Find the Smallest Divisor Given a Threshold (M)
題目
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 * 10^41 <= nums[i] <= 10^6nums.length <= threshold <= 10^6
題意
給定一個整數陣列和一個閾值，要求找到一個最小的除數，使得陣列中每個整數除以這個除數的商之和小于等于閾值，
思路
因為值域固定，可以用二分法夾出答案，
代碼實作
Java
class Solution {
public int smallestDivisor(int[] nums, int threshold) {
int left = 1, right = 1000000;
while (left < right) {
int mid = (right - left) / 2 + left;
int sum = 0;
for (int num : nums) {
sum += Math.ceil(1.0 * num / mid);
}
if (sum <= threshold) {
right = mid;
} else {
left = mid + 1;
}
}
return right;
}
}
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