Two Pointer Technique on Arrays in DSA C# - Interactive Practice

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'

A0

Barray.Length

C-1

D1

Attempts:

3 left

💡 Hint

Common Mistakes

Starting left pointer at 1 instead of 0.

Using array.Length which is out of bounds.

The left pointer should start at index 0, which is the first element of the array.

2fill in blank

medium

Complete the code to initialize the right pointer at the end of the array.

DSA C#

int right = [1];

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

Aarray.Length - 1

B-1

Carray.Length

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Using array.Length which is out of bounds.

Starting right pointer at 0 which is the start, not the end.

The right pointer should start at the last valid index, which is array.Length - 1.

3fill in blank

hard

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

DSA C#

while (left [1] right) {
    // process elements
}

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

A>=

B>

C<

D<=

Attempts:

3 left

💡 Hint

Common Mistakes

Using >= or <= which causes infinite loops or misses the stopping condition.

Using > which reverses the logic.

The loop should run while left is less than right to avoid overlapping pointers.

4fill in blank

hard

Fill both blanks to move the left pointer forward and the right pointer backward.

DSA C#

left [1];
right [2];

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

A++

B--

C+=

D-=

Attempts:

3 left

💡 Hint

Common Mistakes

Using += or -= without specifying the amount.

Using the wrong operator causing pointers to move in wrong directions.

Use ++ to increase left pointer by 1 and -- to decrease right pointer by 1.

5fill in blank

hard

Fill all three blanks to swap elements at left and right pointers and then move both pointers.

DSA C#

int temp = array[[1]];
array[[2]] = array[[3]];

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

Aleft

Bright

Dtemp

Attempts:

3 left

💡 Hint

Common Mistakes

Mixing up left and right indices.

Not using temp to hold the value before overwriting.

Swap elements by storing array[left] in temp, then assign array[right] to array[left].

Practice

1. What is the main idea behind the Two Pointer Technique when used on arrays?

easy

A. Use two indices to traverse the array from different ends or positions simultaneously

B. Use one pointer to traverse and another to store values

C. Sort the array first and then use binary search

D. Use recursion to split the array into two parts

Solution

1. Step 1: Understand the two pointer approach

   The technique uses two pointers (indices) to scan the array from different positions, often from start and end. This helps solve problems efficiently by reducing the number of iterations compared to nested loops.

2. Final Answer:

   Use two indices to traverse the array from different ends or positions simultaneously -> Option A

3. Quick Check:

   Two pointers move from different ends = D [OK]

Hint: Two pointers move from start and end to meet in middle [OK]

Common Mistakes:

• Thinking it always requires sorting first
• Using only one pointer instead of two
• Confusing with recursion or binary search

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

easy

A. int left = -1, right = arr.Length + 1;

B. int left = 1, right = arr.Length;

C. int left = arr.Length, right = 0;

D. int left = 0, right = arr.Length - 1;

Solution

1. Step 1: Understand array indexing and initialize pointers correctly

   Arrays start at index 0 and end at index Length - 1. Left pointer starts at 0 (first element), right pointer at Length - 1 (last element).

2. Final Answer:

   int left = 0, right = arr.Length - 1; -> Option D

3. Quick Check:

   Start at 0 and end at Length-1 = A [OK]

Hint: Start left at 0, right at length minus one [OK]

Common Mistakes:

• Starting right pointer at arr.Length (out of bounds)
• Starting left pointer at 1 instead of 0
• Using negative indices

3. What will be the output of the following C# code?

int[] arr = {1, 2, 3, 4, 5, 6};
int left = 0, right = arr.Length - 1;
int sum = 0;
while (left < right) {
    sum += arr[left] + arr[right];
    left++;
    right--;
}
Console.WriteLine(sum);

medium

A. 21

B. 18

C. 20

D. 15

Solution

1. Step 1: Trace the loop iterations

   Pairs added: (1+6)=7, (2+5)=7, (3+4)=7; sum = 7+7+7 = 21. Loop runs while left < right, so stops after these pairs.

2. Final Answer:

   21 -> Option A

3. Quick Check:

   Sum of pairs from ends = 21 [OK]

Hint: Sum pairs from ends until pointers meet [OK]

Common Mistakes:

• Adding elements twice
• Loop running when left equals right
• Incorrect sum calculation

4. Identify the error in the following C# code using two pointers to find if array arr contains two numbers that sum to target.

int left = 0, right = arr.Length - 1;
while (left <= right) {
    int sum = arr[left] + arr[right];
    if (sum == target) return true;
    else if (sum < target) left++;
    else right--;
}
return false;

medium

A. The sum calculation is incorrect

B. The pointers should both start at 0

C. The loop condition should be left < right instead of left <= right

D. The return statements are missing a value

Solution

1. Step 1: Analyze loop condition

   Using left <= right allows pointers to be equal, causing sum of same element twice. Change to left < right to avoid using the same element twice.

2. Final Answer:

   The loop condition should be left < right instead of left <= right -> Option C

3. Quick Check:

   Pointers must not cross or equal for two distinct elements [OK]

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

Common Mistakes:

• Allowing pointers to be equal causing wrong sums
• Starting both pointers at 0
• Misplacing return statements

5. Given a sorted array arr and a number k, how can you use the two pointer technique to find if there exists a pair with sum equal to k? Choose the correct approach.

hard

A. Initialize both pointers at start; move both right until sum equals k

B. Initialize two pointers at start and end; move left pointer right if sum < k, else move right pointer left

C. Initialize two pointers at middle; move both outward until sum equals k

D. Sort the array again and use binary search for each element

Solution

1. Step 1: Set pointers and move based on sum

   Left pointer at 0, right pointer at last index. If sum < k, move left right to increase sum; if sum > k, move right left to decrease sum.

2. Final Answer:

   Initialize two pointers at start and end; move left pointer right if sum < k, else move right pointer left -> Option B

3. Quick Check:

   Two pointers adjust to reach target sum efficiently [OK]

Hint: Move pointers inward based on sum comparison [OK]

Common Mistakes:

• Starting both pointers at same side
• Moving pointers in wrong direction
• Sorting array again unnecessarily