1207. Unique Number of Occurrences

Given an array of integers arr, return true if the number of occurrences of each value in the array is unique or false otherwise.

Understand the problem

Two layers. First, count the frequency of each value (dict/Counter). Second, check whether those occurrence counts themselves have any duplicates.

A common duplicate check: compare len(list) with len(set(list)). If equal, nothing was duplicated (a set shrinks when duplicates exist).

Approach

1. Count frequencies with Counter(arr) → key = value, value = occurrences
2. Pull only the occurrence counts via .values()
3. Compare len of those counts with len of their set
4. Equal → no duplicate occurrences → return True; else False

Walkthrough — arr = [1, 2, 2, 1, 1, 3]

1. Counter(arr) → {1: 3, 2: 2, 3: 1}
2. .values() → occurrence counts [3, 2, 1]
3. set([3, 2, 1]) = {1, 2, 3} same length → no duplicates → True

Contrast arr = [1, 2]: Counter → {1: 1, 2: 1} · values = [1, 1] · set shrinks to {1} → False

Try it yourself first

Approach and walkthrough are above — write it yourself, then open the fold below when stuck or ready to compare.

from collections import Counter

def unique_occurrences(arr):
    counts = Counter(arr).values()   # occurrence counts, e.g. [3, 2, 1]
    # if putting values into a set keeps the same length, there were no duplicates
    return len(counts) == len(set(counts))

print(unique_occurrences([1, 2, 2, 1, 1, 3]))  # True
print(unique_occurrences([1, 2]))              # False
print(unique_occurrences([3, 5, 7, 7, 5, 5]))  # True

True
False
True