49 lines
1.6 KiB
Python
49 lines
1.6 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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# Two pointer approach
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left, right = 0, len(heights) - 1
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# Since while scanning the array, at a given index i, we
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# only need to consider the minimum of max_left_heights and
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# max_right_heights, we just need to keep moving from both
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# ends and keep track of the maximum height seen so far. We
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# can then compare these two heights and decide from which
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# side to move next.
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max_left = max_right = 0
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output = 0
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while left <= right:
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# By the formula from the DP approach:
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#
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# min(max_left, max_right) - heights[i]
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#
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# If max_left <= max_right, formula becomes:
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#
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# max_left - heights[i]
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#
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if max_left <= max_right:
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# Shorter form to avoid -ve values
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output += max(max_left - heights[left], 0)
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max_left = max(max_left, heights[left])
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left += 1
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# By the formula from the DP approach:
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#
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# min(max_left, max_right) - heights[i]
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#
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# If max_left > max_right, formula becomes:
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#
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# max_right - heights[i]
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#
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else:
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output += max(max_right - heights[right], 0)
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max_right = max(max_right, heights[right])
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right -= 1
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return output
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