# Trapping Rain Water Leetcode Solution

In this post, we are going to solve theÂ Trapping Rain Water Leetcode SolutionÂ problem of Leetcode. This Leetcode problem is done in many programming languages like C++, Java, and Python.

## Problem

GivenÂ `n`Â nonâ€“negative integers representing an elevation map where the width of each bar isÂ `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`

Now, letâ€™s see the leetcode solution ofÂ Trapping Rain Water Leetcode Solution.

### Trapping Rain Water Leetcode Solution in Python

```class Solution:
def trap(self, height: List[int]) -> int:
n = len(height)
l = [0] * n  # l[i] := max(height[0..i])
r = [0] * n  # r[i] := max(height[i..n))

for i, h in enumerate(height):
l[i] = h if i == 0 else max(h, l[i - 1])

for i, h in reversed(list(enumerate(height))):
r[i] = h if i == n - 1 else max(h, r[i + 1])

return sum(min(l[i], r[i]) - h
for i, h in enumerate(height))
```

### Trapping Rain WaterLeetcode Solutionin CPP

```class Solution {
public:
int trap(vector<int>& height) {
const int n = height.size();
int ans = 0;
vector<int> l(n);  // l[i] := max(height[0..i])
vector<int> r(n);  // r[i] := max(height[i..n))

for (int i = 0; i < n; ++i)
l[i] = i == 0 ? height[i] : max(height[i], l[i - 1]);

for (int i = n - 1; i >= 0; --i)
r[i] = i == n - 1 ? height[i] : max(height[i], r[i + 1]);

for (int i = 0; i < n; ++i)
ans += min(l[i], r[i]) - height[i];

return ans;
}
};
```

### Trapping Rain Water Leetcode Solution in Java

```class Solution {
public int trap(int[] height) {
final int n = height.length;
int ans = 0;
int[] l = new int[n]; // l[i] := max(height[0..i])
int[] r = new int[n]; // r[i] := max(height[i..n))

for (int i = 0; i < n; ++i)
l[i] = i == 0 ? height[i] : Math.max(height[i], l[i - 1]);

for (int i = n - 1; i >= 0; --i)
r[i] = i == n - 1 ? height[i] : Math.max(height[i], r[i + 1]);

for (int i = 0; i < n; ++i)
ans += Math.min(l[i], r[i]) - height[i];

return ans;
}
}
```

Note:Â This problemÂ Trapping Rain Water is generated byÂ LeetcodeÂ but the solution is provided byÂ Chase2learn This tutorial is only forÂ EducationalÂ andÂ LearningÂ purposes.

Sharing Is Caring