875. Koko Eating Bananas
https://leetcode.com/problems/koko-eating-bananas
Koko loves to eat bananas. There are
N piles of bananas, the i-th 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 won't eat any more bananas during this hour.
Koko likes to eat slowly, but still wants to finish eating all the bananas before the guards come back.
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
Note:
1 <= piles.length <= 10^4piles.length <= H <= 10^91 <= piles[i] <= 10^9
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Related problems
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Intuition
Any hour Koko can eat only from 1 pile => Max rate is max in the array, any hourly rate > max in the array is wasteful
Min rate is 1, .. she eats 1 banana per hour, she'll take Sum of all bananas in all piles to eat all
Binary search between 1 and Max in array
isValid function -
If pile[i] <= K => Hours += 1
If pile[i] > K => Hours += 1 + (pile[i] / K)
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Time - O(N * Log(Max Bananas))
Space - O(1)
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Any hour Koko can eat only from 1 pile => Max rate is max in the array, any hourly rate > max in the array is wasteful
Min rate is 1, .. she eats 1 banana per hour, she'll take Sum of all bananas in all piles to eat all
Binary search between 1 and Max in array
isValid function -
If pile[i] <= K => Hours += 1
If pile[i] > K => Hours += 1 + (pile[i] / K)
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Time - O(N * Log(Max Bananas))
Space - O(1)
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