# 11. Container With Most Water

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

# Solution

Approach 1: Greedy -- two pointers.

# Code (Python)

Approach 1:

    def maxArea(self, height: List[int]) -> int:
        # idea: greedy. Maintain two pointers closing inwards. Discard the side with less height to maximize chance of most water with tighter bottoms.
        left, right = 0, len(height) - 1
        max_area = 0
        while left < right:
            max_area = max(max_area, (right - left) * min(height[left], height[right]))
            if height[left] < height[right]:
                left += 1
            else:
                right -= 1
        return max_area

# Code (C++)

Approach 1:

// Two pointers.
class Solution {
public:
    int maxArea(vector<int>& height) {
        int head = 0;
        int tail = height.size() - 1;
        int max = 0;
        while (head < tail)
        {
            max = std::max(max, std::min(height[head], height[tail]) * (tail - head));
            if (height[head] < height[tail])
                head++;
            else
                tail--;
        }
        return max;
    }
};