Trapping Rain Water Leetcode Problem


Question: 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.


Example 1:

Input: height = [0,1,0,2,1,0,1,3,2,1,2,1]
Output: 6
Explanation: The above elevation map (black section) is represented by array [0,1,0,2,1,0,1,3,2,1,2,1]. In this case, 6 units of rain water (blue section) are being trapped.

Example 2:

Input: height = [4,2,0,3,2,5]
Output: 9

 

Constraints:

n == height.length
1 <= n <= 2 * 104
0 <= height[i] <= 105



Solution: Java

class Solution {
    public int trap(int[] height) {
        int left = 0, right = height.length - 1;
        int result = 0;
        int leftMax = 0, rightMax = 0;
        while(left <= right) {
            if(height[left] <= height[right]) {
                if(height[left] > leftMax)
                    leftMax = height[left];
                else
                    result += leftMax - height[left];
                left++;
            } else {
                if(height[right] > rightMax)
                    rightMax = height[right];
                else
                    result += rightMax - height[right];
                right--;
            }
        }
        return result;
    }
}


Solution Python:
  def trap(self, height: List[int]) -> int:
    l = 0
    r = len(height) - 1
    maxLeft = height[l]
    maxRight = height[r]
    rain = 0
    while l < r:
        if maxLeft < maxRight:
            l += 1
            maxLeft = max(maxLeft, height[l])
            rain += maxLeft - height[l]
        else:
            r -= 1
            maxRight = max(maxRight, height[r])
            rain += maxRight - height[r]
    
    return rain





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