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Unique Number of Occurrences

Difficulty: Easy

Problem Description

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

Key Insights

  • Count the occurrences of each element in the array.
  • Use a data structure to track the frequency of each occurrence count.
  • If any occurrence count is repeated, return false; otherwise, return true.

Space and Time Complexity

Time Complexity: O(n) - We traverse the array a couple of times (once to count occurrences and once to check counts). Space Complexity: O(n) - We use additional space for counting occurrences and storing them.

Solution

To solve the problem, we can use a hash table (or dictionary) to count the occurrences of each number in the input array. After counting, we will use another hash table to track the frequency of these counts. If any count appears more than once, we return false. If all counts are unique, we return true.

Code Solutions

def uniqueOccurrences(arr):
    from collections import Counter
    
    # Count the occurrences of each number in arr
    count = Counter(arr)
    
    # Get the frequencies of the counts
    occurrences = set(count.values())
    
    # Check if the number of unique occurrences equals the total number of counts
    return len(occurrences) == len(count)

# Example usage
print(uniqueOccurrences([1,2,2,1,1,3]))  # Output: True
print(uniqueOccurrences([1,2]))           # Output: False
print(uniqueOccurrences([-3,0,1,-3,1,1,1,-3,10,0]))  # Output: True
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