<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 isEatingSpeedValid(vector<int>& piles, int k, int h){

//         int hoursSpent = 0;
//         for(int i=0; i<piles.size(); i++){
//             hoursSpent += piles[i]/k;
//             if((piles[i] % k) != 0){
//                 hoursSpent += 1;
//             }
//             if(hoursSpent > h)  return false;  
//         }
//         return true;
//     }

//     int minEatingSpeed(vector<int>& piles, int h) { 

//         int start = 1;
//         int end = INT_MIN;

//         for(int i=0; i<piles.size(); i++){
//             if(piles[i] > end){
//                 end = piles[i];
//             }
//         }

//         //   Apply binary search on possible values of K -> (1,max(piles))

//         int ans = -1;

//         while(start <= end){
//             int mid = start + (end-start)/2;
//             if(isEatingSpeedValid(piles, mid, h)){
//                 ans = mid;
//                 end = mid-1;
//             }
//             else{
//                 start = mid+1;
//             }
//         }
//         return ans;
//     }
// };







//       Brute Force Solution     TC = O(max(Arr) * N)         TLE occur

// class Solution {
// public:

//     int totalHours(vector<int>& piles, int x){
        
//         int ans = 0;

//         for(int i=0; i<piles.size(); i++){
//             if(piles[i] % x == 0)                     //  take seal value
//                 ans += piles[i] /x;
//             else
//                 ans += ((piles[i] / x) + 1);
//         }
//         return ans;
//     }
    
//     int minEatingSpeed(vector<int>& piles, int h) { 
        
//         int n = piles.size();
//         priority_queue<int> arr;        //  for finding max of arr

//         for(int i=0; i<n; i++){
//             arr.push(piles[i]);
//         }
//         int maxi = arr.top();

//         for(int i=1; i<=maxi; i++){
//             int requireTime = totalHours(piles, i);

//             if(requireTime <= h){
//                 return i;
//             }
//         }
//         return 0;
//     }
// };








//            Optimal Solution            TC = O(N) * log(max element)

class Solution {
public:

    bool totalHours(vector<int>& piles, int mid, int h){
        
        int ans = 0;
        for(int i=0; i<piles.size(); i++){
            ans += piles[i] / mid;
            if((piles[i] % mid) != 0)                     //  take seal value
                ans += 1;
            
            if(ans > h)  return false;
        }
        return true;
    }
    
    int minEatingSpeed(vector<int>& piles, int h) { 
        
        int n = piles.size();
        priority_queue<int> arr;        //  for finding max of arr

        for(int i=0; i<n; i++){
            arr.push(piles[i]);
        }
        int maxi = arr.top();
 
//    Apply Binary Search

        int s = 1, e = maxi, ans = maxi;
        int mid = (s+e)/2;

        while(s <= e){
            mid = (s+e)/2;
            if(totalHours(piles, mid, h)){
                ans = mid;
                e = mid - 1;
            }
            else
                s = mid + 1;
        }
        return ans;
    }
};
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