Trapping Rain Water – LeetCode Solution Java , Python 3, Python 2 , C , C++, Best and Optimal Solutions , All you need.

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

C++ Trapping Rain Water LeetCode Solution

class Solution { // 4 ms, faster than 89.31%

public:

int trap(vector<int>& height) {

int n = height.size();

vector<int> leftMax(n), rightMax(n);

for (int i = 1; i < n; ++i)

leftMax[i] = max(height[i-1], leftMax[i-1]);

for (int i = n-2; i >= 0; –i)

rightMax[i] = max(height[i+1], rightMax[i+1]);
int ans = 0;

for (int i = 0; i < n; ++i) {

int waterLevel = min(leftMax[i], rightMax[i]);

if (waterLevel >= height[i]) ans += waterLevel – height[i];

}

return ans;

}

};

Java Trapping Rain Water LeetCode Solution

class Solution { // 1 ms, faster than 85.57%

public int trap(int[] height) {

int n = height.length;

int[] leftMax = new int[n], rightMax = new int[n];

for (int i = 1; i < n; ++i)

leftMax[i] = Math.max(height[i-1], leftMax[i-1]);

for (int i = n-2; i >= 0; –i)

rightMax[i] = Math.max(height[i+1], rightMax[i+1]);
int ans = 0;

for (int i = 0; i < n; ++i) {

int waterLevel = Math.min(leftMax[i], rightMax[i]);

if (waterLevel >= height[i]) ans += waterLevel – height[i];

}

return ans;

}

}
 

Python 3 Trapping Rain Water LeetCode Solution

class Solution: # 52 ms, faster than 81.89%

def trap(self, height: List[int]) -> int:

n = len(height)

maxLeft, maxRight = [0] * n, [0] * n
for i in range(1, n):

maxLeft[i] = max(height[i-1], maxLeft[i-1])

for i in range(n-2, -1, -1):

maxRight[i] = max(height[i+1], maxRight[i+1])
ans = 0

for i in range(n):

waterLevel = min(maxLeft[i], maxRight[i])

if waterLevel >= height[i]:

ans += waterLevel – height[i]

return ans

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