Introduction

In this blog post, we will explore how to solve the LeetCode problem "1250" 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

Given an array nums of positive integers. Your task is to select some subset of nums, multiply each element by an integer and add all these numbers. The array is said to be good if you can obtain a sum of 1 from the array by any possible subset and multiplicand. Return True if the array is good otherwise return False. Example 1: Input: nums = [12,5,7,23] Output: true Explanation: Pick numbers 5 and 7. 53 + 7(-2) = 1 Example 2: Input: nums = [29,6,10] Output: true Explanation: Pick numbers 29, 6 and 10. 291 + 6(-3) + 10*(-1) = 1 Example 3: Input: nums = [3,6] Output: false Constraints: 1 <= nums.length <= 10^5 1 <= nums[i] <= 10^9

Explanation

Here's the breakdown of the solution:

• Bézout's Identity: The core concept revolves around Bézout's identity. An array is "good" if and only if the greatest common divisor (GCD) of its elements is 1. This is because Bézout's identity states that for any two integers a and b, there exist integers x and y such that ax + by = gcd(a, b). This extends to multiple integers: you can obtain the GCD as a linear combination of the elements. If the GCD is 1, you can obtain 1.

• GCD Calculation: Efficiently calculate the GCD of all numbers in the array. We can iteratively compute the GCD using the Euclidean algorithm, starting with the first two elements and then incorporating the remaining elements one by one.

• Check if GCD is 1: After computing the GCD of the entire array, we simply check if the result is equal to 1. If it is, the array is good; otherwise, it's not.

• Time & Space Complexity: The time complexity is O(N * log(max(nums))), where N is the length of the array and log(max(nums)) is the maximum number of steps required by the Euclidean algorithm (since the arguments to gcd reduce significantly with each step). The space complexity is O(1).

Code

    def gcd(a, b):
    """
    Calculates the greatest common divisor of two integers using the Euclidean algorithm.
    """
    while(b):
        a, b = b, a % b
    return a

def is_good_array(nums):
    """
    Determines if an array is "good" based on whether the GCD of its elements is 1.
    """
    result = nums[0]
    for i in range(1, len(nums)):
        result = gcd(result, nums[i])
        if result == 1:
            return True  # Optimization: If GCD becomes 1, no need to continue

    return result == 1