Why Two Pointer Technique Beats Brute Force in DSA C++ - 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 C++

int left = [1];

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

An

B1

C-1

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Starting left pointer at 1 instead of 0.

Using negative index which is invalid.

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 C++

int right = [1] - 1;

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

An

B0

C1

Dsize

Attempts:

3 left

💡 Hint

Common Mistakes

Setting right pointer to n instead of n-1.

Using 0 which points to the start.

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

3fill in blank

hard

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

DSA C++

while ([1] < right) {

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

Aleft

Bright

Cleft + 1

Dright - 1

Attempts:

3 left

💡 Hint

Common Mistakes

Using right < right which is always false.

Using expressions that skip valid pairs.

The loop runs while left pointer is less than right pointer.

4fill in blank

hard

Fill both blanks to move pointers correctly: increment left and decrement right.

DSA C++

if (sum < target) {
    left[1];
} else {
    right[2];
}

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

A++

B--

C+=

D-=

Attempts:

3 left

💡 Hint

Common Mistakes

Using -- on left pointer which moves it backward incorrectly.

Using ++ on right pointer which moves it forward incorrectly.

Increment left pointer when sum is less than target, decrement right pointer otherwise.

5fill in blank

hard

Fill all three blanks to create a dictionary comprehension that maps words to their lengths if length is greater than 3.

DSA C++

auto result = std::unordered_map<std::string, int>{};
for (const auto& word : words) {
    if (word.size() [1] 3) {
        result[[2]] = [3];
    }
}

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

A>

Bword

Cword.size()

D<

Attempts:

3 left

💡 Hint

Common Mistakes

Using < instead of > causing wrong filtering.

Using incorrect key or value in map assignment.

Check if word size is greater than 3, then map word to its size.

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 array once, avoiding double loops [OK]

Common Mistakes:

• Thinking two pointer always sorts array
• Assuming two pointer uses more memory
• Confusing recursion with two pointer

2. Which of the following is the correct syntax to initialize two pointers at the start and end of a vector v in C++?

easy

A. int left = 1, right = v.size();

B. int left = 0, right = v.size() - 1;

C. int left = v.size(), right = 0;

D. int left = -1, right = v.size() + 1;

Solution

1. Step 1: Understand vector indexing

   Vectors in C++ are zero-indexed, so first element is at 0 and last at size()-1.

2. Step 2: Check pointer initialization

   Setting left = 0 and right = v.size() - 1 correctly points to start and end.

3. Final Answer:

   int left = 0, right = v.size() - 1; -> Option B

4. Quick Check:

   Start=0, End=size()-1 [OK]

Hint: Start pointer at 0, end pointer at size()-1 [OK]

Common Mistakes:

• Starting pointers at 1 instead of 0
• Using size() as index (out of range)
• Negative or invalid pointer initialization

3. What is the output of the following C++ code snippet using two pointers?

std::vector<int> v = {1, 2, 3, 4, 5};
int left = 0, right = 4;
while (left < right) {
  std::cout << v[left] + v[right] << " ";
  left++;
  right--;
}

medium

A. 6 6

B. 5 7 9

C. 1 2 3 4 5

D. 9 7 5

Solution

1. Step 1: Trace pointer values and sums

   Pairs: v[0]+v[4]=1+5=6, v[1]+v[3]=2+4=6, then left=2, right=2 (loop stops since left < right is false).

2. Step 2: Output sums until pointers meet

   Outputs are 6 6.

3. Final Answer:

   6 6 -> Option A

4. Quick Check:

   Sum pairs from ends = 6 6 [OK]

Hint: Sum elements at left and right pointers until they meet [OK]

Common Mistakes:

• Including sum when pointers are equal
• Confusing pointer increments
• Printing elements instead of sums

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

std::vector<int> v = {1, 3, 5, 7, 9};
int left = 0, right = v.size() - 1;
while (left < right) {
  if (v[left] + v[right] == 10) {
    std::cout << "Found";
    break;
  } else if (v[left] + v[right] > 10) {
    left++;
  } else {
    right--;
  }
}

medium

A. Loop condition should be left <= right

B. Vector is not sorted before two pointer use

C. Pointers are moved in wrong directions for sum comparison

D. Output statement is missing a newline

Solution

1. Step 1: Understand pointer movement logic

   If sum > 10, we should move right pointer left to reduce sum, not left pointer right.

2. Step 2: Identify correct pointer moves

   Correct moves: if sum > target, right-- ; if sum < target, left++.

3. Final Answer:

   Pointers are moved in wrong directions for sum comparison -> Option C

4. Quick Check:

   Sum > target -> move right pointer left [OK]

Hint: Move right pointer left if sum too big, left pointer right if too small [OK]

Common Mistakes:

• Swapping pointer moves for sum comparisons
• Using wrong loop condition
• Ignoring vector sorting requirement

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

hard

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

B. Use nested loops to check all pairs for sum 8

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

D. Use recursion to check pairs from both ends

Solution

1. Step 1: Initialize two pointers

   Set left pointer at index 0 and right pointer at last index (6).

2. Step 2: Move pointers based on sum comparison

   If sum of elements at pointers is less than 8, move left pointer right; if more, move right pointer left; if equal, record pair and move both pointers.

3. Step 3: Continue until pointers cross

   Stop when left pointer is no longer less than right pointer.

4. Final Answer:

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

5. Quick Check:

   Two pointer moves inward checking sums [OK]

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

Common Mistakes:

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