0011 - Container With Most Water (Medium)

Problem Link

https://leetcode.com/problems/container-with-most-water/

Problem Statement

You are given an integer array height of length n. There are n vertical lines drawn such that the two endpoints of the ith line are (i, 0) and (i, height[i]).

Find two lines that together with the x-axis form a container, such that the container contains the most water.

Return the maximum amount of water a container can store.

Notice that you may not slant the container.

Example 1:

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

Input: height = [1,1]
Output: 1

Constraints:

• n == height.length
• 2 <= n <= 10^5
• 0 <= height[i] <= 10^4

Approach 1: Two Pointers

We know that the area is $width * height$. Given two lines with different heights, we should only take the minimum one and the width is simply their distance. We use two pointers to track the lines, every time we calculate the area and move either one line. If the left line is shorter, then move the left one because we want to keep the maximum one, and vice versa.

C++

Written by @wingkwong

class Solution {
public:
    int maxArea(vector<int>& height) {
        int ans = 0, l = 0, r = height.size() - 1;
        while (l < r) {
            // calculate the current area
            ans = max(ans, min(height[l], height[r]) * (r - l));
            // keep the max one and move the min one
            if (height[l] < height[r]) l++;
            else r--;
        }
        return ans;
    }
};

Java

Written by @ganajayant

class Solution {
    public int maxArea(int[] height) {
        int max = 0;
        int i = 0;
        int j = height.length - 1;
        while (i < j) {
            int leasth = Math.min(height[i], height[j]);
            int width = (j - i);
            if (max < leasth * width) {
                max = leasth * width;
            }
            if (height[i] > height[j]) {
                j--;
            } else {
                i++;
            }
        }
        return max;
    }
}

JavaScript

Written by @MithunPrabhu777

const maxArea = (height) => {
    let [left, right, max] = [0, height.length - 1, 0];

    while (left < right) {
        max = Math.max(max, Math.min(height[left], height[right]) * (right - left));
        if (height[right] > height[left]) {
            left += 1;
        } else {
            right -= 1;
        }
    }

    return max;
};

Python

Written by @radojicic23

def maxArea(height):
    res = 0  
    l, r = 0, len(height) - 1 

    while l < r:
        area = (r - l) * min(height[r], height[l])
        res = max(res, area)
        
        if height[l] < height[r]: 
            l += 1 
        else: 
            r -= 1
    
    return res