Class Solution
- java.lang.Object
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- g1201_1300.s1283_find_the_smallest_divisor_given_a_threshold.Solution
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public class Solution extends Object
1283 - Find the Smallest Divisor Given a Threshold.Medium
Given an array of integers
nums
and an integerthreshold
, we will choose a positive integerdivisor
, divide all the array by it, and sum the division’s result. Find the smallestdivisor
such that the result mentioned above is less than or equal tothreshold
.Each result of the division is rounded to the nearest integer greater than or equal to that element. (For example:
7/3 = 3
and10/2 = 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
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Constructor Summary
Constructors Constructor Description Solution()
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Method Summary
All Methods Instance Methods Concrete Methods Modifier and Type Method Description int
smallestDivisor(int[] nums, int threshold)
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