text: # Leetcode Trapping Rain Water problem solution
  
  In this Leetcode Trapping Rain Water problem solution, we have 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.
  
  ## Problem solution in Python.
  
  ```class Solution:
  def trap(self, height: List[int]) -> int:
  if not height: return 0
  
  water, n = 0, len(height)
  max_left, max_right = [None] * n, [None] * n
  max_left[0], max_right[-1] = height[0], height[-1]
  
  for i in range(1, n):
  max_left[i] = max(max_left[i-1], height[i])
  
  for i in range(n-2, 0, -1):
  max_right[i] = max(max_right[i+1], height[i])
  water += min(max_left[i], max_right[i]) - height[i]
  
  return water
  ```
  
  ## Problem solution in Java.
  
  ```class Solution {
  public int trap(int[] height) {
  int l=0;
  int r=height.length-1;
  while(l<r && height[l]<height[l+1])
  l++;
  while(l<r && height[r]<height[r-1])
  r--;
  int ans=0;
  
  while(l<r){
  int left=height[l];
  int right=height[r];
  if(left<=right){
  while(l<r && left>=height[l]){
  ans+=left-height[l];
  l++;
  }}
  else{
  while(l<r && right>=height[r]){
  ans+=right-height[r];
  r--;
  }}
  }
  return ans;
  }
  }
  ```
  
  ## Problem solution in C++.
  
  ```int trap(vector<int>& height) {
  
  if (height.size()==0) return 0;
  
  vector<int> heightcopy;
  
  heightcopy=height;
  
  int i,j;
  int Level;
  
  i=0;
  j=height.size()-1;
  Level=0;
  
  while (i<j) {
  if (heightcopy[i]<Level) { heightcopy[i]=Level; i++; }
  else if (heightcopy[i]==Level) i++;
  else if (heightcopy[j]<Level) { heightcopy[j]=Level; j--; }
  else if (heightcopy[j]==Level) j--;
  else if (heightcopy[i]>Level && heightcopy[j]>Level) {
  Level++;
  
  }
  
  }
  
  int sumTotal = accumulate(heightcopy.begin(), heightcopy.end(), 0);
  int sumHeight = accumulate(height.begin(),height.end(),0);
  
  return sumTotal-sumHeight;
  
  }
  ```
  
  ## Problem solution in C.
  
  ```int trap(int* height, int heightSize){
  int leftMax=0, rightMax=0, left=0, right=heightSize-1, result=0, max;
  
  while(left < right) {
  if(height[left]>leftMax) leftMax = height[left];
  if(height[right]>rightMax) rightMax = height[right];
  if(leftMax<rightMax) {
  if((leftMax-height[left]) > 0) result += (leftMax-height[left]);
  left++;
  } else {
  if((rightMax-height[right]) > 0) result += (rightMax-height[right]);
  right--;
  }
  }
  
  return result;
  }
  ```
explanation: The document provides a problem statement and multiple solutions in different programming languages (Python, Java, C++, and C) for the LeetCode "Trapping Rain Water" problem. The problem statement describes a scenario where an elevation map is given, and the task is to compute how much water can be trapped after raining. The document does not contain a question or an answer in the form of a query or response but rather presents a problem and its solutions.
distilabel_metadata.content: Evaluate the user given text extract and determine if it contains a math question and optionally and answer to this question.
  
  Output rules:
  
  1. Briefly describe the document and mention if it contains a question or a question and answer.
  2. Finalize the output with: Label: no-question,question,question-and-answer,multiple-question,multiple-question-and-answer
distilabel_metadata.role: system
distilabel_metadata.content: # Leetcode Trapping Rain Water problem solution
  
  In this Leetcode Trapping Rain Water problem solution, we have 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.
  
  ## Problem solution in Python.
  
  ```class Solution:
  def trap(self, height: List[int]) -> int:
  if not height: return 0
  
  water, n = 0, len(height)
  max_left, max_right = [None] * n, [None] * n
  max_left[0], max_right[-1] = height[0], height[-1]
  
  for i in range(1, n):
  max_left[i] = max(max_left[i-1], height[i])
  
  for i in range(n-2, 0, -1):
  max_right[i] = max(max_right[i+1], height[i])
  water += min(max_left[i], max_right[i]) - height[i]
  
  return water
  ```
  
  ## Problem solution in Java.
  
  ```class Solution {
  public int trap(int[] height) {
  int l=0;
  int r=height.length-1;
  while(l<r && height[l]<height[l+1])
  l++;
  while(l<r && height[r]<height[r-1])
  r--;
  int ans=0;
  
  while(l<r){
  int left=height[l];
  int right=height[r];
  if(left<=right){
  while(l<r && left>=height[l]){
  ans+=left-height[l];
  l++;
  }}
  else{
  while(l<r && right>=height[r]){
  ans+=right-height[r];
  r--;
  }}
  }
  return ans;
  }
  }
  ```
  
  ## Problem solution in C++.
  
  ```int trap(vector<int>& height) {
  
  if (height.size()==0) return 0;
  
  vector<int> heightcopy;
  
  heightcopy=height;
  
  int i,j;
  int Level;
  
  i=0;
  j=height.size()-1;
  Level=0;
  
  while (i<j) {
  if (heightcopy[i]<Level) { heightcopy[i]=Level; i++; }
  else if (heightcopy[i]==Level) i++;
  else if (heightcopy[j]<Level) { heightcopy[j]=Level; j--; }
  else if (heightcopy[j]==Level) j--;
  else if (heightcopy[i]>Level && heightcopy[j]>Level) {
  Level++;
  
  }
  
  }
  
  int sumTotal = accumulate(heightcopy.begin(), heightcopy.end(), 0);
  int sumHeight = accumulate(height.begin(),height.end(),0);
  
  return sumTotal-sumHeight;
  
  }
  ```
  
  ## Problem solution in C.
  
  ```int trap(int* height, int heightSize){
  int leftMax=0, rightMax=0, left=0, right=heightSize-1, result=0, max;
  
  while(left < right) {
  if(height[left]>leftMax) leftMax = height[left];
  if(height[right]>rightMax) rightMax = height[right];
  if(leftMax<rightMax) {
  if((leftMax-height[left]) > 0) result += (leftMax-height[left]);
  left++;
  } else {
  if((rightMax-height[right]) > 0) result += (rightMax-height[right]);
  right--;
  }
  }
  
  return result;
  }
  ```
distilabel_metadata.role: user
distilabel_metadata.raw_output_q_a_detector_0: The document provides a problem statement and multiple solutions in different programming languages (Python, Java, C++, and C) for the LeetCode "Trapping Rain Water" problem. The problem statement describes a scenario where an elevation map is given, and the task is to compute how much water can be trapped after raining. The document does not contain a question or an answer in the form of a query or response but rather presents a problem and its solutions.
  
  Label: no-question