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In this post, you will find the solution for the **Trapping Rain Water** **in C++, Java & Python-LeetCode problem**. We are providing the **correct and tested solutions** to coding problems present on **LeetCode**. If you are not able to solve any problem, then you can take help from our Blog/website.

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** Link for the Problem** – Trapping Rain Water– LeetCode Problem

Trapping Rain Water– LeetCode Problem

**Problem:**

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:6Explanation: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 * 10`

^{4}`0 <= height[i] <= 10`

^{5}

Trapping Rain Water– LeetCode Solutions

class Solution { public: int trap(vector<int>& height) { if (height.empty()) return 0; int ans = 0; int l = 0; int r = height.size() - 1; int maxL = height[l]; int maxR = height[r]; while (l < r) if (maxL < maxR) { ans += maxL - height[l]; maxL = max(maxL, height[++l]); } else { ans += maxR - height[r]; maxR = max(maxR, height[--r]); } return ans; } };

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

class Solution: def trap(self, height: List[int]) -> int: if not height: return 0 ans = 0 l = 0 r = len(height) - 1 maxL = height[l] maxR = height[r] while l < r: if maxL < maxR: ans += maxL - height[l] l += 1 maxL = max(maxL, height[l]) else: ans += maxR - height[r] r -= 1 maxR = max(maxR, height[r]) return ans