Trapping Rain Water - LeetCode Daily Challenge

Problem Statement

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

Try here before watching the video.

Video Solution

Java Code

class Solution {
    public int trap(int[] height) {
        int left = 0, right = height.length - 1, leftMax = 0, rightMax = 0;
        int ans = 0;

        while(left < right) {
            if(height[left] <= height[right]) {
                if(leftMax < height[left]) {
                    leftMax = height[left];
                } else {
                    ans += (leftMax - height[left]);
                }
                left += 1;
            } else {
                if(rightMax <=height[right]) {
                    rightMax = height[right];
                } else {
                    ans += (rightMax - height[right]);
                }
                right -= 1;
            }
        }
        return ans;
    }
}

C++ Code

class Solution {
public:
    int trap(vector<int>& height) {
        int left = 0, right = height.size() - 1, leftMax = 0, rightMax = 0;
        int ans = 0;

        while(left < right) {
            if(height[left] <= height[right]) {
                if(leftMax < height[left]) {
                    leftMax = height[left];
                } else {
                    ans += (leftMax - height[left]);
                }
                left += 1;
            } else {
                if(rightMax <= height[right]) {
                    rightMax = height[right];
                } else {
                    ans += (rightMax - height[right]);
                }
                right -= 1;
            }
        }
        return ans;
    }
};

Python Code

class Solution:
    def trap(self, height: List[int]) -> int:
        left, right = 0, len(height) - 1
        leftMax, rightMax = 0, 0
        ans = 0

        while left < right:
            if height[left] <= height[right]:
                if leftMax < height[left]:
                    leftMax = height[left]
                else:
                    ans += leftMax - height[left]
                left += 1
            else:
                if rightMax <= height[right]:
                    rightMax = height[right]
                else:
                    ans += rightMax - height[right]
                right -= 1

        return ans

Javascript Code

var trap = function(height) {
    let left = 0, right = height.length - 1;
    let leftMax = 0, rightMax = 0;
    let ans = 0;

    while (left < right) {
        if (height[left] <= height[right]) {
            if (leftMax < height[left]) {
                leftMax = height[left];
            } else {
                ans += leftMax - height[left];
            }
            left++;
        } else {
            if (rightMax <= height[right]) {
                rightMax = height[right];
            } else {
                ans += rightMax - height[right];
            }
            right--;
        }
    }

    return ans;
};

Go Code

func trap(height []int) int {
    left, right := 0, len(height)-1
    leftMax, rightMax := 0, 0
    ans := 0

    for left < right {
        if height[left] <= height[right] {
            if leftMax < height[left] {
                leftMax = height[left]
            } else {
                ans += leftMax - height[left]
            }
            left++
        } else {
            if rightMax <= height[right] {
                rightMax = height[right]
            } else {
                ans += rightMax - height[right]
            }
            right--
        }
    }

    return ans
}

Complexity Analysis

Time Complexity: O(N)
Space Complexity: O(1)