1283. Find the Smallest Divisor Given a Threshold

题目

Given an array of integers nums and an integer threshold, we will choose a positive integer divisor and divide all the array by it and sum the result of the division. Find the smallest divisor such that the result mentioned above is less than or equal to threshold.

Each result of division is rounded to the nearest integer greater than or equal to that element. (For example: 7/3 = 3 and 10/2 = 5).

It is guaranteed 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 to 7 (1+1+2+3) and if the divisor is 5 the sum will be 5 (1+1+1+2).

Example 2:

Input: nums = [2,3,5,7,11], threshold = 11
Output: 3

Example 3:

Input: nums = [19], threshold = 5
Output: 4

Constraints:

• 1 <= nums.length <= 5 * 10^4
• 1 <= nums[i] <= 10^6
• nums.length <= threshold <= 10^6

题目大意

给你一个整数数组 nums 和一个正整数 threshold  ，你需要选择一个正整数作为除数，然后将数组里每个数都除以它，并对除法结果求和。请你找出能够使上述结果小于等于阈值 threshold 的除数中 最小 的那个。每个数除以除数后都向上取整，比方说 7/3 = 3 ， 10/2 = 5 。题目保证一定有解。

提示：

• 1 <= nums.length <= 5 * 10^4
• 1 <= nums[i] <= 10^6
• nums.length <= threshold <= 10^6

解题思路

• 给出一个数组和一个阈值，要求找到一个除数，使得数组里面每个数和这个除数的商之和不超过这个阈值。求除数的最小值。
• 这一题是典型的二分搜索的题目。根据题意，在 [1, 1000000] 区间内搜索除数，针对每次 mid，计算一次商的累加和。如果和比 threshold 小，说明除数太大，所以缩小右区间；如果和比 threshold 大，说明除数太小，所以缩小左区间。最终找到的 low 值就是最求的最小除数。

代码

func smallestDivisor(nums []int, threshold int) int {
	low, high := 1, 1000000
	for low < high {
		mid := low + (high-low)>>1
		if calDivisor(nums, mid, threshold) {
			high = mid
		} else {
			low = mid + 1
		}
	}
	return low
}

func calDivisor(nums []int, mid, threshold int) bool {
	sum := 0
	for i := range nums {
		if nums[i]%mid != 0 {
			sum += nums[i]/mid + 1
		} else {
			sum += nums[i] / mid
		}
	}
	if sum <= threshold {
		return true
	}
	return false
}