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 <= 104piles.length <= h <= 1091 <= 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;
}
}
Input
Output
#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
Input
Output