Container With Most Water in DSA Javascript - Build from Scratch

Container With Most Water

📖 Scenario: Imagine you have a row of vertical lines on a graph. Each line's height is given by a number in a list. You want to find two lines that, together with the x-axis, form a container that holds the most water.This is like choosing two walls to hold water between them. The water amount depends on the shorter wall and the distance between the walls.

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

📋 What You'll Learn

Create an array called heights with the exact values: [1, 8, 6, 2, 5, 4, 8, 3, 7]

Create two variables called left and right to point to the start and end of the array

Use a while loop to move left and right pointers to find the max water container

Print the maximum water amount stored in a variable called maxArea

💡 Why This Matters

🌍 Real World

This problem models situations where you want to find the best boundaries to hold or contain something, like water or other resources, using minimal space.

💼 Career

Understanding two-pointer techniques and array traversal is important for coding interviews and solving optimization problems efficiently.

1

Create the heights array

Create an array called heights with these exact values: [1, 8, 6, 2, 5, 4, 8, 3, 7]

DSA Javascript

// Create the array heights with the given values
// Your code here

Hint

Use const heights = [...] to create the array with the exact numbers.

2

Set up pointers for left and right

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

DSA Javascript

const heights = [1, 8, 6, 2, 5, 4, 8, 3, 7];
// Create left and right pointers
// Your code here

Hint

Use let left = 0; and let right = heights.length - 1; to set pointers.

3

Find the maximum water container

Create a variable called maxArea and set it to 0. Use a while loop that runs while left is less than right. Inside the loop, calculate the area between left and right lines, update maxArea if the area is bigger, then move left or right pointer inward depending on which height is smaller.

DSA Javascript

const heights = [1, 8, 6, 2, 5, 4, 8, 3, 7];
let left = 0;
let right = heights.length - 1;
// Calculate maxArea using a while loop
// Your code here

Hint

Use Math.min to find the smaller height. Update maxArea if current area is bigger. Move the pointer of the smaller height inward.

4

Print the maximum water area

Print the value of maxArea using console.log(maxArea).

DSA Javascript

const heights = [1, 8, 6, 2, 5, 4, 8, 3, 7];
let left = 0;
let right = heights.length - 1;
let maxArea = 0;

while (left  maxArea) {
        maxArea = area;
    }
    if (heights[left] < heights[right]) {
        left++;
    } else {
        right--;
    }
}
// Print maxArea
// Your code here

Hint

Use console.log(maxArea) to show the result.

Practice

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

easy

A. The maximum area formed between two lines representing container walls.

B. The minimum height in the array.

C. The sum of all heights in the array.

D. The difference between the tallest and shortest lines.

Solution

1. Step 1: Understand the problem statement

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

2. Step 2: Identify what to maximize

   The area between two lines is width (distance between lines) times the height (minimum of the two lines). We want the maximum such area.

3. Final Answer:

   The maximum area formed between two lines representing container walls. -> Option A

4. Quick Check:

   Max area = max water container [OK]

Hint: Look for two lines maximizing min height * distance [OK]

Common Mistakes:

• Thinking it's about sum of heights
• Confusing max height with max area
• Ignoring the distance between lines

2. Which of the following JavaScript code snippets correctly initializes two pointers for the Container With Most Water problem?

easy

A. let left = 1, right = heights.length;

B. let left = -1, right = heights.length + 1;

C. let left = heights.length, right = 0;

D. let left = 0, right = heights.length - 1;

Solution

1. Step 1: Identify correct pointer initialization

   We start with left pointer at index 0 and right pointer at last index (length - 1).

2. Step 2: Check each option

   let left = 0, right = heights.length - 1; correctly sets left = 0 and right = heights.length - 1. Others use invalid indices.

3. Final Answer:

   let left = 0, right = heights.length - 1; -> Option D

4. Quick Check:

   Start pointers at array ends [OK]

Hint: Start pointers at array ends: 0 and length-1 [OK]

Common Mistakes:

• Using invalid indices outside array bounds
• Starting both pointers at same position
• Confusing length with last index

3. What is the output of the following JavaScript code?

function maxArea(height) {
  let left = 0, right = height.length - 1;
  let max = 0;
  while (left < right) {
    const area = Math.min(height[left], height[right]) * (right - left);
    max = Math.max(max, area);
    if (height[left] < height[right]) {
      left++;
    } else {
      right--;
    }
  }
  return max;
}

console.log(maxArea([1,8,6,2,5,4,8,3,7]));

medium

A. 56

B. 49

C. 64

D. 72

Solution

1. Step 1: Trace the two-pointer approach

   Start with left=0 (height=1), right=8 (height=7), area = min(1,7)*8=8. Max=8.

2. Step 2: Move pointers and update max

   Since height[left]

3. Final Answer:

   49 -> Option B

4. Quick Check:

   Max area for input = 49 [OK]

Hint: Move pointer at smaller height inward to find max area [OK]

Common Mistakes:

• Calculating area incorrectly using max height
• Not updating max area properly
• Moving wrong pointer in the loop

4. Identify the bug in this JavaScript function for Container With Most Water:

function maxArea(height) {
  let left = 0, right = height.length - 1;
  let max = 0;
  while (left < right) {
    const area = Math.min(height[left], height[right]) * (right - left);
    max = Math.max(max, area);
    if (height[left] < height[right]) {
      left++;
    } else {
      right++;
    }
  }
  return max;
}

medium

A. The right pointer should decrement (right--) instead of increment (right++).

B. The area calculation should use max instead of min height.

C. The max variable should be initialized to Infinity.

D. The loop condition should be left < right, not left <= right.

Solution

1. Step 1: Check pointer movement logic

   The right pointer is incorrectly incremented (right++), which moves it out of array bounds.

2. Step 2: Correct pointer movement

   When height[left] >= height[right], right pointer should move left (right--) to find a potentially larger area.

3. Final Answer:

   The right pointer should decrement (right--) instead of increment (right++). -> Option A

4. Quick Check:

   Right pointer moves left to reduce width [OK]

Hint: Right pointer moves left (decrement) to find max area [OK]

Common Mistakes:

• Using right++ instead of right--
• Loop condition allowing left == right
• Incorrect area calculation using max height

5. Given an array of heights, how would you modify the Container With Most Water algorithm to also return the indices of the two lines forming the container with the maximum water?

hard

A. Sort the array and return indices of the tallest two lines.

B. Return the first two indices always, as they form the max container.

C. Track indices when updating max area and return them along with max area.

D. Use a nested loop to check all pairs and return the last pair found.

Solution

1. Step 1: Add variables to track indices

   Initialize variables to store left and right indices when max area updates.

2. Step 2: Update indices when max area changes

   Inside the loop, when a new max area is found, save current left and right indices.

3. Step 3: Return max area and indices

   Return an object or array containing max area and the two indices after loop ends.

4. Final Answer:

   Track indices when updating max area and return them along with max area. -> Option C

5. Quick Check:

   Save indices on max update for output [OK]

Hint: Save indices when max area updates during pointer moves [OK]

Common Mistakes:

• Returning fixed indices without checking area
• Sorting array loses original indices
• Using nested loops increases time complexity