Trapping Rain Water Leetcode Problem
Question: 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
Solution: Java
class Solution {
public int trap(int[] height) {
int left = 0, right = height.length - 1;
int result = 0;
int leftMax = 0, rightMax = 0;
while(left <= right) {
if(height[left] <= height[right]) {
if(height[left] > leftMax)
leftMax = height[left];
else
result += leftMax - height[left];
left++;
} else {
if(height[right] > rightMax)
rightMax = height[right];
else
result += rightMax - height[right];
right--;
}
}
return result;
}
}
Solution Python:
def trap(self, height: List[int]) -> int:
l = 0
r = len(height) - 1
maxLeft = height[l]
maxRight = height[r]
rain = 0
while l < r:
if maxLeft < maxRight:
l += 1
maxLeft = max(maxLeft, height[l])
rain += maxLeft - height[l]
else:
r -= 1
maxRight = max(maxRight, height[r])
rain += maxRight - height[r]
return rain
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