Problem Statement

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).

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 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

Example 3:

Input: nums = [21212,10101,12121], threshold = 1000000
Output: 1

Example 4:

Input: nums = [2,3,5,7,11], threshold = 11
Output: 3

Constraints:

• 1 <= nums.length <= 5 * 104
• 1 <= nums[i] <= 106
• nums.length <= threshold <= 106

Problem Link

Find the Smallest Divisor Given a Threshold - LeetCode

Code

/*
Explanation:
Binary search the result.
If the sum > threshold, the divisor is too small.
If the sum <= threshold, the divisor is big enough.

Complexity
Time O(NlogM), where M = max(A)
Space O(1)
*/
class Solution {
public:
    int divide(int n,int d)
    {
        if(n%d==0)
            return n/d;
        
        return (n/d)+1;
    }
    int smallestDivisor(vector<int>& nums, int threshold) {
        
        int left = 1;
        int right = 1e9;
        
        while(left<right)
        {
            int mid = (left+right)>>1;//proposed divisor
            int sum = 0;
            for(int ele:nums)
                sum+=divide(ele,mid);
            
            if(sum>threshold)
                left = mid+1;
            else
                right = mid;
            
        }
        return left;
    }
};