1283. Find the Smallest Divisor Given a Threshold #

Problem #

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

Problem Summary #

Given an integer array nums and a positive integer threshold, you need to choose a positive integer as the divisor, then divide each number in the array by it and sum the division results. Find the smallest divisor among those that can make the above result less than or equal to the threshold threshold. Each number is rounded up after division by the divisor; for example, 7/3 = 3, 10/2 = 5. The problem guarantees that there will be an answer.

Notes:

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

Solution Ideas #

• Given an array and a threshold, find a divisor such that the sum of the quotients of each number in the array divided by this divisor does not exceed the threshold. Find the minimum value of the divisor.
• This problem is a typical binary search problem. According to the problem statement, search for the divisor in the interval [1, 1000000]. For each mid, compute the accumulated sum of the quotients once. If the sum is smaller than threshold, it means the divisor is too large, so shrink the right interval; if the sum is larger than threshold, it means the divisor is too small, so shrink the left interval. The final low value found is the desired smallest divisor.

Code #

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
}