Solving the Container With Most Water Problem

The "Container With Most Water" problem is a classic example of applying the two-pointer technique to find an optimal solution efficiently. This problem is common in coding interviews and challenges your understanding of array manipulation.

The Problem

You are given an array height, representing the height of vertical lines on a chart. The task is to find the two lines that, together with the x-axis, form a container that can hold the maximum amount of water.

The Solution

Here's a straightforward JavaScript implementation:

The solution uses two pointers, left and right, to traverse the array from both ends. The maxArea variable keeps track of the maximum area found so far. The while loop continues until the two pointers meet. The width variable represents the width of the container. The currentArea variable represents the area of the container formed by the two pointers. The Math.min function is used to find the minimum height of the two pointers. The Math.max function is used to update the maxArea variable. The if statement checks if the height of the left pointer is less than the height of the right pointer. If so, the left pointer is moved one step to the right. Otherwise, the right pointer is moved one step to the left.

Breaking Down the Solution

• Two-Pointer Technique: Start with two pointers, one at the beginning (left) and one at the end (right) of the array. These pointers represent potential boundaries of the water container.

• Calculate Area: At each step, calculate the area formed by the lines at the left and right pointers. The area is determined by the distance between the pointers (width) and the height of the shorter line.

• Update Max Area: Keep track of the maximum area encountered so far. If the current area is larger than the maxArea, update maxArea.

• Move Pointers: Move the pointer pointing to the shorter line towards the other pointer. This is because moving the shorter line could potentially find a taller line and thus increase the area.

• Repeat: Continue the process until the left and right pointers meet, meaning all potential pairs have been evaluated.

• Return Result: The maxArea at the end of the iteration process will be the maximum amount of water the container can store.

Conclusion

This problem is an excellent demonstration of the two-pointer technique's effectiveness in solving array-based challenges. It particularly highlights how optimal solutions often involve comparing pairs of elements and adjusting strategies based on their comparison. The "Container With Most Water" problem is not just about understanding arrays, but also about grasping the significance of element positions and how they relate to an optimal solution.

Rust Solution

Rust's index arithmetic forces a cast the JS version never needs: height.len() - 1 produces an unsigned index, so the width must go through (right - left) as i32 to become a signed i32 before it multiplies the heights. The shorter wall comes from height[left].min(height[right]), where min is an inherent method called on the i32 value itself rather than a free Math.min. The parameter height: Vec<i32> is taken by value, every reassigned binding needs let mut, and the function yields its result as the trailing max_area expression with no return keyword.

Go Solution

Go folds its while loop into for left < right — for is the language's only loop keyword. There is no Math.min here: the shorter wall is picked out longhand into var h int via an explicit if height[left] < height[right] / else, and that same comparison is then repeated to advance the pointers. := short declarations infer each local's type from its initializer (maxArea := 0, right := len(height) - 1), the argument is a []int slice sized by the builtin len, and left++ / right-- are statements rather than expressions.