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# 875. Koko Eating Bananas

## Problem Statement

<br>

Koko loves to eat bananas. There are `n` piles of bananas, the `ith` pile has `piles[i]` bananas. The guards have gone and will come back in `h` hours.

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

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

Return *the minimum integer* `k` *such that she can eat all the bananas within* `h` *hours*.

&#x20;

**Example 1:**

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

**Example 2:**

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

**Example 3:**

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

&#x20;

**Constraints:**

* `1 <= piles.length <= 104`
* `piles.length <= h <= 109`
* `1 <= piles[i] <= 109`

## Intuition

```
Approach:
He can eat a maximum of max(array) bananas
range is 1->max(array)

Do BS on this answer space to find the min
```

### Links

<https://leetcode.com/problems/koko-eating-bananas/description/>

### Video Links

### Approach 1:

```
```

{% code title="C++" lineNumbers="true" %}

```cpp
class Solution {
public:
    int minEatingSpeed(vector<int>& piles, int h) {        
        int size=piles.size(), ans, max=-1;
        
        for(int i=0;i<size;i++){
            if(max<piles[i])
                max=piles[i];
        }

        if(size==h) return max;
        if(size==1){
            return ceil(piles[0]/double(h));
        }

        int low=1, high=max;

        while(low<=high){
            int mid=low+(high-low)/2;
            double count=0;

            for(int i=0; i<size; i++){
                // Replacement for ceil function
                count += (piles[i] + mid -1)/mid;
            }

            if(count>h){
                low=mid+1;
            }
            else { //count<=h
                high=mid-1;
                ans = mid;
            }
        }

        return ans;
    }
};
```

{% endcode %}

### Approach 2:

```
```

{% code title="C++" lineNumbers="true" %}

```cpp
```

{% endcode %}

### Approach 3:

```
```

{% code title="C++" lineNumbers="true" %}

```cpp
```

{% endcode %}

### Approach 4:

```
```

{% code title="C++" lineNumbers="true" %}

```cpp
```

{% endcode %}

### Similar Problems

###