Trapping Rain Water LeetCode Programming Solutions | LeetCode Problem Solutions in C++, Java, & Python [💯Correct]

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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: 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

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