Find the Smallest Divisor Given a Threshold

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  • ByteDance Interview Questions

Video Solution

Problem Statement

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.

Sample Test Cases

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 * 10^4
  • 1 <= nums[i] <= 10^6
  • nums.length <= threshold <= 10^6

Problem Solution

As we know that the maximum value of divisor will always be less than the maximum value (max_val) present in the array.

To get the divisor we can traverse through the range of 1 to max_val using binary search algorithm and for each value of divisor we will find the sum by dividing each element of the array and take the ceil of the answers and check if the sum calculated is less than the given threshold or not .

Binary search to get the correct divisor: if the trial sum
a) <= threshold, divisor >= solution, decrease right bound;
b) > threshold, divisor < solution, increase left bound.

Complexity Analysis

Time Complexity:O(nlogN) where N is the range of the max of nums, n=nums.size().

Space Complexity: O(1) no extra data structure used to store the values.

Code Implementation

#include<bits/stdc++.h>
using namespace std;

int smallestDivisor(vector<int>& nums, int threshold) {
        sort(nums.begin(),nums.end());

          int l=1,r=nums[nums.size()-1];
        int ans=0;
        while(l<=r){
            int mid=l+(r-l)/2;
            long long int sum=0;
            for(int i=0;i<nums.size();i++){
                if(nums[i]%mid==0){
                    sum+=(nums[i]/mid);
                } else{
                    sum+=(nums[i]/mid)+1;
                }
            }
            if(sum>threshold){
                l=mid+1;
            } else{
                ans=mid;
                r=mid-1;
            }
        }
        return ans;
    }
int main()
{
    vector<int>nums;
    nums.push_back(1);
    nums.push_back(2);
    nums.push_back(5);
    nums.push_back(9);
    int threshold=6;

    int x=smallestDivisor(nums,threshold);
    cout<<x;
}

 

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