How do you solve Unique Number of Occurrences on LeetCode?

Unique Number of Occurrences (LeetCode #1207) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Using a hash map can significantly reduce the time complexity.

What is the brute force solution for Unique Number of Occurrences?

The brute force approach for Unique Number of Occurrences is Brute Force, with O(n²) time complexity and O(1) space complexity. In this approach, we will count the occurrences of each number and then check if any counts are repeated. This is straightforward but inefficient because we will be repeatedly counting occurrences. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Unique Number of Occurrences?

Unique Number of Occurrences uses the following concepts: Array, Hash Table. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Unique Number of Occurrences?

Always clarify the problem and constraints before jumping to code. Explain your thought process and approach before writing code. Test your solution with edge cases to ensure correctness.

What are common mistakes when solving Unique Number of Occurrences?

Not using a hash map and counting occurrences repeatedly. Forgetting to check if a count already exists in the set.

#1207

Unique Number of Occurrences

Easy

ArrayHash TableHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

Using a hash map allows us to efficiently count occurrences in a single pass, and then we can check for uniqueness using a set, which is much faster.

⚙️

Algorithm

3 steps

1Step 1: Use a hash map to count occurrences of each number in the array.
2Step 2: Use a set to track the counts and check for duplicates.
3Step 3: If a count already exists in the set, return false; otherwise, add it to the set.

solution.py9 lines

1def uniqueOccurrences(arr):
2    from collections import Counter
3    count_map = Counter(arr)
4    counts = set()
5    for count in count_map.values():
6        if count in counts:
7            return False
8        counts.add(count)
9    return True

ℹ

Complexity note: The time complexity is O(n) because we traverse the array once to count occurrences and then traverse the hash map to check for uniqueness. The space complexity is O(n) due to the hash map and set used to store counts.

1Using a hash map can significantly reduce the time complexity.
2Checking for uniqueness can be efficiently done using a set.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.