When it comes to technical interviews, the Container With Most Water problem stands out as a classic programming challenge. It demands a smart combination of problem-solving, algorithm optimization, and array traversal. In this article, we will explore different approaches to find the max area that can be formed between two vertical lines on a graph.

Understanding the Problem

You are given an array where each element represents the height of vertical lines drawn on the x-axis. Your goal is to select two lines that, along with the x-axis, create a water container with the maximum area.
The area depends on the height and width between the two lines:

• Height = the shorter of the two lines.
• Width = the distance between the two lines (indices difference).

Thus, the max area is determined by the formula:
Area = height × width

Naive Approach: Brute Force

The simplest way to solve this programming challenge is by checking every pair of vertical lines:

1. Calculate the area formed between every pair.
2. Keep track of the max area found.

However, this approach involves nested loops leading to O(n²) time complexity, making it inefficient for large arrays.

Efficient Solution: Two Pointers Technique

To achieve algorithm optimization, we use the two pointers method. This method significantly reduces the time complexity to O(n) by intelligently navigating the array traversal:

How the Two Pointers Work:

• Place one pointer at the start (left) and another at the end (right) of the array.
• Calculate the area between the two vertical lines.
• Move the pointer that points to the shorter line towards the center.
• Repeat the process while updating the max area at each step.

By doing this, we ensure an efficient solution without missing any potential larger containers.

Step-by-Step Example

Consider the height array: [1,8,6,2,5,4,8,3,7]

• Start with left = 0 and right = 8.
• Calculate the area:
  min(1,7) * (8-0) = 1*8 = 8
• Move left pointer since height at left is smaller.
• Continue recalculating the area at each move until pointers meet.

The two pointers ensure that all promising combinations are evaluated quickly.

Python Code Implementation

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def maxArea(height):
    left, right = 0, len(height) - 1
    max_area = 0
    
    while left < right:
        current_area = min(height[left], height[right]) * (right - left)
        max_area = max(max_area, current_area)
        
        if height[left] < height[right]:
            left += 1
        else:
            right -= 1
            
    return max_area

# Example Usage
heights = [1,8,6,2,5,4,8,3,7]
print(maxArea(heights))  # Output: 49

This solution offers algorithm optimization by minimizing unnecessary calculations.

Important Points to Remember

• Always choose the height of the shorter line.
• Move the pointer from the side of the smaller line to maximize area chances.
• Focus on efficient solution strategies when dealing with large datasets.

Time and Space Complexity

• Time Complexity: O(n) because each element is visited at most once.
• Space Complexity: O(1) as we use only constant extra space.

Real-World Applications

This type of problem-solving can be extended to real-world scenarios like:

• Optimizing storage tanks design.
• Maximizing usable space between barriers.
• Efficient layout planning in software and engineering.

Conclusion

The Container With Most Water is an excellent programming challenge to test your skills in array traversal, algorithm optimization, and using the two pointers strategy. Understanding how height and width interplay, along with smart pointer movement, allows you to find an efficient solution easily.
Mastering this problem improves your readiness for many interview and problem-solving scenarios!