary uses a linear space complexity and dominates the stack space of the list and set.

Code

class Solution:
    def uniqueOccurrences(self, arr: List[int]) -> bool:
        occurrences = {}

        for num in arr:
            occurrences[num] = occurrences.get(num, 0) + 1
        
        return len(list(occurrences.values())) == len(set(occurrences.values()))