How do you solve Prime Pairs With Target Sum on LeetCode?

Prime Pairs With Target Sum (LeetCode #2761) 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 the Sieve of Eratosthenes allows for efficient prime checking.

What is the time complexity of Prime Pairs With Target Sum?

The optimal solution for Prime Pairs With Target Sum runs in O(n log log n) time and uses O(n) space. The Sieve of Eratosthenes runs in O(n log log n) time, and we use O(n) space to store the prime numbers.

What is the brute force solution for Prime Pairs With Target Sum?

The brute force approach for Prime Pairs With Target Sum is Brute Force, with O(n²) time complexity and O(1) space complexity. In this approach, we will check every possible pair of numbers (x, y) to see if they are both prime and if their sum equals n. This is straightforward but inefficient for larger values of n. The optimal Optimal Solution approach improves this to O(n log log n).

What algorithm or data structure is used to solve Prime Pairs With Target Sum?

Prime Pairs With Target Sum uses the following concepts: Array, Math, Enumeration, Number Theory. The recommended patterns to study are: Sieve of Eratosthenes, Two Pointers.

What are the interview tips for Prime Pairs With Target Sum?

Explain your thought process clearly when discussing prime numbers. Mention the Sieve of Eratosthenes as a potential optimization early on. Be cautious about edge cases, such as when n is very small.

What are common mistakes when solving Prime Pairs With Target Sum?

Not using a prime-checking method efficiently, leading to timeouts. Overlooking the condition x <= y, which can lead to duplicate pairs.

#2761

Prime Pairs With Target Sum

Medium

ArrayMathEnumerationNumber 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)

By using the Sieve of Eratosthenes, we can efficiently find all prime numbers up to n. Then, we can check pairs of primes that sum to n in linear time, significantly improving performance.

⚙️

Algorithm

3 steps

1Step 1: Use the Sieve of Eratosthenes to generate a list of all prime numbers up to n.
2Step 2: Store these primes in a set for O(1) lookup time.
3Step 3: Iterate through the list of primes and for each prime x, check if (n - x) is also prime. If so, add the pair [x, n - x] to the result.

solution.py18 lines

1# Full working Python code
2def sieve_of_eratosthenes(n):
3    is_prime = [True] * (n + 1)
4    is_prime[0], is_prime[1] = False, False
5    for i in range(2, int(n**0.5) + 1):
6        if is_prime[i]:
7            for j in range(i * i, n + 1, i):
8                is_prime[j] = False
9    return [i for i in range(n + 1) if is_prime[i]]
10
11def prime_pairs(n):
12    primes = sieve_of_eratosthenes(n)
13    prime_set = set(primes)
14    result = []
15    for x in primes:
16        if (n - x) in prime_set:
17            result.append([x, n - x])
18    return result

ℹ

Complexity note: The Sieve of Eratosthenes runs in O(n log log n) time, and we use O(n) space to store the prime numbers.

1Using the Sieve of Eratosthenes allows for efficient prime checking.
2Only check pairs where x <= n/2 to avoid duplicates.

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