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Trapping rain water

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Given n non-negative integers representing an elevation map where the width of each bar is 1, compute how much water it can trap after raining.

#dynamic programming method
class Solution(object):
    def trap(self, height):
        :type height: List[int]
        :rtype: int
        if height == []:
            return 0

        n = len(height)
        max_left = [0]* n
        max_right = [0]* n
        ##print(max_left, height)
        max_left[0] = 0
        max_right[n-1] = 0
        for i in range (1, n):
            max_left[i] = max(max_left[i - 1], height[i-1])
        for i in range(n-2, -1, -1):
            max_right[i] = max(max_right[i + 1], height[i + 1])

        output = 0
        print(max_left, max_right)
        for i in range(n):
            lower_boundary = min(max_left[i], max_right[i])
            max_trap_at_i = lower_boundary - height[i]
            if max_trap_at_i > 0:
                output += max_trap_at_i
        return output

Two Pointer method:

class Solution:
    def trap(self, height: List[int]) -> int:
        ans = 0
        l,r = 0 , len(height) -1
        l_max, r_max = 0,0
        while l < r:
		# case 1: lower_bound is from left.
            if height[l] < height[r]:
                if height[l] >= l_max:
                    l_max = height[l]
                    ans += l_max - height[l]
                l += 1
			# case 2: the lower_bound is from right
                if height[r] >= r_max:
                    r_max = height[r]
                    ans += r_max - height[r]
                r -= 1
        return ans

Posted by Jamie Meyer 4 months ago

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