Container With Most Water in DSA Typescript - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of Two Pointer Approach for Container With Most Water

What is the output of the following TypeScript code that finds the maximum water container area?

DSA Typescript

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

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

A 48

B 56

C 49

D 36

💡 Hint

Think about the widest container with the shortest height limiting the water.

The maximum area is 49, formed between the lines at index 1 (height 8) and index 8 (height 7), width 7, area 7*7=49.

🧠 Conceptual

intermediate

Why Two Pointer Approach Works for Container With Most Water

Why does the two pointer approach correctly find the maximum water container area?

A Because moving the pointer with smaller height can potentially find a taller line and increase area.

B Because moving the pointer with larger height always increases the width and area.

C Because it checks all pairs of lines exhaustively.

D Because it sorts the heights first to find the largest area.

💡 Hint

Consider which pointer limits the current area and why moving it might help.

The smaller height limits the area, so moving that pointer might find a taller line and increase area. Moving the larger height pointer won't help increase area.

🔧 Debug

advanced

Identify the Error in Container With Most Water Implementation

What error does the following TypeScript code produce when run?

DSA Typescript

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

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

A Returns 49 correctly

B Returns 0 because width becomes negative

C Infinite loop because pointers never cross

D Throws runtime error due to invalid index

💡 Hint

Check the loop condition and simulate the execution with the given input.

The code returns 49 correctly. The loop condition 'left <= right' leads to an extra iteration when left == right (width=0, area=0), but maxArea is unaffected since Math.max ignores 0. No negative width occurs as the loop terminates properly.

❓ Predict Output

advanced

Output of Modified Container With Most Water Code with Early Return

What is the output of this TypeScript code that attempts to find max water container but returns early?

DSA Typescript

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

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

A 36

B 56

C 49

D 8

💡 Hint

Look at when the function returns inside the loop.

The function returns immediately after first comparison if height[left] < height[right], so it returns area between index 0 and 8: width 8, height 1, area 8.

🚀 Application

expert

Maximum Water Container Area for Large Input

Given an array of 100,000 elements where height[i] = i % 1000 + 1, what is the maximum water container area?

A 99999

B 99000000

C 50000000

D 49950000

💡 Hint

The maximum height is 1000 at indices i where i % 1000 === 999. Find the maximum distance between such indices.

Positions of height 1000: 999, 1999, ..., 99999. The widest pair is index 999 and 99999, width = 99999 - 999 = 99000, area = 99000 * 1000 = 99000000.

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