1283. Find the Smallest Divisor Given a Threshold
https://leetcode.com/problems/find-the-smallest-divisor-given-a-threshold/
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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
Intuition
Sort the array, lower divisor = 1, highest divisor = last element in array
binary search on midpoint
If Valid => save and search for better ans => lower range => hi = mid - 1
else lo = mid + 1
Valid => sum of Math.ceil(quotient) <= threshold
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Time - O(N log N)
Space - O(1)
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| class Solution { | |
| public int smallestDivisor(int[] nums, int threshold) { | |
| Arrays.sort(nums); | |
| // Valid divisors between 1, and max elem | |
| int lo = 1, hi = nums[nums.length - 1]; | |
| int ans = 1; | |
| // binary search | |
| while (lo <= hi) { | |
| int mid = lo + (hi - lo) / 2; | |
| if (isValid(nums, mid, threshold)) { | |
| // save valid and look for smaller divisor | |
| ans = mid; | |
| hi = mid - 1; | |
| } else { | |
| // sum is greater => reduce quotient sum => explore bigger divisor | |
| lo = mid + 1; | |
| } | |
| } | |
| return ans; | |
| } | |
| private boolean isValid(int[] nums, int div, int threshold) { | |
| int ans = 0; | |
| for (int num: nums) { | |
| ans += (int) Math.ceil(num * 1.0 / div); | |
| } | |
| return ans <= threshold; | |
| } | |
| } |
