<p>Given an array of integers <code>nums</code> and an integer <code>threshold</code>, we will choose a positive integer <code>divisor</code>, divide all the array by it, and sum the division&#39;s result. Find the <strong>smallest</strong> <code>divisor</code> such that the result mentioned above is less than or equal to <code>threshold</code>.</p>

<p>Each result of the division is rounded to the nearest integer greater than or equal to that element. (For example: <code>7/3 = 3</code> and <code>10/2 = 5</code>).</p>

<p>The test cases are generated so&nbsp;that there will be an answer.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums = [1,2,5,9], threshold = 6
<strong>Output:</strong> 5
<strong>Explanation:</strong> 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). 
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums = [44,22,33,11,1], threshold = 5
<strong>Output:</strong> 44
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= nums.length &lt;= 5 * 10<sup>4</sup></code></li>
	<li><code>1 &lt;= nums[i] &lt;= 10<sup>6</sup></code></li>
	<li><code>nums.length &lt;= threshold &lt;= 10<sup>6</sup></code></li>
</ul>

class Solution {
public:
    int sumOfD(vector<int>& nums,int mid){
        int sum=0;
        int n=nums.size();
        for(int i=0;i<n;i++){
            sum=sum+ceil((double)nums[i]/(double)mid);
        }
        return sum;
    }
    int smallestDivisor(vector<int>& nums, int threshold) {
        int low=1,high=*max_element(nums.begin(),nums.end());
        while(low<=high){
            int mid=(low+high)/2;
            if(sumOfD(nums,mid)<=threshold){
                high=mid-1;
            }else low=mid+1;
        }
        return low;
    }
};
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