How do you solve Closest Prime Numbers in Range on LeetCode?

Closest Prime Numbers in Range (LeetCode #2523) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log log n) time complexity and O(n) space complexity. Key insight: Using efficient algorithms like the Sieve of Eratosthenes can drastically reduce computation time.

What is the brute force solution for Closest Prime Numbers in Range?

The brute force approach for Closest Prime Numbers in Range is Brute Force, with O(n²) time complexity and O(1) space complexity. We can find all prime numbers in the given range by checking each number individually. This approach is straightforward but inefficient, as we will check each number for primality multiple times. The optimal Optimal Solution approach improves this to O(n log log n).

What algorithm or data structure is used to solve Closest Prime Numbers in Range?

Closest Prime Numbers in Range uses the following concepts: Math, Number Theory. The recommended patterns to study are: Sieve of Eratosthenes, Two Pointers.

What are the interview tips for Closest Prime Numbers in Range?

Explain your thought process clearly and ask clarifying questions about the problem. Discuss potential edge cases before jumping into coding. Optimize your solution step-by-step, explaining why each improvement is necessary.

What are common mistakes when solving Closest Prime Numbers in Range?

Not considering edge cases where there are fewer than two primes in the range. Overlooking the need to return the smallest num1 when multiple pairs have the same difference.

#2523

Closest Prime Numbers in Range

Medium

MathNumber TheorySieve of EratosthenesTwo Pointers

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n log log n)Space O(n)

Using the Sieve of Eratosthenes allows us to efficiently find all prime numbers up to 'right'. Then, we can simply iterate through the list of primes to find the closest pair.

⚙️

Algorithm

3 steps

1Step 1: Use the Sieve of Eratosthenes to find all prime numbers up to 'right'.
2Step 2: Store the prime numbers in a list.
3Step 3: Iterate through the list of primes to find the pair with the minimum difference.

solution.py25 lines

1# Full working Python code
2
3def sieve_of_eratosthenes(n):
4    is_prime = [True] * (n + 1)
5    is_prime[0], is_prime[1] = False, False
6    for i in range(2, int(n**0.5) + 1):
7        if is_prime[i]:
8            for j in range(i * i, n + 1, i):
9                is_prime[j] = False
10    return [i for i in range(n + 1) if is_prime[i]]
11
12def closest_primes(left, right):
13    primes = sieve_of_eratosthenes(right)
14    filtered_primes = [p for p in primes if p >= left]
15    min_diff = float('inf')
16    ans = [-1, -1]
17    for i in range(len(filtered_primes) - 1):
18        diff = filtered_primes[i + 1] - filtered_primes[i]
19        if diff < min_diff:
20            min_diff = diff
21            ans = [filtered_primes[i], filtered_primes[i + 1]]
22    return ans
23
24# Example usage
25print(closest_primes(10, 19))

ℹ

Complexity note: The Sieve of Eratosthenes runs in O(n log log n) time, which is efficient for finding all primes up to 'right'. The space complexity is O(n) due to the storage of the prime boolean array.

1Using efficient algorithms like the Sieve of Eratosthenes can drastically reduce computation time.
2Filtering primes after generating them allows for easy pair comparison.

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