LeetCode: Container With Most Water

Posted on November 13, 2017July 26, 2020 by braindenny

Get the most water from containers

Similar Problems:

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.

The above vertical lines are represented by array [1,8,6,2,5,4,8,3,7]. In this case, the max area of water (blue section) the container can contain is 49.

Example:

Input: [1,8,6,2,5,4,8,3,7]
Output: 49

Github: code.dennyzhang.com

Credits To: leetcode.com

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

Solution: Two Pointer

## https://code.dennyzhang.com/container-with-most-water
## Basic Ideas: two pointer
##
##   Two pointer: 
##      From left to right
##      From right to left
##
##   If height[left]<height[right], move right doesn't make sense
##   Likewise for the other one
##
## Complexity: Time O(n), Space O(1)
class Solution:
    def maxArea(self, height: List[int]) -> int:
        n = len(height)
        left, right = 0, n-1
        res = 0
        while left<right:
            res = max(res, min(height[left], height[right])*(right-left))
            if height[left]<height[right]:
                left += 1
            else:
                right -= 1
        return res

Solution: Two Pointer

// 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

// 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
}

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