- Difficulty: Medium
- Tags: LeetCode, Medium, Array, Binary Search, leetcode-1283, O(logn), O(1)
Problem
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