Trapping Rain Water - https://leetcode.com/explore/interview/card/facebook/5/round-1-phone-interview/295/

gistfile1.txt

class Solution {
    public int trap(int[] height) {
      int count = 0;
      
      int left = 0;
      int right = height.length - 1;
      
      int maxLeft = 0;
      int maxRight = 0;
      while (left < right) {
        
        if (height[left] < height[right]) {
          if (height[left] > maxLeft) {
            maxLeft = height[left];
          } else {
            // We have trap
            count += maxLeft - height[left];
          }
          left++;
        } else {
          if (height[right] > maxRight) {
            maxRight = height[right];
          } else {
            // we have a trap 
            count += maxRight - height[right];
          }
          
          right--;
        }
      }
      
      return count;
    }
}

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


The above elevation map 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. Thanks Marcos for contributing this image!

Example:

Input: [0,1,0,2,1,0,1,3,2,1,2,1]
Output: 6