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'

Aarray_length

B0

C-1

D1

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 C

int right = [1] - 1;

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

Aarray_length

Barray_length + 1

C0

D1

Attempts:

3 left

💡 Hint

Common Mistakes

Setting right pointer to array_length (out of bounds).

Setting right pointer to 0.

The right pointer starts at the last index, which is array_length - 1.

3fill in blank

hard

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

DSA C

while ([1] < right) { /* process */ }

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

Aright

Bleft + 1

C0

Dleft

Attempts:

3 left

💡 Hint

Common Mistakes

Using right < right in condition.

Using left + 1 < right which skips valid pairs.

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

4fill in blank

hard

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

DSA C

if (sum < target) {
    [1]++;
} else {
    [2]--;
}

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

Aleft

Bright

Csum

Dtarget

Attempts:

3 left

💡 Hint

Common Mistakes

Moving sum or target instead of pointers.

Moving pointers in wrong direction.

If sum is less than target, move left pointer forward to increase sum. Otherwise, move right pointer backward to decrease sum.

5fill in blank

hard

Fill the blanks to correctly handle sum comparisons, checking equality first.

DSA C

if (sum == target) {
    // found pair
} else if (sum [1] target) {
    [2]++;
} else {
    [3]--;
}

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

A<

Bleft

C>

Dright

Attempts:

3 left

💡 Hint

Common Mistakes

Using '>' in the else if.

Wrong pointer movements.

Check equality first. If sum < target, increment left to increase sum. Else (sum > target), decrement right to decrease sum.

Practice

1. Why does the two pointer technique often perform better than brute force in problems involving arrays or strings?

easy

A. Because it uses extra memory to store intermediate results.

B. Because it uses recursion to explore all possibilities.

C. Because it sorts the array first to make searching faster.

D. Because it reduces the number of nested loops by using two indices moving towards each other or in one direction.

Solution

1. Step 1: Understand brute force approach

   Brute force often uses nested loops to check all pairs or combinations, leading to higher time complexity.

2. Step 2: Analyze two pointer technique advantage

   Two pointers use two indices that move intelligently, avoiding unnecessary checks and reducing complexity.

3. Final Answer:

   Because it reduces the number of nested loops by using two indices moving towards each other or in one direction. -> Option D

4. Quick Check:

   Two pointers reduce nested loops = A [OK]

Hint: Two pointers move smartly to avoid nested loops [OK]

Common Mistakes:

• Thinking two pointers always require sorting
• Confusing two pointers with recursion
• Assuming extra memory is always used

2. Which of the following is the correct syntax to initialize two pointers `left` and `right` for an array `arr` of size `n` in C?

easy

A. int left = 1, right = n;

B. int left = n, right = 0;

C. int left = 0, right = n - 1;

D. int left = n - 1, right = 0;

Solution

1. Step 1: Identify valid pointer initialization

   For two pointers at array ends, left starts at 0 and right at n-1 (last index).

2. Step 2: Check each option

   int left = 0, right = n - 1; correctly sets left=0 and right=n-1. Others have invalid indices out of range.

3. Final Answer:

   int left = 0, right = n - 1; -> Option C

4. Quick Check:

   Array indices start at 0, last index is n-1 = B [OK]

Hint: Start left at 0 and right at n-1 for array ends [OK]

Common Mistakes:

• Starting indices at 1 instead of 0
• Setting right pointer beyond array size
• Swapping left and right initialization

3. What is the output of the following C code snippet using two pointers to find if a sorted array contains two numbers that sum to 9?

int arr[] = {1, 2, 4, 4, 5, 6};
int left = 0, right = 5;
int found = 0;
while (left < right) {
    int sum = arr[left] + arr[right];
    if (sum == 9) {
        found = 1;
        break;
    } else if (sum < 9) {
        left++;
    } else {
        right--;
    }
}
printf("%d\n", found);

medium

A. 1

B. 0

C. 6

D. Compilation error

Solution

1. Step 1: Trace the two pointer movement

   Start left=0 (1), right=5 (6), sum=7 < 9, move left to 1 (2). Then sum=2+6=8 < 9, left=2 (4). Sum=4+6=10 > 9, right=4 (5). Sum=4+5=9 found!

2. Step 2: Confirm found flag and output

   Since sum==9 found, found=1 and loop breaks. printf prints 1.

3. Final Answer:

   1 -> Option A

4. Quick Check:

   Two pointers find sum 9 = 1 [OK]

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

Common Mistakes:

• Not updating pointers correctly
• Confusing sum comparison logic
• Assuming array is unsorted

4. The following code attempts to find if any two numbers in a sorted array sum to 10 using two pointers. What is the bug?

int arr[] = {1, 3, 5, 7, 9};
int left = 0, right = 4;
int found = 0;
while (left <= right) {
    int sum = arr[left] + arr[right];
    if (sum == 10) {
        found = 1;
        break;
    } else if (sum < 10) {
        left++;
    } else {
        right--;
    }
}
printf("%d\n", found);

medium

A. The loop condition should be left < right, not left <= right.

B. The sum comparison should be with 9, not 10.

C. The pointers should start at left=1 and right=3.

D. The array must be unsorted for two pointer to work.

Solution

1. Step 1: Analyze loop condition

   Using left <= right allows pointers to be equal, causing sum of same element twice, which is invalid for two distinct elements.

2. Step 2: Correct loop condition

   Changing to left < right ensures pointers do not cross or point to same element, avoiding incorrect sums.

3. Final Answer:

   The loop condition should be left < right, not left <= right. -> Option A

4. Quick Check:

   Loop must stop before pointers meet = C [OK]

Hint: Use left < right to avoid using same element twice [OK]

Common Mistakes:

• Allowing pointers to be equal in loop
• Changing target sum incorrectly
• Starting pointers inside array instead of ends

5. Given a sorted array of integers, how can the two pointer technique be adapted to find all unique pairs that sum to a target value without duplicates? Choose the best approach.

hard

A. Use a hash map to store counts and ignore two pointer technique.

B. Use two pointers and skip duplicate values by moving pointers past equal elements after finding a pair.

C. Sort the array again after each pair is found to remove duplicates.

D. Use nested loops to check all pairs and store pairs in a set to avoid duplicates.

Solution

1. Step 1: Understand duplicate pairs problem

   When array has repeated elements, two pointer may find same pair multiple times.

2. Step 2: Skip duplicates after finding a pair

   After finding a pair, move left pointer forward while next element equals current, and similarly move right pointer backward while next equals current, to avoid duplicates.

3. Final Answer:

   Use two pointers and skip duplicate values by moving pointers past equal elements after finding a pair. -> Option B

4. Quick Check:

   Skip duplicates by advancing pointers after pair found = D [OK]

Hint: Skip duplicates by moving pointers past equal elements [OK]

Common Mistakes:

• Re-sorting array repeatedly
• Using nested loops instead of two pointers
• Ignoring duplicates causing repeated pairs