<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 {
public:
    bool isPossible(vector<int> piles,int h,long long m)
    {
        int i=0;
        while(i<piles.size())
        {
            long long t=piles[i]/m;
            if((piles[i]%m)>0)
                t++;
            h=h-t;
            i++;
            if(h<0)
                return false;
        }
        return true;

    }
    int minEatingSpeed(vector<int>& piles, int h) {
        long long total=0;
        for(int i: piles)
            total+=i;
        int minimum=total/h;
        if(total%h >0)
            minimum++;
    
        if(isPossible(piles,h,minimum))
            return minimum;
        long long l=minimum+1,r=total;
        while(l<r)
        {
            long long m=l+(r-l)/2;
            if(isPossible(piles,h,m))
            {
                r=m;
            }
            else
            {
                l=m+1;
            }
        }
        //cout<<l<<" "<<r<<endl;
        return r;
    }
};
29000