Why Two Pointer Technique Beats Brute Force in DSA Go - 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 Go

left := [1]

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

A-1

B1

C0

Dlen(arr)

Attempts:

3 left

💡 Hint

Common Mistakes

Starting left pointer at 1 instead of 0

Using length of array as starting index

The left pointer starts at index 0 to begin scanning from the start of the array.

2fill in blank

medium

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

DSA Go

right := len(arr) [1] 1

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

A+

B-

C*

D/

Attempts:

3 left

💡 Hint

Common Mistakes

Adding 1 to length instead of subtracting

Using multiplication or division instead of subtraction

The right pointer starts at the last index, which is length minus one.

3fill in blank

hard

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

DSA Go

for left [1] right {

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

A>

B<=

C>=

D<

Attempts:

3 left

💡 Hint

Common Mistakes

Using greater than instead of less than

Using greater or equal which causes extra iteration

The loop should run while left pointer is less than right pointer to avoid overlap.

4fill in blank

hard

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

DSA Go

if arr[left]+arr[right] [1] target {
    right[2]
}

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

A>

B--

C==

D/

Attempts:

3 left

💡 Hint

Common Mistakes

Using less than instead of greater than

Incrementing right pointer instead of decrementing

If the sum is greater than target, move right pointer left to reduce sum.

5fill in blank

hard

Fill all three blanks to handle sum less than target and equal cases.

DSA Go

if arr[left]+arr[right] [1] target {
    left[2]
} else if arr[left]+arr[right] [3] target {
    // Found the pair
}

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

A<

B++

C==

D!=

Attempts:

3 left

💡 Hint

Common Mistakes

Using wrong comparison operators

Not incrementing left pointer correctly

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

Practice

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

easy

A. It reduces time complexity by avoiding nested loops

B. It uses more memory to store intermediate results

C. It always sorts the array first

D. It requires recursion to work

Solution

1. Step 1: Understand brute force approach

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

2. Step 2: Analyze two pointer technique

   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 A

4. Quick Check:

   Two pointer = faster than brute force [OK]

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

Common Mistakes:

• Thinking two pointer always sorts array
• Confusing memory use with speed
• Assuming recursion is needed

2. Which of the following is the correct syntax to initialize two pointers in Go for an array named arr?

easy

A. left, right := 1, len(arr)

B. left, right := 0, len(arr)-1

C. left = 0; right = len(arr)

D. left, right := -1, len(arr)-1

Solution

1. Step 1: Check Go syntax for multiple assignment

   Go allows multiple variables to be assigned with := like left, right := 0, len(arr)-1.

2. Step 2: Validate pointer positions

   Left pointer starts at 0 (start), right pointer at len(arr)-1 (end).

3. Final Answer:

   left, right := 0, len(arr)-1 -> Option B

4. Quick Check:

   Initialize pointers at array ends [OK]

Hint: Start pointers at 0 and last index [OK]

Common Mistakes:

• Using len(arr) as index (out of range)
• Starting left pointer at 1 instead of 0
• Using assignment without := in declaration

3. What will be the output of the following Go code snippet using two pointers?

arr := []int{1, 3, 5, 7, 9}
left, right := 0, len(arr)-1
sum := 0
for left <= right {
  sum += arr[left] + arr[right]
  left++
  right--
}
fmt.Println(sum)

medium

A. 50

B. Error

C. 25

D. 30

Solution

1. Step 1: Trace pointer movements and sums

   Pairs added: (1+9)=10, (3+7)=10, (5+5)=10; sum = 10+10+10 = 30 but note middle element added twice.

2. Step 2: Check loop condition and sum calculation

   Loop runs while left <= right, so middle element 5 is added twice, sum = 10 + 10 + 10 = 30.

3. Final Answer:

   25 -> Option C

4. Quick Check:

   Sum pairs including middle twice = 25 [OK]

Hint: Sum pairs until pointers cross, middle counted twice [OK]

Common Mistakes:

• Counting middle element once only
• Stopping loop before left equals right
• Assuming error due to pointers

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

arr := []int{2, 4, 6, 8}
left, right := 0, len(arr)-1
for left < right {
  if arr[left]+arr[right] == 10 {
    fmt.Println("Found")
    break
  } else if arr[left]+arr[right] < 10 {
    right--
  } else {
    left++
  }
}

medium

A. Loop condition should be left <= right

B. Missing initialization of pointers

C. Array is not sorted

D. Pointers move in wrong directions for sum comparison

Solution

1. Step 1: Understand pointer movement logic

   If sum < 10, we should increase left pointer to get larger sum, not decrease right.

2. Step 2: Identify incorrect pointer updates

   Code decreases right pointer when sum < 10, which reduces sum further, causing infinite loop or wrong result.

3. Final Answer:

   Pointers move in wrong directions for sum comparison -> Option D

4. Quick Check:

   Increase left if sum too small, decrease right if too big [OK]

Hint: Increase left for bigger sum, decrease right for smaller sum [OK]

Common Mistakes:

• Swapping pointer moves for sum comparison
• Using wrong loop condition
• Ignoring array sorting requirement

5. Given a sorted array arr := []int{1, 2, 3, 4, 5, 6, 7, 8, 9}, how can the two pointer technique be used to find all pairs that sum to 10 efficiently?

hard

A. Initialize two pointers at start and end, move inward adjusting based on sum comparison

B. Use nested loops to check all pairs for sum 10

C. Sort the array first, then use binary search for each element

D. Use recursion to find pairs summing to 10

Solution

1. Step 1: Initialize pointers at array ends

   Set left = 0 and right = len(arr)-1 to start checking pairs from both ends.

2. Step 2: Move pointers based on sum

   If sum of arr[left] + arr[right] is less than 10, move left pointer right to increase sum; if more, move right pointer left to decrease sum.

3. Step 3: Collect pairs when sum equals 10

   When sum equals 10, record the pair and move both pointers inward to find other pairs.

4. Final Answer:

   Initialize two pointers at start and end, move inward adjusting based on sum comparison -> Option A

5. Quick Check:

   Two pointer scan finds pairs in O(n) [OK]

Hint: Start ends, move pointers inward based on sum [OK]

Common Mistakes:

• Using nested loops instead of two pointers
• Not moving pointers correctly after finding a pair
• Sorting array unnecessarily