<p>Koko loves to eat bananas. There are <code>n</code> piles of bananas, the <code>i<sup>th</sup></code> pile has <code>piles[i]</code> bananas. The guards have gone and will come back in <code>h</code> hours.</p>

<p>Koko can decide her bananas-per-hour eating speed of <code>k</code>. Each hour, she chooses some pile of bananas and eats <code>k</code> bananas from that pile. If the pile has less than <code>k</code> bananas, she eats all of them instead and will not eat any more bananas during this hour.</p>

<p>Koko likes to eat slowly but still wants to finish eating all the bananas before the guards return.</p>

<p>Return <em>the minimum integer</em> <code>k</code> <em>such that she can eat all the bananas within</em> <code>h</code> <em>hours</em>.</p>

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

<pre>
<strong>Input:</strong> piles = [3,6,7,11], h = 8
<strong>Output:</strong> 4
</pre>

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

<pre>
<strong>Input:</strong> piles = [30,11,23,4,20], h = 5
<strong>Output:</strong> 30
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> piles = [30,11,23,4,20], h = 6
<strong>Output:</strong> 23
</pre>

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

<ul>
	<li><code>1 &lt;= piles.length &lt;= 10<sup>4</sup></code></li>
	<li><code>piles.length &lt;= h &lt;= 10<sup>9</sup></code></li>
	<li><code>1 &lt;= piles[i] &lt;= 10<sup>9</sup></code></li>
</ul>

class Solution {
private:
    int sorts(vector<int>& piles){
        sort(piles.rbegin(),piles.rend()) ;
        return piles[0] ;
    }
    long long int ans(vector<int>& piles,int mid){
        long long int sum = 0 ;
        for (int i = 0 ; i<piles.size(); ++i){
            if(piles[i]%mid==0) sum+=(piles[i]/mid) ;
            else sum+=((piles[i]/mid)+1) ;
        }
        return sum ;
    }
public:
    int minEatingSpeed(vector<int>& piles, int h) {
        int low = 1, high = sorts(piles), x = INT_MAX ;
        while(low<=high){
            int mid = low+(high-low)/2 ;
            if(ans(piles,mid)<=h){
                high = mid-1 ;
                x = min(x,mid) ;
            } 
            else low = mid+1 ;
        }
        return x ;
    }
};
25100