49 lines
1.6 KiB
Python
49 lines
1.6 KiB
Python
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class Solution:
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def minEatingSpeed(self, piles: List[int], h: int) -> int:
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'''
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NOTE: Do read the problem carefully!
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Absolute min speed (k) at which Koko can eat bananas would be 1
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and max speed would be max(piles). Of course, these might not
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be correct and might help Koko finish eating before `h` hours.
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So, we need to check each k from 1:max(piles) and figure out the
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min k that lets us consume <= h
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We can iterate over each k value and check or we can do binary
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search and figure out least k.
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'''
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start, end = 1, max(piles)
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k = end
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while start <= end:
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mid = (start + end) // 2
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total = 0
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for p in piles:
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# If pile has 3 bananas and k = 2, then 3 / 2 = 1.5
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# but we don't want fractional values, we will consider
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# next greatest integer which'd be ceil(p / mid)
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total += math.ceil(p / mid)
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# k is valid
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if total <= h:
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# Update k if mid is smaller
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if mid < k:
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k = mid
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end = mid - 1
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# If mid starts to go bigger than k, then
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# we can't possibly find any better results
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# so just break
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else:
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break
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# k is too small 👉 total is large so we need
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# to move `start`
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else:
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start = mid + 1
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return k
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