Introduction

In this blog post, we will explore how to solve the LeetCode problem "3326" 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. Any positive divisor of a natural number x that is strictly less than x is called a proper divisor of x. For example, 2 is a proper divisor of 4, while 6 is not a proper divisor of 6. You are allowed to perform an operation any number of times on nums, where in each operation you select any one element from nums and divide it by its greatest proper divisor. Return the minimum number of operations required to make the array non-decreasing. If it is not possible to make the array non-decreasing using any number of operations, return -1. Example 1: Input: nums = [25,7] Output: 1 Explanation: Using a single operation, 25 gets divided by 5 and nums becomes [5, 7]. Example 2: Input: nums = [7,7,6] Output: -1 Example 3: Input: nums = [1,1,1,1] Output: 0 Constraints: 1 <= nums.length <= 105 1 <= nums[i] <= 106

Explanation

Here's the breakdown of the solution:

• Greedy Approach: Process the array from right to left. If the current element is smaller than the previous element, repeatedly divide the current element by its greatest proper divisor until it's no longer smaller or it becomes 1.
• Greatest Proper Divisor Optimization: Precompute the greatest proper divisor for each number up to the maximum value in nums to avoid repeated calculations.

• Impossibility Check: If an element becomes 1 and is still smaller than the previous element, it's impossible to make the array non-decreasing, so return -1.

• Runtime Complexity: O(N + M), where N is the length of nums and M is the maximum value in nums. Storage Complexity: O(M).

Code

    def min_operations(nums):
    n = len(nums)
    max_val = max(nums)
    greatest_divisor = [0] * (max_val + 1)
    for i in range(2, max_val + 1):
        if greatest_divisor[i] == 0:
            for j in range(i, max_val + 1, i):
                greatest_divisor[j] = j // i

    operations = 0
    for i in range(n - 2, -1, -1):
        while nums[i] > nums[i + 1]:
            if nums[i] == 1:
                return -1
            nums[i] //= greatest_divisor[nums[i]]
            operations += 1

    return operations