42. Trapping Rain Water LeetCode Solution

In this guide we will provide 42. Trapping Rain Water LeetCode Solution with best time and space complexity. The solution to Trapping Rain Water problem is provided in various programming languages like C++, Java and python. This will be helpful for you if you are preparing for placements, hackathon, interviews or practice purposes. The solutions provided here are very easy to follow and with detailed explanations.

Problem Statement

Complexity Analysis

Trapping Rain Water solution in C++
Trapping Rain Water soution in Java
Trapping Rain Water solution Python
Additional Resources

42. Trapping Rain Water LeetCode Solution image

Problem Statement of 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

Complexity Analysis

Time Complexity: O(n)
Space Complexity: O(n)

42. Trapping Rain Water LeetCode Solution in C++

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;
  }
};
/* code provided by PROGIEZ */

42. 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;
  }
}
// code provided by PROGIEZ

42. 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))
 #code by PROGIEZ