# Solving Leetcode Interviews in Seconds with AI: Find the Smallest Divisor Given a Threshold


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "1283" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like [Chatmagic](https://www.chatmagic.app), 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 of integers nums and an integer threshold, we will choose a positive integer divisor, divide all the array by it, and sum the division's result. Find the smallest divisor such that the result mentioned above is less than or equal to threshold. Each result of the division is rounded to the nearest integer greater than or equal to that element. (For example: 7/3 = 3 and 10/2 = 5). The test cases are generated so that there will be an answer.   Example 1:  Input: nums = [1,2,5,9], threshold = 6 Output: 5 Explanation: We can get a sum to 17 (1+2+5+9) if the divisor is 1.  If the divisor is 4 we can get a sum of 7 (1+1+2+3) and if the divisor is 5 the sum will be 5 (1+1+1+2).   Example 2:  Input: nums = [44,22,33,11,1], threshold = 5 Output: 44    Constraints:  1 <= nums.length <= 5 * 104 1 <= nums[i] <= 106 nums.length <= threshold <= 106  

	# Explanation
	Here's the solution approach, complexity analysis, and the Python code:

*   **Binary Search:** The core idea is to use binary search over the possible range of divisors (from 1 to the maximum element in `nums`).
*   **Calculate Sum:** For each divisor considered in the binary search, calculate the sum of ceiling divisions (using `math.ceil`).
*   **Adjust Search Range:** Based on whether the calculated sum is greater or less than/equal to the `threshold`, adjust the binary search range accordingly.

*   **Runtime Complexity:** O(N log M), where N is the length of `nums` and M is the maximum value in `nums`.
*   **Storage Complexity:** O(1)

	
	# Code
	```python
	import math

def smallestDivisor(nums, threshold):
    """
    Finds the smallest divisor such that the sum of ceiling divisions is less than or equal to threshold.

    Args:
        nums: A list of integers.
        threshold: An integer representing the threshold.

    Returns:
        The smallest divisor.
    """

    left = 1
    right = max(nums)
    ans = right  # Initialize with the maximum possible divisor

    while left <= right:
        mid = (left + right) // 2
        sum_val = 0
        for num in nums:
            sum_val += math.ceil(num / mid)

        if sum_val <= threshold:
            ans = mid
            right = mid - 1  # Try to find a smaller divisor
        else:
            left = mid + 1  # Need a larger divisor

    return ans
	```