## 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 the line i is at (i, ai) and (i, 0). Find two lines, which, together with the x-axis forms a container, such that the container contains the most water.

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

Example 3:

Input: height = [4,3,2,1,4]
Output: 16

Example 4:

Input: height = [1,2,1]
Output: 2

Constraints:

• n == height.length
• 2 <= n <= 105
• 0 <= height[i] <= 104

Solution

Start by evaluating the widest container, using the first and the last line. All other possible containers are less wide, so to hold more water, they need to be higher. Thus, after evaluating that widest container, skip lines at both ends that don't support a higher height. Then evaluate that new container we arrived at. Repeat until there are no more possible containers left.

int maxArea(vector<int>& height){
    int water = 0;
    int i = 0, j = height.size()-1;
    while(i<j){
        int h = min(height[i],height[j]);
        water = max(water,(j-i)*h);
        while(height[i] <= h && i<j) i++;
        while(height[j] <= h && i<j) j--;
    }
    return water;
}