The “Find the Smallest Divisor Given a Threshold” problem is a classic binary search question that helps you practice optimization under constraints. <\/p>\n\n\n\n
In this blog, we\u2019ll explain the problem, walk you through both brute force and optimal solutions, and make everything easy to understand with code comments, dry runs, and clear explanations.<\/p>\n\n\n\n
Here’s the [Problem Link<\/span><\/mark><\/a><\/strong>] to begin with.<\/p>\n\n\n\n Given an array of integers Each result of the division is rounded to the nearest integer greater than or equal to that element. (For example: The test cases are generated so that there will be an answer.<\/p>\n\n\n\n Example 1:<\/strong><\/p>\n\n\n\n Example 2:<\/strong><\/p>\n\n\n\n Constraints:<\/strong><\/p>\n\n\n\n Let\u2019s solve the problem in the most straightforward way:<\/p>\n\n\n\n This approach is easy to understand but not efficient for large numbers.<\/p>\n\n\n\n
\n\n\n\nnums<\/code> and an integer threshold<\/code>, we will choose a positive integer divisor<\/code>, divide all the array by it, and sum the division’s result. Find the smallest<\/strong> divisor<\/code> such that the result mentioned above is less than or equal to threshold<\/code>.<\/p>\n\n\n\n7\/3 = 3<\/code> and 10\/2 = 5<\/code>).<\/p>\n\n\n\nInput:<\/strong> nums = [1,2,5,9], threshold = 6
Output:<\/strong> 5
Explanation:<\/strong> 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). <\/pre>\n\n\n\nInput:<\/strong> nums = [44,22,33,11,1], threshold = 5
Output:<\/strong> 44<\/pre>\n\n\n\n\n
1 <= nums.length <= 5 * 104<\/sup><\/code><\/li>\n\n\n\n1 <= nums[i] <= 106<\/sup><\/code><\/li>\n\n\n\nnums.length <= threshold <= 106<\/sup><\/code><\/li>\n<\/ul>\n\n\n
\n\n\n\nBrute Force Solution (Try Every Possible Divisor)<\/h2>\n\n\n\n
Intuition and Approach<\/h3>\n\n\n\n
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Try every possible divisor from 1 up to the largest number in the array.<\/li>\n\n\n\n
For each divisor, calculate the total sum (rounded up for each element). If the sum is within the threshold, return the divisor.<\/li>\n\n\n\n
We check every possible divisor, so we\u2019re sure to find the smallest one that works.<\/li>\n<\/ol>\n\n\n\nCode Implementation<\/h3>\n\n\n\n