## Algorithm

Problem Name: 875. Koko Eating Bananas

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.

Example 1:

```Input: piles = [3,6,7,11], h = 8
Output: 4
```

Example 2:

```Input: piles = [30,11,23,4,20], h = 5
Output: 30
```

Example 3:

```Input: piles = [30,11,23,4,20], h = 6
Output: 23
```

Constraints:

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

## Code Examples

### #1 Code Example with Java Programming

```Code - Java Programming```

``````
class Solution {
public int minEatingSpeed(int[] piles, int h) {
int start = 1;
int end = 0;
for (int pile : piles) {
end = Math.max(end, pile);
}
while (start  < = end) {
int mid = start + (end - start) / 2;
if (isPossible(piles, mid, h)) {
end = mid - 1;
} else {
start = mid + 1;
}
}
return start;
}

private boolean isPossible(int[] piles, int k, int h) {
int numOfHours = 0;
for (int pile : piles) {
numOfHours += Math.ceil((double) pile / k);
}
return numOfHours  < = h;
}
}
``````
Copy The Code &

Input

cmd
piles = [3,6,7,11], h = 8

Output

cmd
4

### #2 Code Example with Python Programming

```Code - Python Programming```

``````
class Solution:
def minEatingSpeed(self, piles, H):
piles.sort()
l, r = 1, max(piles)
while l <= r:
mid = (l + r) // 2
h = sum(math.ceil(p / mid) for p in piles)
if h > H:
l = mid + 1
elif h < H:
r = mid - 1
else:
return mid
return l
``````
Copy The Code &

Input

cmd
piles = [3,6,7,11], h = 8

Output

cmd
4