Container With Most Water

Background Knowledge

The "Container With Most Water" problem is a classic example of a problem that can be solved using the two pointers technique. This technique is commonly used in array and string problems where we need to find a pair of elements that satisfy certain conditions. The key idea behind the two pointers technique is to maintain two pointers, typically one at the start and one at the end of the array, and move them towards each other based on certain conditions.

In this problem, we are given an array of non-negative integers representing the heights of vertical lines. The goal is to find two lines that together with the x-axis form a container holding the most water. The amount of water that can be held by a container formed by two lines is determined by the minimum height of the two lines and the distance between them. This can be calculated using the formula $w = \min(h_1, h_2) \times d$, where $w$ is the amount of water, $h_1$ and $h_2$ are the heights of the two lines, and $d$ is the distance between them.

To solve this problem, we need to have a good understanding of how to calculate the area of a container formed by two lines and how to use the two pointers technique to find the pair of lines that maximizes this area. We also need to consider how to efficiently move the pointers to find the optimal solution.

Algorithm/Approach

The algorithm pattern used to solve this type of problem is the two pointers technique. The general approach is to initialize two pointers, one at the start and one at the end of the array, and then move them towards each other based on certain conditions. In this case, we will move the pointers based on the height of the lines and the distance between them.

Step-by-Step Strategy

To implement the solution, we can follow these steps:

• Initialize two pointers, left and right, to the start and end of the array, respectively.
• Calculate the area of the container formed by the lines at the left and right pointers.
• Move the pointer that points to the shorter line towards the other pointer.
• Repeat the calculation and movement steps until the left and right pointers meet.
• Keep track of the maximum area found during the process.

Common Pitfalls

Some common pitfalls to watch out for when implementing the solution include:

• Not correctly calculating the area of the container formed by the two lines.
• Not moving the pointers correctly based on the height of the lines and the distance between them.
• Not keeping track of the maximum area found during the process.

Time & Space Complexity

The expected time complexity for this problem is $O(n)$, where $n$ is the number of lines, since we are scanning the array once. The expected space complexity is $O(1)$, since we are only using a constant amount of space to store the pointers and the maximum area.