5. Trapping Rain Water

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

Approach

• maintain max for both side and think this whatever is the highest if we minus the current then that much water can be there for that graph
• Time: O(n) Space: O(1)

class Solution {
    public int trap(int[] height) {
        int l = 0, r = height.length - 1;
        int res = 0, lm = height[l], rm = height[r];
        while (l < r) {
            if (lm < rm) {
                l++;
                lm = Math.max(lm,height[l]);
                res += lm - height[l];
            } else {
                r--;
                rm = Math.max(rm,height[r]);
                res += rm - height[r];
            }
        }
        return res;
    }
}