Introduction

In this blog post, we will explore how to solve the LeetCode problem "2584" 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 a 0-indexed integer array nums of length n. A split at an index i where 0 <= i <= n - 2 is called valid if the product of the first i + 1 elements and the product of the remaining elements are coprime. For example, if nums = [2, 3, 3], then a split at the index i = 0 is valid because 2 and 9 are coprime, while a split at the index i = 1 is not valid because 6 and 3 are not coprime. A split at the index i = 2 is not valid because i == n - 1. Return the smallest index i at which the array can be split validly or -1 if there is no such split. Two values val1 and val2 are coprime if gcd(val1, val2) == 1 where gcd(val1, val2) is the greatest common divisor of val1 and val2. Example 1: Input: nums = [4,7,8,15,3,5] Output: 2 Explanation: The table above shows the values of the product of the first i + 1 elements, the remaining elements, and their gcd at each index i. The only valid split is at index 2. Example 2: Input: nums = [4,7,15,8,3,5] Output: -1 Explanation: The table above shows the values of the product of the first i + 1 elements, the remaining elements, and their gcd at each index i. There is no valid split. Constraints: n == nums.length 1 <= n <= 104 1 <= nums[i] <= 106

Explanation

Here's the breakdown of the approach, complexity, and the Python code:

• High-Level Approach:

  • Precompute prime factorization for each number in nums and store it.
  • Iterate through possible split indices. For each index, maintain the prime factors present in the left and right products using dictionaries.
  • Check if the left and right prime factor sets have any common prime factors (i.e., their GCD is 1). If so, return the index.

• Complexity:

  • Runtime: O(N * sqrt(M)), where N is the length of nums and M is the maximum value in nums.
  • Storage: O(M), predominantly due to storing prime factorizations and tracking prime factors on the left and right.

Code

    import math

def smallest_index(nums):
    n = len(nums)

    def prime_factorization(num):
        factors = {}
        d = 2
        while d * d <= num:
            while num % d == 0:
                factors[d] = factors.get(d, 0) + 1
                num //= d
            d += 1
        if num > 1:
            factors[num] = factors.get(num, 0) + 1
        return factors

    prime_factors = [prime_factorization(num) for num in nums]

    for i in range(n - 1):
        left_factors = {}
        for j in range(i + 1):
            for factor, count in prime_factors[j].items():
                left_factors[factor] = left_factors.get(factor, 0) + count

        right_factors = {}
        for j in range(i + 1, n):
            for factor, count in prime_factors[j].items():
                right_factors[factor] = right_factors.get(factor, 0) + count

        coprime = True
        for factor in left_factors:
            if factor in right_factors:
                coprime = False
                break

        if coprime:
            return i

    return -1