# 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;
}
};