Why Two Pointer Technique Beats Brute Force in DSA Javascript - Test Your Knowledge

Practice - 5 Tasks

1fill in blank

easy

Complete the code to initialize the left pointer at the start of the array.

DSA Javascript

let left = [1];

Drag options to blanks, or click blank then click option'

A1

B0

C-1

Darray.length

Attempts:

3 left

💡 Hint

Common Mistakes

Starting left pointer at 1 instead of 0

Using array length as starting index

The left pointer starts at index 0, the beginning of the array.

2fill in blank

medium

Complete the code to move the right pointer to the end of the array.

DSA Javascript

let right = array[1] - 1;

Drag options to blanks, or click blank then click option'

A.count

B.size

C.length

D.length()

Attempts:

3 left

💡 Hint

Common Mistakes

Using .size or .count which are not valid for arrays

Using .length() as a function

In JavaScript, array length is accessed with .length. The last index is length - 1.

3fill in blank

hard

Fix the error in the while loop condition to run while left is less than right.

DSA Javascript

while (left [1] right) {
  // logic here
}

Drag options to blanks, or click blank then click option'

A<

B<=

C>

D>=

Attempts:

3 left

💡 Hint

Common Mistakes

Using >= or > which causes incorrect loop behavior

Using <= which allows pointers to overlap

The loop should run while left < right to avoid overlapping pointers.

4fill in blank

hard

Fill both blanks to correctly move pointers based on sum comparison.

DSA Javascript

if (array[left] + array[right] [1] target) {
  left [2];
}

Drag options to blanks, or click blank then click option'

A<

B>

C++

D--

Attempts:

3 left

💡 Hint

Common Mistakes

Using > instead of < for comparison

Decrementing left pointer instead of incrementing

If sum is less than target, move left pointer right to increase sum.

5fill in blank

hard

Fill all three blanks to complete the two pointer sum check logic.

DSA Javascript

if (array[left] + array[right] === target) {
  return [[1], [2]];
} else if (array[left] + array[right] [3] target) {
  left++;
} else {
  right--;
}

Drag options to blanks, or click blank then click option'

Aleft

Bright

C<

D>

Attempts:

3 left

💡 Hint

Common Mistakes

Swapping left and right in return

Using > instead of < in comparison

Return indices when sum matches target. Move left pointer if sum is less, else move right pointer.

Practice

1. What is the main advantage of the two pointer technique over brute force in array problems?

easy

A. It always sorts the array before processing

B. It uses more memory to store intermediate results

C. It reduces time complexity by avoiding nested loops

D. It requires recursion to work properly

Solution

1. Step 1: Understand brute force time complexity

   Brute force often uses nested loops, leading to O(n²) time complexity.

2. Step 2: Analyze two pointer technique efficiency

   Two pointers move through the array in a single pass, reducing time complexity to O(n).

3. Final Answer:

   It reduces time complexity by avoiding nested loops -> Option C

4. Quick Check:

   Two pointer = faster than brute force [OK]

Hint: Two pointers scan once, brute force scans twice [OK]

Common Mistakes:

• Thinking two pointer uses more memory
• Assuming two pointer always sorts array
• Believing recursion is required

2. Which of the following is the correct syntax to initialize two pointers at the start and end of an array arr in JavaScript?

easy

A. const left = 0, right = arr.length - 1;

B. let left = arr[0], right = arr[arr.length];

C. var left = 1, right = arr.length;

D. const left = arr.length, right = 0;

Solution

1. Step 1: Identify valid pointer initialization

   Pointers should be indices, so left = 0 and right = arr.length - 1 are correct.

2. Step 2: Check invalid options

   Options B, C, and D use incorrect indices or values outside array bounds.

3. Final Answer:

   const left = 0, right = arr.length - 1; -> Option A

4. Quick Check:

   Start at 0 and end at length-1 [OK]

Hint: Array indices start at 0, end at length-1 [OK]

Common Mistakes:

• Using arr.length as index (out of bounds)
• Starting pointers at wrong indices
• Confusing values with indices

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

const arr = [1, 2, 3, 4, 5];
let left = 0, right = arr.length - 1;
let sumPairs = [];
while (left < right) {
  sumPairs.push(arr[left] + arr[right]);
  left++;
  right--;
}
console.log(sumPairs);

medium

A. [6, 6, 6]

B. [6, 6]

C. [5, 7, 9]

D. [1, 5]

Solution

1. Step 1: Trace pointer movements and sums

   Pairs: (1+5)=6, (2+4)=6, (3+3) stops because left == right, so only two pairs added.

2. Step 2: Check loop condition and sums stored

   Loop runs while left < right, so only two sums are pushed: [6, 6].

3. Final Answer:

   [6, 6] -> Option B

4. Quick Check:

   Sum pairs from ends = [6, 6] [OK]

Hint: Two pointers meet in middle, sum pairs once [OK]

Common Mistakes:

• Including middle element sum when pointers meet
• Miscounting number of pairs
• Confusing pointer update order

4. Identify the bug in the following two pointer code that tries to find if a pair sums to 10:

const arr = [1, 3, 5, 7, 9];
let left = 0, right = arr.length - 1;
while (left <= right) {
  if (arr[left] + arr[right] === 10) {
    console.log('Found');
    break;
  } else if (arr[left] + arr[right] < 10) {
    left++;
  } else {
    right++;
  }
}

medium

A. Loop condition should be left < right

B. Left pointer should decrement, not increment

C. Array is not sorted

D. Right pointer should decrement, not increment

Solution

1. Step 1: Analyze pointer updates

   If sum is greater than 10, right pointer should move left (decrement) to reduce sum, not increment.

2. Step 2: Check loop condition and array

   Loop condition is fine; array is sorted; only pointer update is wrong.

3. Final Answer:

   Right pointer should decrement, not increment -> Option D

4. Quick Check:

   Decrease right pointer to reduce sum [OK]

Hint: Decrease right pointer when sum too big [OK]

Common Mistakes:

• Incrementing right pointer instead of decrementing
• Changing loop condition unnecessarily
• Ignoring array sorting requirement

5. Given a sorted array arr and a target sum k, which approach best finds all unique pairs that sum to k efficiently?

Choose the best method that beats brute force.

hard

A. Use two pointers starting at both ends, move inward based on sum comparison

B. Use nested loops to check every pair for sum equals k

C. Sort the array and use binary search for each element's complement

D. Use a hash map to store elements and check complements in one pass

Solution

1. Step 1: Compare approaches for efficiency

   Nested loops (B) are O(n²), binary search (C) is O(n log n), hash map (A) is O(n) but uses extra space.

2. Step 2: Two pointer technique advantage

   Two pointers (D) scan array once in O(n) time with O(1) space, ideal for sorted arrays.

3. Final Answer:

   Use two pointers starting at both ends, move inward based on sum comparison -> Option A

4. Quick Check:

   Two pointer = best for sorted arrays [OK]

Hint: Two pointers scan once, no extra space needed [OK]

Common Mistakes:

• Choosing nested loops despite inefficiency
• Ignoring space complexity of hash map
• Using binary search unnecessarily