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

Practice - 5 Tasks

1fill in blank

easy

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

DSA Python

left = [1]

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

A0

B-1

C1

Dlen(arr)

Attempts:

3 left

💡 Hint

Common Mistakes

Starting the pointer at 1 instead of 0

Using negative indices incorrectly

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

2fill in blank

medium

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

DSA Python

right = [1]

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

A0

Blen(arr)

Clen(arr) - 1

D-1

Attempts:

3 left

💡 Hint

Common Mistakes

Using len(arr) which is out of range

Starting at 0 which is the first element

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 Python

while [1] < right:

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

Aright - 1

Bright

Cleft + 1

Dleft

Attempts:

3 left

💡 Hint

Common Mistakes

Using right < right which is always false

Using left + 1 which is incorrect syntax

The loop should run while the left pointer is less than the right pointer.

4fill in blank

hard

Fill both blanks to move pointers correctly: increment left if sum is less, decrement right if sum is more.

DSA Python

if current_sum [1] target:
    left [2] 1

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

A<

B>

C+=

D-=

Attempts:

3 left

💡 Hint

Common Mistakes

Using '>' instead of '<' in comparison

Using '-=' to move left pointer forward

If the current sum is less than the target, move the left pointer forward by adding 1.

5fill in blank

hard

Fill all three blanks to complete the two pointer sum check: compare, move left, or move right.

DSA Python

if current_sum == target:
    return True
elif current_sum [1] target:
    left [2] 1
else:
    right [3] 1

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

A<

B+=

C-=

D==

Attempts:

3 left

💡 Hint

Common Mistakes

Using '==' in comparison instead of '<'

Mixing up '+=' and '-=' for pointers

If current_sum is less than target, move left pointer forward; else move right pointer backward.

Practice

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

easy

A. It requires recursion to work properly

B. It uses more memory to store intermediate results

C. It always sorts the array before processing

D. It reduces time complexity by avoiding nested loops

Solution

1. Step 1: Understand brute force approach

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

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 D

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 the array
• Confusing memory use with speed
• Assuming recursion is needed

2. Which of the following is the correct syntax to initialize two pointers at the start and end of a list named arr in Python?

easy

A. left = 0; right = len(arr) - 1

B. left = 1; right = len(arr)

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

D. left = -1; right = len(arr) + 1

Solution

1. Step 1: Identify valid indices in Python list

   Python lists start at index 0 and end at len(arr) - 1.

2. Step 2: Check pointer initialization

   Left pointer should be 0 (start), right pointer should be len(arr) - 1 (end).

3. Final Answer:

   left = 0; right = len(arr) - 1 -> Option A

4. Quick Check:

   Start = 0, End = length-1 [OK]

Hint: Start pointer at 0, end pointer at length minus one [OK]

Common Mistakes:

• Starting pointers outside valid index range
• Using 1-based indexing in Python
• Setting right pointer to len(arr) instead of len(arr)-1

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

def two_sum_sorted(arr, target):
    left, right = 0, len(arr) - 1
    while left < right:
        current_sum = arr[left] + arr[right]
        if current_sum == target:
            return [left, right]
        elif current_sum < target:
            left += 1
        else:
            right -= 1
    return [-1, -1]

print(two_sum_sorted([1, 3, 4, 5, 7, 10], 9))

medium

A. [0, 5]

B. [1, 3]

C. [2, 3]

D. [-1, -1]

Solution

1. Step 1: Trace the execution step-by-step

   arr = [1,3,4,5,7,10]
   left=0 (1), right=5 (10), sum=11 >9 -> right=4
   left=0 (1), right=4 (7), sum=8 <9 -> left=1
   left=1 (3), right=4 (7), sum=10 >9 -> right=3
   left=1 (3), right=3 (5), sum=8 <9 -> left=2
   left=2 (4), right=3 (5), sum=9 ==9 -> return [2, 3]

2. Final Answer:

   [2, 3] -> Option C

3. Quick Check:

   arr[2] + arr[3] = 4 + 5 = 9 [OK]

Hint: Move pointers based on sum compared to target [OK]

Common Mistakes:

• Not updating pointers correctly
• Returning wrong indices
• Confusing array values with indices

4. Find the error in this two pointer code snippet that tries to find if a pair sums to target in a sorted list:

def has_pair(arr, target):
    left, right = 0, len(arr)
    while left < right:
        if arr[left] + arr[right] == target:
            return True
        elif arr[left] + arr[right] < target:
            left += 1
        else:
            right -= 1
    return False

medium

A. left should start at 1 instead of 0

B. right should be initialized to len(arr) - 1

C. The while condition should be left <= right

D. The function should return indices instead of True/False

Solution

1. Step 1: Check pointer initialization

   right is set to len(arr), which is out of valid index range (max index is len(arr)-1).

2. Step 2: Correct pointer initialization

   right should be len(arr) - 1 to point to last element.

3. Final Answer:

   right should be initialized to len(arr) - 1 -> Option B

4. Quick Check:

   Valid indices = 0 to length-1 [OK]

Hint: Right pointer must be last valid index, not length [OK]

Common Mistakes:

• Using len(arr) as index causes IndexError
• Changing left pointer start incorrectly
• Misusing while loop condition

5. Given a sorted array arr and a target sum, how does the two pointer technique efficiently find all unique pairs that add up to the target compared to brute force?

hard

A. By moving pointers inward and skipping duplicates, it avoids checking all pairs

B. By sorting the array multiple times to find pairs quickly

C. By using nested loops to check every possible pair

D. By using recursion to split the array and find pairs

Solution

1. Step 1: Understand brute force approach

   Brute force checks every pair with nested loops, O(n²) time.

2. Step 2: Explain two pointer efficiency

   Two pointers start at ends, move inward based on sum comparison, skipping duplicates to avoid repeats, achieving O(n) time.

3. Final Answer:

   By moving pointers inward and skipping duplicates, it avoids checking all pairs -> Option A

4. Quick Check:

   Two pointer skips pairs, brute force checks all [OK]

Hint: Skip duplicates and move pointers inward for efficiency [OK]

Common Mistakes:

• Thinking sorting happens multiple times
• Confusing recursion with two pointer
• Assuming nested loops are used in two pointer