Introduction

In this blog post, we will explore how to solve the LeetCode problem "3115" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

You are given an integer array nums. Return an integer that is the maximum distance between the indices of two (not necessarily different) prime numbers in nums. Example 1: Input: nums = [4,2,9,5,3] Output: 3 Explanation: nums[1], nums[3], and nums[4] are prime. So the answer is |4 - 1| = 3. Example 2: Input: nums = [4,8,2,8] Output: 0 Explanation: nums[2] is prime. Because there is just one prime number, the answer is |2 - 2| = 0. Constraints: 1 <= nums.length <= 3 * 105 1 <= nums[i] <= 100 The input is generated such that the number of prime numbers in the nums is at least one.

Explanation

Here's the approach, complexity, and code:

• Identify Primes: Create a boolean array to efficiently mark prime numbers within the given range (1 to 100). This is done using the Sieve of Eratosthenes.
• Find Prime Indices: Iterate through the input array nums, and store the indices where prime numbers occur.

• Calculate Maximum Distance: If there are multiple prime indices, calculate the difference between the largest and smallest index to get the maximum distance. Handle the edge case when there's only one prime index by returning 0.

• Time Complexity: O(n), where n is the length of nums. Sieve of Eratosthenes takes O(1) since the maximum value is 100. Iterating through nums takes O(n).

• Space Complexity: O(1). The isPrime array has a fixed size (101), and we use a list to store the prime indices, which in the worst case will be O(n), where n is the length of nums. However, since there will only be prime numbers between 1 to 100 in nums, and the number of prime numbers between 1 to 100 is fixed, the complexity can also be considered as O(1).

Code

    def max_distance_between_primes(nums):
    """
    Calculates the maximum distance between the indices of two prime numbers in nums.

    Args:
        nums: A list of integers.

    Returns:
        The maximum distance between the indices of two prime numbers.
    """

    # Sieve of Eratosthenes to find primes up to 100
    isPrime = [True] * 101
    isPrime[0] = isPrime[1] = False
    for i in range(2, int(100**0.5) + 1):
        if isPrime[i]:
            for j in range(i*i, 101, i):
                isPrime[j] = False

    prime_indices = []
    for i, num in enumerate(nums):
        if isPrime[num]:
            prime_indices.append(i)

    if len(prime_indices) < 2:
        return 0
    else:
        return prime_indices[-1] - prime_indices[0]