The two-pointer technique

Maximizing Container Area with Two Pointers in Go

In algorithms that work with arrays or lists, the two-pointer technique stands out for its efficiency. In this article, we will apply it to the classic "Container With Most Water" problem, which seeks the largest area between two vertical lines on a graph.

Problem Description

Given an array of non-negative integers representing the heights of vertical lines, find the pair of lines that, together with the x-axis, form the container with the largest area.

Example

Consider the array height = [1, 8, 6, 2, 5, 4, 8, 3, 7]. The objective is to determine which two lines generate the maximum area.

Two Pointers Technique

The technique uses two pointers, one at the beginning and one at the end of the array, moving them iteratively towards the center to find the optimal solution.

Step by Step

1. Startup:

   • maxArea is initialized to 0, storing the largest area found so far.
   • Two pointers, l (left) and r (right), are positioned at the beginning and end of the array, respectively.

2. Iteration:

   • The loop continues as long as l is less than r.
   • The area between the lines in l and r is calculated as min(height[l], height[r]) * (r - l).
   • maxArea is updated if the calculated area is larger.

3. Movement of Pointers:

   • To optimize the search, the pointer pointing to the smaller line is moved:
     • If height[l] , increment <code>l.
     • Otherwise, decrement r.

4. Return:

   • When l and r intersect, the loop ends, and maxArea contains the maximum area.

Detailed Example

Let's analyze the array height = [1, 8, 6, 2, 5, 4, 8, 3, 7]:

1. Startup:

   • maxArea = 0
   • l = 0 (height 1), r = 8 (height 7)

2. First Iteration:

   • Area: min(1, 7) * (8 - 0) = 8
   • maxArea = max(0, 8) = 8
   • Move l (because height[l] )

3. Second Iteration:

   • l = 1 (height 8), r = 8 (height 7)
   • Area: min(8, 7) * (8 - 1) = 49
   • maxArea = max(8, 49) = 49
   • Move r

...and so on, repeating the process until the hands meet.

The end result will be maxArea = 49.

Go solution

Follow the Go code implementing the technique:

package maxarea

func maxArea(height []int) int {
    maxArea := 0
    l, r := 0, len(height)-1

    for l < r {
        area := min(height[l], height[r]) * (r - l)
        maxArea = max(maxArea, area)
        if height[l] < height[r] {
            l++
        } else {
            r--
        }
    }
    return maxArea
}

func min(a, b int) int {
    if a < b {
        return a
    }
    return b
}

func max(a, b int) int {
    if a > b {
        return a
    }
    return b
}

Conclusion

The two pointer technique offers an efficient solution to problems involving arrays. In the case of "Container With Most Water", it guarantees linear time complexity, making it an ideal approach.

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