In this post, we are going to solve the 11. Container With Most Water problem of Leetcode. This problem 11. Container With Most Water is a Leetcode medium level problem. Let’s see code, 11. Container With Most Water.

Problem

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 <= 105
• 0 <= height[i] <= 104

Now, let’s see the code of 11. Container With Most Water – Leetcode Solution.

Container With Most Water – Leetcode Solution

11. Container With Most Water – Solution in Java

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

11. Container With Most Water – Solution in C++

class Solution {
public:
    int maxArea(vector<int>& height) {
        int i=0; int j=height.size()-1;
        int ans = 0;
        while(i < j){
            ans = max(ans,min(height[i],height[j])*(j-i));
            if(height[i]<height[j])i++;
            else if(height[i]>height[j])j--;
            else {
                i++;
                j--;
            }
        }
        return ans;
    }
};

11. Container With Most Water – Solution in Python

class Solution:
    def maxArea(self, height: List[int]) -> int:
        p1 = 0  
        p2 = len(height) - 1 
        max_area = 0
        while p1 != p2:
            if height[p1] > height[p2]:
                area = height[p2] * (p2 - p1) 
                p2 -= 1 
            else:
                area = height[p1] * (p2 - p1)
                p1 += 1
            if area > max_area: max_area = area 
        return max_area

Note: This problem 11. Container With Most Water is generated by Leetcode but the solution is provided by CodingBroz. This tutorial is only for Educational and Learning purpose.