How do you solve Check if Any Element Has Prime Frequency on LeetCode?

Check if Any Element Has Prime Frequency (LeetCode #3591) 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: Prime numbers are only greater than 1 and have no divisors other than 1 and themselves.

What is the time complexity of Check if Any Element Has Prime Frequency?

The optimal solution for Check if Any Element Has Prime Frequency runs in O(n) time and uses O(n) space. Counting frequencies is O(n) and checking primes for each frequency is efficient due to direct access in the hash map.

What is the brute force solution for Check if Any Element Has Prime Frequency?

The brute force approach for Check if Any Element Has Prime Frequency is Brute Force, with O(n²) time complexity and O(1) space complexity. Check the frequency of each element by counting occurrences, then check if any frequency is prime. This is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Check if Any Element Has Prime Frequency?

Check if Any Element Has Prime Frequency uses the following concepts: Array, Hash Table, Math, Counting, Number Theory. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Check if Any Element Has Prime Frequency?

Explain your thought process clearly. Discuss edge cases and constraints. Practice coding on a whiteboard or shared document.

What are common mistakes when solving Check if Any Element Has Prime Frequency?

Not checking for edge cases like empty arrays. Confusing prime checking logic.

#3591

Check if Any Element Has Prime Frequency

Easy

ArrayHash TableMathCountingNumber TheoryHash 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)

Use a hash map to count frequencies efficiently, then check for primes in a single pass. This reduces unnecessary computations.

⚙️

Algorithm

3 steps

1Step 1: Count the frequency of each element using a hash map.
2Step 2: Check if any frequency is prime using a helper function.
3Step 3: Return true if any frequency is prime, otherwise return false.

solution.py13 lines

1def is_prime(n):
2    if n <= 1:
3        return False
4    for i in range(2, int(n**0.5) + 1):
5        if n % i == 0:
6            return False
7    return True
8
9def prime_frequency(nums):
10    freq = {}
11    for num in nums:
12        freq[num] = freq.get(num, 0) + 1
13    return any(is_prime(count) for count in freq.values())

ℹ

Complexity note: Counting frequencies is O(n) and checking primes for each frequency is efficient due to direct access in the hash map.

1Prime numbers are only greater than 1 and have no divisors other than 1 and themselves.
2Using a hash map allows for efficient frequency counting.

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