Get the most water from containers

Similar Problems:

- Series: Trapping Rain & Follow-up
- CheatSheet: Leetcode For Code Interview
- Tag: #trappingrain, #twopointer

Given n non-negative integers a1, a2, …, an, where each represents a point at coordinate (i, ai). n vertical lines are drawn such that the two endpoints of line i is at (i, ai) and (i, 0). Find two lines, which together with x-axis forms a container, such that the container contains the most water.

Note: You may not slant the container and n is at least 2.

Github: code.dennyzhang.com

Credits To: leetcode.com

Leave me comments, if you have better ways to solve.

**Key Observations:**

By moving shorter end pointer further doesn't eliminate the final answer

- Solution: Two Pointer

// Blog link: https://code.dennyzhang.com/container-with-most-water // Basic Ideas: Two pointer // Start with l, r = 0, len(l)-1 // Then move the pointer of shorter value // Let's check the case that both pointers are the same the value. // We won't find a better solution with only one edge of left pointer or right pointer. // Thus we can move either pointer // Complexity: Time O(n), Space O(1) func maxArea(height []int) int { l, r := 0, len(height)-1 res, v := 0, 0 for l<r { if height[l] <= height[r] { v = height[l]*(r-l) l++ } else { v = height[r]*(r-l) r-- } if v>res { res = v } } return res }

- Solution: Two Pointer + Pruning

// Blog link: https://code.dennyzhang.com/container-with-most-water // Basic Ideas: Two pointer // Start with l, r = 0, len(l)-1 // Then move the pointer of shorter value // Let's check the case that both pointers are the same the value. // We won't find a better solution with only one edge of left pointer or right pointer. // Thus we can move either pointer // Complexity: Time O(n), Space O(1) func maxArea(height []int) int { l, r := 0, len(height)-1 res, v := 0, 0 for l<r { lmax, rmax := height[l], height[r] if height[l] <= height[r] { v = height[l]*(r-l) } else { v = height[r]*(r-l) } if v>res { res = v } if height[l] <= height[r] { for l<r && height[l]<=lmax { l++ } } else { for l<r && height[r]<=rmax { r-- } } } return res }