<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:
    int ansforone( long long mid , vector<long long> & piles){
        long long finale = 0;
        // mid = 3
        // 7 8 9 10 11
        for( int i = 0 ; i < piles.size(); i++){
            finale += (piles[i] + mid - 1) / mid; // 
            // if( piles[i]%mid == 0) finale += (long long)piles[i]/mid;
            // else  finale += (long long)piles[i]/mid + 1;
        }
        return finale;
    }
    int minEatingSpeed(vector<int>& piles, int h) {
        vector<long long> pile;
        for (int x: piles) pile.push_back(1LL * x);
        long long n = piles.size();
        long long sum = n+1;
        long long lo = 1;
        long long hi = 1E10;
        // for( int i = 0 ; i < n ; i++) if( hi < piles[i]) hi = piles[i];
        // hi += 10;
        while( lo <= hi ){ // while (lo + 1 < hi)
            long long mid = lo + (hi - lo)/2;
            if( ansforone( mid , pile) <= h){
                sum = mid;
                hi = mid - 1;
            }
            else lo = mid + 1;
        }
        cout << hi - 1 << '\n';
        return sum;

    }
};
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