LeetCode-1283. Find the Smallest Divisor Given a Threshold

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

题解：

标准二分搜索。

class Solution {
public:
    int calDiv(vector<int>& nums, int div) {
        int sum = 0;
        for (int i = 0; i < nums.size(); i++) {
            int num = 0;
            if (nums[i] % div == 0) {
                num = nums[i] / div;
            }
            else {
                num = nums[i] / div + 1;
            }
            sum += num;
        }
        return sum;
    }
    int smallestDivisor(vector<int>& nums, int threshold) {
        int n = nums.size();
        if (n == 0) {
            return 0;
        }
        int left = 1, right = threshold;
        for (int i = 0; i < n; i++) {
            right = max(right, nums[i]);
        }
        while (left < right) {
            int mid = (left + right) / 2;
            int sum = calDiv(nums, mid);
            if (sum <= threshold) {
                right = mid;
            }
            else {
                left = mid + 1;
            }
        }
        return left;
    }
};

最后

以上就是愤怒百合最近收集整理的关于LeetCode-1283. Find the Smallest Divisor Given a Threshold的全部内容，更多相关LeetCode-1283.内容请搜索靠谱客的其他文章。