7 Steps to Efficiently Solve the Problem”Container with Most Water: A Comprehensive Guide to Solving the Problem Efficiently

The “Container with Most Water” problem is a popular algorithmic challenge often found in coding interviews, competitive programming, and computer science exercises. It focuses on finding two lines on a graph that can contain the maximum amount of water. This problem is an excellent example of using two-pointer techniques to achieve an optimal solution.

In this article, we’ll explore what the problem entails, how to break it down, and efficient strategies to solve it, along with code implementation and examples.

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Problem Statement: What is the Container with Most Water Problem?

Imagine you are given an array height consisting of n non-negative integers, where each integer represents the height of a vertical line on the x-axis. The objective is to find two lines such that they, along with the x-axis, form a container that can hold the most water. You need to return the maximum area of water that can be contained by any such pair of lines.

Mathematical Representation:

• Each element in the array represents the height of a line at that index.
• The area of water contained between any two lines i and j (where i < j) is determined by:
  [ \text{Area} = \text{min}(\text{height}[i], \text{height}[j]) \times (j – i) ]
  The width of the container is (j - i), and the height is the minimum of the two heights at indices i and j.

Example:

1. Input:
   height = [1, 8, 6, 2, 5, 4, 8, 3, 7] Output: 49 Explanation:
   The two lines at indices 1 and 8 (heights 8 and 7) form the container that can hold the maximum water, with an area of:
   [ \text{Area} = \min(8, 7) \times (8 – 1) = 7 \times 7 = 49 ]
2. Input:
   height = [1, 1] Output: 1 Explanation:
   The two lines at indices 0 and 1 form a container with an area of 1.

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Key Observations and Constraints

1. Maximizing the Area:
   To find the maximum area, you need to find the widest possible container with the greatest height.
2. Choosing Lines Efficiently:
   If you use a brute force approach by comparing all pairs of lines, the time complexity will be O(n^2), which is inefficient for large inputs.
3. Optimal Solution with Two Pointers:
   The most efficient way to solve this problem is by using a two-pointer technique.

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Two-Pointer Technique: The Optimal Approach

Why Use Two Pointers?

The key insight is that the area formed by the container is determined by the shorter line. Moving the pointer associated with the shorter line inward might lead to a better area, as it increases the possibility of finding a taller line.

Steps to Solve Using Two Pointers

1. Initialize Two Pointers:

• left pointer at the beginning (index 0).
• right pointer at the end (index n-1).

1. Calculate Area:

• Calculate the area between the two lines at left and right.
• The area is given by:
  [ \text{Area} = \min(\text{height}[left], \text{height}[right]) \times (right – left) ]

1. Update the Maximum Area:

• Keep track of the maximum area found so far.

1. Move Pointers Based on Heights:

• If height[left] < height[right], move left pointer inward (left++) to find a potentially taller line.
• Otherwise, move the right pointer inward (right--).

1. Repeat Until Pointers Meet:

• Continue the process until left is not less than right.

Time Complexity

• The algorithm runs in O(n) time complexity because each pointer is moved exactly once from one end to the other.
• The space complexity is O(1) as no additional space is used.

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Code Implementation

Here’s a Python implementation of the two-pointer approach:

def maxArea(height):
    left = 0
    right = len(height) - 1
    max_area = 0

    while left < right:
        # Calculate the area
        current_area = min(height[left], height[right]) * (right - left)
        max_area = max(max_area, current_area)

        # Move the pointer based on the height
        if height[left] < height[right]:
            left += 1
        else:
            right -= 1

    return max_area

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

Explanation:

1. Initialization:

• left is at the start (0), and right is at the end (n-1).

1. Calculate Current Area:

• Compute the area between the two lines at left and right.
• Update max_area if the current_area is greater.

1. Adjust Pointers:

• Move the pointer pointing to the shorter line inward.

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Edge Cases to Consider

1. All Heights are the Same:
   If all heights in the array are equal, the maximum area will be determined by the farthest distance between left and right.
2. Height Array is of Length 2:
   If the height array contains only two elements, the area will be simply min(height[0], height[1]).
3. Single Line or Empty Array:
   If the array has less than two lines, the maximum area is 0 as no container can be formed.

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Understanding the Optimality of the Two-Pointer Technique

The two-pointer technique works efficiently because:

• Maximizing Width First: The widest container is considered first, and the height is then adjusted by moving the pointer associated with the shorter line.
• Ensuring Best Possible Area: By always moving the pointer of the shorter line, you aim to find a potentially taller line that could increase the area.

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Real-World Applications

1. Optimizing Storage in Containers:
   This problem is conceptually similar to finding the optimal way to store liquids in containers to maximize capacity.
2. Landscape Planning:
   If you imagine each line as a building in a skyline, this approach could help in understanding how to maximize the space between buildings for certain construction constraints.
3. Stock Market Visualization:
   The concept can be visualized in stock price charts where the height of the line represents stock prices over time, and you need to find the maximum difference (or profit) between any two days.

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FAQs

1. Why do we use two pointers for this problem?
The two-pointer approach is efficient because it maximizes the width of the container first and then tries to maximize the height by moving the pointer pointing to the shorter line. This approach reduces the time complexity to O(n).

2. Can we use a brute force solution to solve this problem?
Yes, a brute force solution is possible by comparing all pairs of lines, but it would have a time complexity of O(n^2), which is inefficient for larger inputs.

3. What happens if all heights are equal?
If all heights are equal, the maximum area will be determined by the farthest distance between the two lines (right - left).

4. What is the minimum number of lines required to form a container?
At least two lines are required to form a container. If there are fewer than two lines, no container can be formed, and the maximum area will be 0.

5. How does this problem relate to the sliding window technique?
While both the two-pointer and sliding window techniques involve moving pointers, they are used differently. In this problem, two pointers are used to optimize the container size by considering both width and height simultaneously.

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1. LeetCode – Container With Most Water Problem
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5. YouTube – Visual Explanation of Container with Most Water
   YouTube – Container With Most Water Explanation
   A visual and comprehensive explanation of the problem, perfect for those who learn better through video content.