Maximum water in container problem

We want to find out how much water can be held between two sticks. The heights of our sticks are:

height = [1, 2, 4, 3]
Indexes ( 0, 1, 2, 3)

We use two pointers: one at the left end and one at the right end.
Formula for water:
area = min(height[left], height[right]) × (right - left)

Step 1:

left = 0 , right = 3

• height[left] = 1
• height[right] = 3
min(height[left], height[right]) = min(1,3) = 1
width = right - left = 3 - 0 = 3
area = 1 × 3 = 3
  Max area so far = 3
  Since height[left] < height[right], move left → left = 1.

Step 2:

left = 1 , right = 3

• height[left] = 2
• height[right] = 3
min(height[left], height[right]) = min(2,3) = 2
width = 3 - 1 = 2
area = 2 × 2 = 4
  Max area so far = 4
  Since height[left] < height[right], move left → left = 2.

Step 3:

left = 2 , right = 3

• height[left] = 4
• height[right] = 3
min(height[left], height[right]) = min(4,3) = 3
width = 3 - 2 = 1
area = 3 × 1 = 3
  Max area still = 4
  Since height[left] > height[right], move right → right = 2.

Step 4:

left = 2 , right = 2
Both pointers meet, so stop.

Final Answer:

The maximum water area = 4, between left = 1 and right = 3

• height[left] = 2
• height[right] = 3

Golang implementation of the problem

package main

import (
	"fmt"
)

func maxArea(height []int) int {
	left := 0
	right := len(height) - 1
	maxArea := 0

	for left < right {
		// take the smaller height between left and right
		h := height[left]
		if height[right] < h {
			h = height[right]
		}

		// width = distance between left and right
		width := right - left
		area := h * width

		// update maximum area if needed
		if area > maxArea {
			maxArea = area
		}

		// move the pointer with smaller height
		if height[left] < height[right] {
			left++
		} else {
			right--
		}
	}

	return maxArea
}

func main() {
	height := []int{1, 2, 4, 3}
	fmt.Println("Maximum Area:", maxArea(height)) // Output: 4
}