Container With Most Water in DSA Typescript

Mental Model

We want to find two lines that hold the most water between them by checking pairs from the edges inward.

Analogy: Imagine two walls on a beach holding water between them; moving the shorter wall inward might find a taller wall to hold more water.

Height array: [1, 8, 6, 2, 5, 4, 8, 3, 7]
Positions:  0  1  2  3  4  5  6  7  8
Pointers: ↑                             ↑

Dry Run Walkthrough

Input: height = [1, 8, 6, 2, 5, 4, 8, 3, 7]

Goal: Find two lines that together with the x-axis form a container holding the most water.

Step 1: Calculate area between pointers at positions 0 and 8

Positions: 0(1) -> ... -> 8(7)
Area = min(1,7) * (8 - 0) = 1 * 8 = 8

Why: Start with widest container to get a baseline area.

Step 2: Move left pointer from 0 to 1 because height at 0 is smaller

Positions: 1(8) -> ... -> 8(7)
Area = min(8,7) * (8 - 1) = 7 * 7 = 49

Why: Moving the shorter line might find a taller line to increase area.

Step 3: Move right pointer from 8 to 7 because height at 8 is smaller

Positions: 1(8) -> ... -> 7(3)
Area = min(8,3) * (7 - 1) = 3 * 6 = 18

Why: Try to find a taller line on the right side.

Step 4: Move right pointer from 7 to 6 because height at 7 is smaller

Positions: 1(8) -> ... -> 6(8)
Area = min(8,8) * (6 - 1) = 8 * 5 = 40

Why: Found a taller line, calculate area.

Step 5: Move left pointer from 1 to 2 because height at 1 equals height at 6

Positions: 2(6) -> ... -> 6(8)
Area = min(6,8) * (6 - 2) = 6 * 4 = 24

Why: Move left pointer inward to try for bigger area.

Step 6: Move left pointer from 2 to 3 because height at 2 is smaller

Positions: 3(2) -> ... -> 6(8)
Area = min(2,8) * (6 - 3) = 2 * 3 = 6

Why: Try to find taller line on left side.

Step 7: Move left pointer from 3 to 4 because height at 3 is smaller

Positions: 4(5) -> ... -> 6(8)
Area = min(5,8) * (6 - 4) = 5 * 2 = 10

Why: Keep moving left pointer inward.

Step 8: Move left pointer from 4 to 5 because height at 4 is smaller

Positions: 5(4) -> ... -> 6(8)
Area = min(4,8) * (6 - 5) = 4 * 1 = 4

Why: Try next possible container.

Step 9: Move left pointer from 5 to 6; pointers meet, stop

Pointers at same position 6(8), no container

Why: No more pairs to check.

Result:

Positions 1(8) and 8(7) form the container with max area 49
Final container: 8 -> ... -> 7

Annotated Code

DSA Typescript

class ContainerWithMostWater {
  static 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;
      if (currentArea > maxArea) {
        maxArea = currentArea;
      }

      if (height[left] < height[right]) {
        left++; // move left pointer inward
      } else {
        right--; // move right pointer inward
      }
    }

    return maxArea;
  }
}

// Driver code
const height = [1, 8, 6, 2, 5, 4, 8, 3, 7];
const result = ContainerWithMostWater.maxArea(height);
console.log(result);

while (left < right) {

loop until pointers meet to check all pairs

const width = right - left;

calculate distance between pointers

const currentHeight = Math.min(height[left], height[right]);

container height limited by shorter line

const currentArea = width * currentHeight;

calculate current container area

if (currentArea > maxArea) { maxArea = currentArea; }

update max area if current is bigger

if (height[left] < height[right]) { left++; } else { right--; }

move pointer at shorter line inward to find taller line

Output

49

Complexity Analysis

Time: O(n) because we move pointers inward only once each, checking all pairs efficiently

Space: O(1) because we use only a few variables regardless of input size

vs Alternative: Naive approach checks all pairs with O(n^2) time, this two-pointer method is much faster

Edge Cases

Empty array

Returns 0 as no container can be formed

DSA Typescript

while (left < right) {

Array with one element

Returns 0 as no container can be formed

DSA Typescript

while (left < right) {

All heights equal

Returns max area as height * (length - 1)

DSA Typescript

if (currentArea > maxArea) { maxArea = currentArea; }

When to Use This Pattern

When you see a problem asking for max area between two lines in an array, use two pointers from edges moving inward to find max container efficiently.

Common Mistakes

Mistake: Moving the pointer at the taller line instead of the shorter line
Fix: Always move the pointer at the shorter line inward to potentially find a taller line and increase area

Summary

Finds the maximum water container formed by two lines in an array.

Use when you need to maximize area between two vertical lines efficiently.

The key is moving the pointer at the shorter line inward to find a taller line and larger area.

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