How do you solve Maximize Count of Distinct Primes After Split on LeetCode?

Maximize Count of Distinct Primes After Split (LeetCode #3569) 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: Preprocessing primes allows for efficient checks.

What is the time complexity of Maximize Count of Distinct Primes After Split?

The optimal solution for Maximize Count of Distinct Primes After Split runs in O(n) time and uses O(n) space. Preprocessing primes takes O(n log log n), and each query is handled in O(n) due to prefix/suffix counts.

What is the brute force solution for Maximize Count of Distinct Primes After Split?

The brute force approach for Maximize Count of Distinct Primes After Split is Brute Force, with O(n²) time complexity and O(1) space complexity. For each query, we check every possible split point and count distinct primes in both parts. This is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Maximize Count of Distinct Primes After Split?

Maximize Count of Distinct Primes After Split uses the following concepts: Array, Math, Segment Tree, Number Theory. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Maximize Count of Distinct Primes After Split?

Explain your thought process clearly. Discuss trade-offs between brute force and optimal solutions. Practice similar problems to build confidence.

What are common mistakes when solving Maximize Count of Distinct Primes After Split?

Forgetting to update counts after each query. Not handling edge cases with small arrays.

#3569

Maximize Count of Distinct Primes After Split

Hard

ArrayMathSegment TreeNumber 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 sieve to preprocess primes and maintain counts of distinct primes efficiently. This allows quick updates and calculations for each query.

⚙️

Algorithm

3 steps

1Step 1: Preprocess all primes up to max(nums) using the Sieve of Eratosthenes.
2Step 2: Maintain a count of distinct primes in prefix and suffix using two arrays.
3Step 3: For each query, update the number and recalculate distinct counts, then find the maximum distinct primes across all splits.

solution.py17 lines

1def maxDistinctPrimes(nums, queries):
2    from sympy import isprime
3    max_num = max(nums)
4    primes = [False] * (max_num + 1)
5    for i in range(2, max_num + 1):
6        if isprime(i): primes[i] = True
7    prefix_count = [0] * len(nums)
8    suffix_count = [0] * len(nums)
9    for i in range(len(nums)):
10        prefix_count[i] = prefix_count[i-1] + (1 if primes[nums[i]] and (i == 0 or nums[i] != nums[i-1]) else 0)
11    for i in range(len(nums)-1, -1, -1):
12        suffix_count[i] = suffix_count[i+1] + (1 if primes[nums[i]] and (i == len(nums)-1 or nums[i] != nums[i+1]) else 0)
13    results = []
14    for idx, val in queries:
15        nums[idx] = val
16        results.append(max(prefix_count[k-1] + suffix_count[k] for k in range(1, len(nums))))
17    return results

ℹ

Complexity note: Preprocessing primes takes O(n log log n), and each query is handled in O(n) due to prefix/suffix counts.

1Preprocessing primes allows for efficient checks.
2Maintaining prefix and suffix counts optimizes distinct prime calculations.

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