Class Solution


  • public class Solution
    extends Object
    1283 - Find the Smallest Divisor Given a Threshold\. Medium Given an array of integers `nums` and an integer `threshold`, we will choose a positive integer `divisor`, divide all the array by it, and sum the division's result. Find the **smallest** `divisor` such that the result mentioned above is less than or equal to `threshold`. Each result of the division is rounded to the nearest integer greater than or equal to that element. (For example: `7/3 = 3` and `10/2 = 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
    • Constructor Detail

      • Solution

        public Solution()
    • Method Detail

      • smallestDivisor

        public int smallestDivisor​(int[] nums,
                                   int threshold)