How do you solve Complete Prime Number on LeetCode?

Complete Prime Number (LeetCode #3765) 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: Primality check is crucial for both prefixes and suffixes.

What is the time complexity of Complete Prime Number?

The optimal solution for Complete Prime Number runs in O(n) time and uses O(n) space. We check each prefix and suffix efficiently, leading to linear checks for n digits.

What is the brute force solution for Complete Prime Number?

The brute force approach for Complete Prime Number is Brute Force, with O(n²) time complexity and O(1) space complexity. Check every prefix and suffix of the number to see if they are prime. This method is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Complete Prime Number?

Complete Prime Number uses the following concepts: Math, Enumeration, Number Theory. The recommended patterns to study are: Number Theory, String Manipulation.

What are the interview tips for Complete Prime Number?

Explain your thought process clearly. Discuss edge cases upfront. Optimize your solution after brute force.

What are common mistakes when solving Complete Prime Number?

Forgetting to check single-digit numbers. Not handling leading zeros in prefixes.

#3765

Complete Prime Number

Medium

MathEnumerationNumber TheoryNumber TheoryString Manipulation

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 single function to check primality and iterate through prefixes and suffixes efficiently.

⚙️

Algorithm

3 steps

1Step 1: Convert the number to a string for easy manipulation.
2Step 2: Use a helper function to check if each prefix and suffix is prime.
3Step 3: Return true only if all checks pass.

solution.py14 lines

1def is_prime(n):
2    if n < 2:
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 is_complete_prime(num):
10    s = str(num)
11    for i in range(len(s)):
12        if not is_prime(int(s[:i + 1])) or not is_prime(int(s[i:])):
13            return False
14    return True

ℹ

Complexity note: We check each prefix and suffix efficiently, leading to linear checks for n digits.

1Primality check is crucial for both prefixes and suffixes.
2Single-digit primes are special cases.

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