Container With Most Water in DSA Go - Build from Scratch

Container With Most Water

📖 Scenario: Imagine you have a row of vertical lines on a flat surface. Each line has a certain height. You want to find two lines that, together with the surface, can hold the most water between them.

🎯 Goal: Build a program that finds the maximum amount of water that can be trapped between two lines from a list of line heights.

📋 What You'll Learn

Create a slice called heights with the exact values: 1, 8, 6, 2, 5, 4, 8, 3, 7

Create two integer variables called left and right to represent the start and end indexes of the slice

Write a loop that moves left and right pointers to find the maximum water container

Print the maximum water amount as an integer

💡 Why This Matters

🌍 Real World

This problem models situations where you want to find the best container or boundary to hold the maximum volume, such as in water storage or land plotting.

💼 Career

Understanding two-pointer techniques and array manipulation is important for coding interviews and real-world software development tasks involving optimization.

1

Create the heights slice

Create a slice of integers called heights with these exact values: 1, 8, 6, 2, 5, 4, 8, 3, 7

DSA Go

# Create the heights slice with the given values
# Your code here

Hint

Use heights := []int{...} to create the slice with the exact numbers.

2

Set left and right pointers

Create two integer variables called left and right. Set left to 0 and right to the last index of heights.

DSA Go

package main

func main() {
    heights := []int{1, 8, 6, 2, 5, 4, 8, 3, 7}
    // Create left and right pointers
    // Your code here
}

Hint

Use left := 0 and right := len(heights) - 1 to set the pointers.

3

Find the maximum water container

Create an integer variable called maxArea and set it to 0. Use a for loop that runs while left < right. Inside the loop, calculate the area between left and right lines, update maxArea if the area is larger, and move the pointer that points to the shorter line.

DSA Go

package main

func main() {
    heights := []int{1, 8, 6, 2, 5, 4, 8, 3, 7}
    left := 0
    right := len(heights) - 1
    // Find maxArea using a loop
    // Your code here
}

Hint

Use a loop with left < right. Calculate area as height * width. Move the pointer with the smaller height.

4

Print the maximum water amount

Use fmt.Println(maxArea) to print the maximum water amount. Remember to import the fmt package at the top.

DSA Go

package main

func main() {
    heights := []int{1, 8, 6, 2, 5, 4, 8, 3, 7}
    left := 0
    right := len(heights) - 1
    maxArea := 0

    for left  maxArea {
            maxArea = area
        }

        if heights[left] < heights[right] {
            left++
        } else {
            right--
        }
    }
    // Print maxArea
    // Your code here
}

Hint

Use fmt.Println(maxArea) to show the result. Don't forget to import fmt.

Practice

1. What does the problem Container With Most Water ask you to find given an array of heights?

easy

A. The sum of all heights

B. The minimum height in the array

C. The maximum area of water that can be contained between two lines

D. The number of lines taller than a given height

Solution

1. Step 1: Understand the problem statement

   The problem asks to find two lines which together with the x-axis form a container that holds the most water.

2. Step 2: Identify what to calculate

   We calculate the area between two lines as the width between them times the smaller height.

3. Final Answer:

   The maximum area of water that can be contained between two lines -> Option C

4. Quick Check:

   Container with most water = maximum area [OK]

Hint: Focus on max area between two heights, not sum or count [OK]

Common Mistakes:

• Confusing max area with sum of heights
• Ignoring the smaller height when calculating area
• Thinking about counting lines instead of area

2. Which of the following Go code snippets correctly declares a slice of heights for the Container With Most Water problem?

easy

A. heights := []int{1, 8, 6, 2, 5, 4, 8, 3, 7}

B. heights := [1, 8, 6, 2, 5, 4, 8, 3, 7]

C. heights := (1, 8, 6, 2, 5, 4, 8, 3, 7)

D. heights := {1, 8, 6, 2, 5, 4, 8, 3, 7}

Solution

1. Step 1: Recall Go slice declaration syntax

   In Go, slices are declared using []type{values}, for example []int{1,2,3}.

2. Step 2: Check each option

   heights := []int{1, 8, 6, 2, 5, 4, 8, 3, 7} uses []int{...} which is correct. Options A, B, C use invalid syntax for slices.

