How do you solve Count Prime-Gap Balanced Subarrays on LeetCode?

Count Prime-Gap Balanced Subarrays (LeetCode #3589) 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: Understanding prime numbers is crucial.

What is the time complexity of Count Prime-Gap Balanced Subarrays?

The optimal solution for Count Prime-Gap Balanced Subarrays runs in O(n) time and uses O(n) space. The sliding window allows us to efficiently count subarrays in linear time.

What is the brute force solution for Count Prime-Gap Balanced Subarrays?

The brute force approach for Count Prime-Gap Balanced Subarrays is Brute Force, with O(n²) time complexity and O(1) space complexity. Check every possible subarray to see if it meets the prime-gap balanced criteria. This is straightforward but inefficient due to repeated calculations. The optimal Optimal Solution approach improves this to O(n).

What are the interview tips for Count Prime-Gap Balanced Subarrays?

Explain your thought process clearly. Discuss edge cases like empty arrays. Practice similar problems to build confidence.

What are common mistakes when solving Count Prime-Gap Balanced Subarrays?

Not identifying all primes correctly. Mismanaging the sliding window boundaries.

#3589

Count Prime-Gap Balanced Subarrays

Medium

ArrayMathQueueSliding WindowNumber TheoryMonotonic QueueHash 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 sliding window to efficiently track prime numbers and their max/min values, reducing the need for repeated calculations.

⚙️

Algorithm

3 steps

1Step 1: Generate a list of primes in the array using the Sieve of Eratosthenes.
2Step 2: Use a sliding window to maintain the current subarray of primes and their max/min values.
3Step 3: Count valid subarrays where the difference between max and min is <= k.

solution.py20 lines

1def count_prime_gap_balanced(nums, k):
2    def sieve(n):
3        is_prime = [True] * (n + 1)
4        for i in range(2, int(n**0.5) + 1):
5            if is_prime[i]:
6                for j in range(i * i, n + 1, i):
7                    is_prime[j] = False
8        return [i for i in range(2, n + 1) if is_prime[i]]
9    primes = sieve(max(nums))
10    prime_set = set(primes)
11    count = 0
12    left = 0
13    prime_window = []
14    for right in range(len(nums)):
15        if nums[right] in prime_set:
16            prime_window.append(nums[right])
17            while len(prime_window) > 1 and max(prime_window) - min(prime_window) > k:
18                prime_window.pop(0)
19            count += len(prime_window) - 1
20    return count

ℹ

Complexity note: The sliding window allows us to efficiently count subarrays in linear time.

1Understanding prime numbers is crucial.
2Sliding window technique optimizes subarray checks.

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