Trapping Rain Water Leetcode Solution

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

Table of Contents

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: 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, let’s see the leetcode solution of 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 Water Leetcode Solution in 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: This problem Trapping Rain Water is generated by Leetcode but the solution is provided by Chase2learn This tutorial is only for Educational and Learning purposes.

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