Container With Most Water in DSA Typescript - Build from Scratch

Container With Most Water

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

🎯 Goal: You will write a program that finds the maximum amount of water that can be trapped between two lines from 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 represent the start and end of the array.

Use a while loop to move left and right pointers and calculate the area between lines.

Print the maximum area found.

💡 Why This Matters

🌍 Real World

This problem models situations like finding the best place to build a container or reservoir between walls or barriers.

💼 Career

Understanding two-pointer techniques and array manipulation is useful in many coding interviews and real-world programming tasks.

1

Create the heights array

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

DSA Typescript

# 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 Typescript

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

Calculate max water area using two pointers

Create a variable called maxArea and set it to 0. Use a while loop with the condition left < right. Inside the loop, calculate the area between heights[left] and heights[right]. Update maxArea if the current area is larger. Move left forward if heights[left] is less than heights[right], otherwise move right backward.

DSA Typescript

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

Hint

Use a while loop and update maxArea inside it. Move left or right based on which height is smaller.

4

Print the maximum water area

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

DSA Typescript

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 heights

B. The minimum height in the array

C. The sum of all heights in the array

D. The number of lines taller than a given height

Solution

1. Step 1: Understand the problem goal

   The problem asks to find two lines that together with the x-axis form a container holding the most water, which means the largest area between two heights.

2. Step 2: Identify what area means here

   The area is calculated by the distance between two lines multiplied by the smaller height of the two lines.

3. Final Answer:

   The maximum area formed between two lines representing heights -> Option A

4. Quick Check:

   Container area = max area between two heights [OK]

Hint: Look for two lines maximizing width and min height [OK]

Common Mistakes:

• Confusing area with sum of heights
• Choosing tallest lines without considering distance
• Ignoring the smaller height between two lines

2. Which of the following TypeScript function signatures correctly represents a function to solve the Container With Most Water problem?

easy

A. function maxArea(height: number[]): number

B. function maxArea(height: string): string

C. function maxArea(height: number): number[]

D. function maxArea(height: boolean[]): boolean

Solution

1. Step 1: Identify input and output types

   The input is an array of numbers representing heights, so type is number[]. The output is a single number representing max area.

2. Step 2: Match function signature

   function maxArea(height: number[]): number matches input as number[] and output as number, which is correct for this problem.

3. Final Answer:

   function maxArea(height: number[]): number -> Option A

4. Quick Check:

   Input array of numbers, output number [OK]

Hint: Input is array of numbers, output is a number [OK]

Common Mistakes:

• Using wrong input type like string or boolean
• Returning array instead of single number
• Mismatching input and output types

3. What will be the output of the following TypeScript code?

function maxArea(height: number[]): number {
  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: Understand the two-pointer approach

   Start with left=0 and right=8 (length-1). Calculate area = min(height[left], height[right]) * (right-left). Move pointer with smaller height inward.

2. Step 2: Trace key iterations

   Initial area: min(1,7)*8=8
   Move left to 1 (height=8), area: min(8,7)*7=49 (max updated)
   Move right to 7 (height=3), area: min(8,3)*6=18
   Move right to 6 (height=8), area: min(8,8)*5=40
   Move right to 5 (height=4), area: min(8,4)*4=16
   Move right to 4 (height=5), area: min(8,5)*3=15
   Move right to 3 (height=2), area: min(8,2)*2=4
   Move right to 2 (height=6), area: min(8,6)*1=6
   Max area found is 49.
   Continue moving left pointer:
   left=2 (height=6), right=2 (height=6) loop ends.
   Actually, the maximum area is found earlier at 49, but let's check if any larger area exists.
   Check area at left=1, right=8: min(8,7)*7=49
   Check area at left=1, right=6: min(8,8)*5=40
   Check area at left=2, right=8: min(6,7)*6=42
   Check area at left=1, right=7: min(8,3)*6=18
   Check area at left=0, right=8: min(1,7)*8=8
   Check area at left=1, right=5: min(8,4)*4=16
   Check area at left=1, right=4: min(8,5)*3=15
   Check area at left=1, right=3: min(8,2)*2=4
   Check area at left=1, right=2: min(8,6)*1=6
   Check area at left=2, right=7: min(6,3)*5=15
   Check area at left=2, right=6: min(6,8)*4=24
   Check area at left=3, right=6: min(2,8)*3=6
   Check area at left=4, right=6: min(5,8)*2=10
   Check area at left=5, right=6: min(4,8)*1=4
   Maximum area is 49.
   However, the known correct answer for this input is 49, but the official LeetCode solution and common knowledge say the max area is 49.
   Therefore, correct answer is 49.

3. Final Answer:

   49 -> Option B

4. Quick Check:

   Max area = 49 [OK]

Hint: Use two pointers, move smaller height inward [OK]

Common Mistakes:

• Calculating area with max height instead of min
• Not updating max area correctly
• Moving wrong pointer in the loop

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

function maxArea(height: number[]): number {
  let left = 0;
  let 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 loop condition should be left < right, not left <= right

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

C. The pointers should move based on height[left] < height[right], not >

D. The max variable should be initialized to -Infinity

Solution

1. Step 1: Check pointer movement logic

   The algorithm moves the pointer with smaller height inward to try larger area. The code moves left pointer if height[left] > height[right], which is incorrect.

2. Step 2: Correct pointer movement condition

   It should move left pointer if height[left] < height[right], else move right pointer. This ensures we discard the smaller height side.

3. Final Answer:

   The pointers should move based on height[left] < height[right], not > -> Option C

4. Quick Check:

   Move pointer of smaller height [OK]

Hint: Move pointer at smaller height side [OK]

Common Mistakes:

• Using wrong comparison operator for pointer movement
• Including equal condition in loop causing extra iteration
• Using max height instead of min for area

5. Given an array of heights, how would you modify the two-pointer approach to find the maximum container area if the container must include the first line (index 0)?

hard

A. Calculate area for all pairs including index 0 using nested loops

B. Fix right pointer at 0 and move left pointer from start to end, calculating max area

C. Use two pointers starting at 1 and end-1, ignoring index 0

D. Fix left pointer at 0 and move right pointer from end to start, calculating max area

Solution

1. Step 1: Understand the constraint

   The container must include the first line at index 0, so left pointer stays fixed at 0.

2. Step 2: Move right pointer from the end towards 0

   Move right pointer from the last index towards 1, calculating area with fixed left pointer at 0, updating max area.

3. Step 3: Avoid nested loops for efficiency

   This approach keeps O(n) time by moving only one pointer, unlike nested loops which are O(n²).

4. Final Answer:

   Fix left pointer at 0 and move right pointer from end to start, calculating max area -> Option D

5. Quick Check:

   Fix one pointer, move other to find max area [OK]

Hint: Fix one pointer, move other to maximize area [OK]

Common Mistakes:

• Moving wrong pointer fixed at 0
• Using nested loops causing slow performance
• Ignoring the fixed pointer constraint