62 lines
1.9 KiB
Python
62 lines
1.9 KiB
Python
# Time: O(N)
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# Space: O(N)
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class Solution:
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def trap(self, heights: List[int]) -> int:
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if len(heights) < 2: return 0
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# For every ith height, we need to calculate the max height
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# on left and max on right of it
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max_lefts, max_rights = [0] * len(heights), [0] * len(heights)
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maxl = 0
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for i in range(len(heights)):
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max_lefts[i] = maxl
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maxl = max(maxl, heights[i])
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maxr = 0
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for j in range(len(heights) - 1, -1, -1):
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max_rights[j] = maxr
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maxr = max(maxr, heights[j])
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print(max_lefts)
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print(max_rights)
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# For every ith height, we can now compute the water output it can hold
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# by doing the following:
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#
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# min(max_height_to_left, max_height_to_right) - height_of_bar
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#
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# If we ignore height_of_bar for a moment, the amount of water that can be
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# held would be constrained by the least heights out of left/right.
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#
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# 3
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# 2 |
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# | |
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# | 0 |
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#
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# In the above case, would be min(2, 3) == 2. But, if we fill the space up with a bar
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# then:
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#
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# 3
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# 2 |
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# | 1 |
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# | | |
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#
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# We can't put 2 water units. We need to subtract the height of the bar from the prev
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# min() which will give us 1.
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#
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# NOTE: We need to do this for every ith height in the array.
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output = 0
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for i, h in enumerate(heights):
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value = min(max_lefts[i], max_rights[i]) - heights[i]
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# It's possible that ith height is very large, causing
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# the expression above to give us -ve value. In this case
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# we just ignore it (i.e, 0 output for this ith height)
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if value > 0:
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output += value
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return output
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