3. Final Answer:

   heights := []int{1, 8, 6, 2, 5, 4, 8, 3, 7} -> Option A

4. Quick Check:

   Slice declaration uses []type{values} [OK]

Hint: Use []int{...} to declare slices in Go [OK]

Common Mistakes:

• Using parentheses or braces instead of [] for slices
• Confusing arrays and slices syntax
• Missing type declaration in slice

3. What is the output of the following Go code snippet that implements the Container With Most Water logic?

package main
import "fmt"
func maxArea(height []int) int {
    left, right := 0, len(height)-1
    maxArea := 0
    for left < right {
        width := right - left
        var h int
        if height[left] < height[right] {
            h = height[left]
            left++
        } else {
            h = height[right]
            right--
        }
        area := width * h
        if area > maxArea {
            maxArea = area
        }
    }
    return maxArea
}
func main() {
    heights := []int{1,8,6,2,5,4,8,3,7}
    fmt.Println(maxArea(heights))
}

medium

A. 49

B. 56

C. 48

D. 64

Solution

1. Step 1: Understand the two-pointer approach

   Start with left=0 and right=8, calculate area and move the pointer with smaller height inward.

2. Step 2: Trace the iterations

   Initial area: min(1,7)*8=8 Move left to 1 Next area: min(8,7)*7=49 Move right to 7 Next area: min(8,3)*6=18 Move right to 6 Next area: min(8,8)*5=40 Move right to 5 Next area: min(8,4)*4=16 Move right to 4 Next area: min(8,5)*3=15 Move right to 3 Next area: min(8,2)*2=4 Move right to 2 Next area: min(8,6)*1=1 Move left to 2 Max area found is 49.

3. Final Answer:

   49 -> Option A

4. Quick Check:

   Max area calculated by two-pointer = 49 [OK]

Hint: Use two pointers moving inward, track max area [OK]

Common Mistakes:

• Calculating area with max height instead of min
• Not moving the pointer with smaller height
• Confusing width calculation

4. Identify the bug in this Go function for Container With Most Water and select the fix:

func maxArea(height []int) int {
    left, right := 0, len(height)-1
    maxArea := 0
    for left <= right {
        width := right - left
        var h int
        if height[left] < height[right] {
            h = height[left]
            left++
        } else {
            h = height[right]
            right--
        }
        area := width * h
        if area > maxArea {
            maxArea = area
        }
    }
    return maxArea
}

medium

A. Use maxArea = area without checking if area > maxArea

B. Initialize maxArea to -1 instead of 0

C. Increment both left and right pointers inside the loop

D. Change loop condition to 'left < right' instead of 'left <= right'

Solution

1. Step 1: Analyze the loop condition

   The loop runs while left <= right, which allows left and right to be equal, causing width to be zero or negative.

2. Step 2: Fix the loop condition

   Changing to left < right ensures width is always positive and valid for area calculation.

3. Final Answer:

   Change loop condition to 'left < right' instead of 'left <= right' -> Option D

4. Quick Check:

   Loop must run while left < right [OK]

Hint: Use strict less than for pointers to avoid zero width [OK]

Common Mistakes:

• Using <= instead of < causing invalid area
• Updating both pointers at once
• Not checking if area is greater before updating maxArea

5. Given the heights slice []int{4, 3, 2, 1, 4}, what is the maximum water container area? Use the two-pointer approach and explain why the maximum area is found at those positions.

hard

A. 12

B. 16

C. 8

D. 6

Solution

1. Step 1: Initialize pointers and calculate areas

   Start with left=0 (height=4), right=4 (height=4), width=4, area=4*4=16.

2. Step 2: Move pointers and check other areas

   Since heights[left] == heights[right], move right pointer left: Now left=0, right=3 (height=1), width=3, area=min(4,1)*3=3 Move right to 2 (height=2), area= min(4,2)*2=4 Move right to 1 (height=3), area= min(4,3)*1=3 Max area remains 16.

3. Final Answer:

   16 -> Option B

4. Quick Check:

   Max area between positions 0 and 4 = 16 [OK]

Hint: Max area often at widest distance with min height [OK]

Common Mistakes:

• Not considering both ends first
• Moving wrong pointer when heights are equal
• Calculating area incorrectly with wrong width