Container With Most Water

The "Container With Most Water" problem is a well-known question in the field of algorithms and data structures, particularly emphasizing the use of the two-pointer technique. This problem focuses on finding an optimal solution by understanding how to efficiently navigate the space of possible solutions. Let's explore this problem from its basic principles.

Problem Statement:

  1. Input: An array of non-negative integers, where each integer represents the height of a line drawn at that point.
  2. Goal: Find two lines, which, together with the x-axis, form a container that would hold the greatest amount of water.

Fundamental Concepts:

  • Array: A linear data structure that holds a collection of elements.
  • Volume Calculation: The amount of water a container can hold is determined by the distance between lines (width) and the height of the shorter line (since the water would overflow over the shorter line).

Solution Approach:

  1. Two-Pointer Technique:

    • Idea: Start with two pointers, one at the beginning and one at the end of the array. The initial container encompasses the entire array.
    • Implementation: Move the pointers closer together, always moving the pointer that points to the shorter line.
    • Rationale: By moving the shorter line, we are trying to find a taller line that can potentially hold more water. Moving the taller line would not help, as the height of the shorter line limits the container's capacity.
  2. Volume Calculation:

    • Formula: Volume = width * height
    • Width: Distance between the two pointers.
    • Height: Height of the shorter line (between the two pointers).
  3. Time and Space Complexity:

    • Time Complexity: O(n), where n is the length of the input array. Each element is visited only once.
    • Space Complexity: O(1), as the solution requires a constant amount of space.

Example Implementation in Python:

def max_area(height):
    Finds the maximum area of water that can be contained.

    height (List[int]): An array of integers where each integer represents the height of a line.

    int: The maximum area of water that can be contained.
    max_area = 0
    left, right = 0, len(height) - 1  # Two pointers

    while left < right:
        # Calculate the area and update max_area if necessary.
        width = right - left
        area = width * min(height[left], height[right])
        max_area = max(max_area, area)

        # Move the pointer pointing to the shorter line.
        if height[left] < height[right]:
            left += 1
            right -= 1

    return max_area

This implementation:

  • Initializes two pointers at the opposite ends of the array.
  • Iterates until the pointers meet, at each step calculating the area formed between the lines at the pointers and updating the maximum area found.
  • Moves the pointer at the shorter line towards the other pointer in each iteration, seeking a potentially taller line that could increase the area.