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# Trapping Rain Water LeetCode Solution

## Problem – 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`

### Trapping Rain Water LeetCode Solution in C++

``````class Solution {
public:
int trap(int A[], int n) {
int left=0; int right=n-1;
int res=0;
int maxleft=0, maxright=0;
while(left<=right){
if(A[left]<=A[right]){
if(A[left]>=maxleft) maxleft=A[left];
else res+=maxleft-A[left];
left++;
}
else{
if(A[right]>=maxright) maxright= A[right];
else res+=maxright-A[right];
right--;
}
}
return res;
}
};``````

### Trapping Rain Water LeetCode Solution in Python

``````class Solution(object):
def trap(self, height):
"""
:type height: List[int]
:rtype: int
"""
leftCursor, rightCursor = 0, len(height)-1
leftMax, rightMax, storedWater = 0, 0, 0

while (leftCursor <= rightCursor):
leftMax = max(leftMax, height[leftCursor])
rightMax = max(rightMax, height[rightCursor])
if leftMax < rightMax:
storedWater += leftMax - height[leftCursor]
leftCursor += 1
else:
storedWater += rightMax - height[rightCursor]
rightCursor -= 1

return storedWater``````

### Trapping Rain Water LeetCode Solution in Java

``````class Solution {
public int trap(int[] height) {
if (height == null || height.length == 0) {
return 0;
}
int left = 0; int right = height.length - 1; // Pointers to both ends of the array.
int maxLeft = 0; int maxRight = 0;

int totalWater = 0;
while (left < right) {
// Water could, potentially, fill everything from left to right, if there is nothing in between.
if (height[left] < height[right]) {
// If the current elevation is greater than the previous maximum, water cannot occupy that point at all.
// However, we do know that everything from maxLeft to the current index, has been optimally filled, as we've
// been adding water to the brim of the last maxLeft.
if (height[left] >= maxLeft) {
// So, we say we've found a new maximum, and look to see how much water we can fill from this point on.
maxLeft = height[left];
// If we've yet to find a maximum, we know that we can fill the current point with water up to the previous
// maximum, as any more will overflow it. We also subtract the current height, as that is the elevation the
// ground will be at.
} else {
totalWater += maxLeft - height[left];
}
// Increment left, we'll now look at the next point.
left++;
// If the height at the left is NOT greater than height at the right, we cannot fill from left to right without over-
// flowing; however, we do know that we could potentially fill from right to left, if there is nothing in between.
} else {
// Similarly to above, we see that we've found a height greater than the max, and cannot fill it whatsoever, but
// everything before is optimally filled
if (height[right] >= maxRight) {
// We can say we've found a new maximum and move on.
maxRight = height[right];
// If we haven't found a greater elevation, we can fill the current elevation with maxRight - height[right]
// water.
} else {
totalWater += maxRight - height[right];
}
// Decrement left, we'll look at the next point.
right--;
}
}
// Return the sum we've been adding to.
}
}``````
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