11. Container With Most Water

Medium

Problem Description

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.

Examples

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
Jump to Solution: Python Java C++ JavaScript C#

Python Solution

class Solution:
    def maxArea(self, height: List[int]) -> int:
        left, right = 0, len(height) - 1
        max_area = 0
        
        while left < right:
            width = right - left
            h = min(height[left], height[right])
            area = width * h
            max_area = max(max_area, area)
            
            if height[left] < height[right]:
                left += 1
            else:
                right -= 1
                
        return max_area

Java Solution

class Solution {
    public int maxArea(int[] height) {
        int left = 0;
        int right = height.length - 1;
        int maxArea = 0;
        
        while (left < right) {
            int width = right - left;
            int h = Math.min(height[left], height[right]);
            int area = width * h;
            maxArea = Math.max(maxArea, area);
            
            if (height[left] < height[right]) {
                left++;
            } else {
                right--;
            }
        }
        
        return maxArea;
    }
}

C++ Solution

class Solution {
public:
    int maxArea(vector& height) {
        int left = 0;
        int right = height.size() - 1;
        int maxArea = 0;
        
        while (left < right) {
            int width = right - left;
            int h = min(height[left], height[right]);
            int area = width * h;
            maxArea = max(maxArea, area);
            
            if (height[left] < height[right]) {
                left++;
            } else {
                right--;
            }
        }
        
        return maxArea;
    }
};

JavaScript Solution

/**
 * @param {number[]} height
 * @return {number}
 */
var maxArea = function(height) {
    let left = 0;
    let right = height.length - 1;
    let maxArea = 0;
    
    while (left < right) {
        const width = right - left;
        const h = Math.min(height[left], height[right]);
        const area = width * h;
        maxArea = Math.max(maxArea, area);
        
        if (height[left] < height[right]) {
            left++;
        } else {
            right--;
        }
    }
    
    return maxArea;
};

C# Solution

public class Solution {
    public int MaxArea(int[] height) {
        int left = 0;
        int right = height.Length - 1;
        int maxArea = 0;
        
        while (left < right) {
            int width = right - left;
            int h = Math.Min(height[left], height[right]);
            int area = width * h;
            maxArea = Math.Max(maxArea, area);
            
            if (height[left] < height[right]) {
                left++;
            } else {
                right--;
            }
        }
        
        return maxArea;
    }
}

Complexity Analysis

Solution Explanation

This solution uses the two-pointer technique:

Key points: