Sort Colors Two Pointer Dutch Flag 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'

Anums.Length

B1

C-1

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Starting left pointer at 1 or negative index.

Initializing left pointer to array length.

The left pointer should start at index 0 to begin sorting from the start.

2fill in blank

medium

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

DSA C#

int right = nums.[1] - 1;

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

ACapacity

BCount

CLength

DSize

Attempts:

3 left

💡 Hint

Common Mistakes

Using Count which is for collections, not arrays.

Using Size or Capacity which are invalid here.

In C#, arrays use Length property to get the number of elements.

3fill in blank

hard

Fix the error in the while loop condition to continue sorting while pointers do not cross.

DSA C#

while ([1] <= right) { /* sorting logic */ }

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

Aleft

Bi

Cright

Dnums

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'left' instead of 'i' in the condition.

Using 'right' on both sides of the condition.

The variable 'i' is used to traverse the array and should be compared with 'right'.

4fill in blank

hard

Fill both blanks to swap elements at indices i and left, then move pointers accordingly.

DSA C#

int temp = nums[[1]]; nums[[2]] = nums[left]; nums[left] = temp; left++; i++;

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

Ai

Bleft

Cright

Dtemp

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'left' or 'right' in the first blank instead of 'i'.

Mixing indices causing wrong swaps.

We swap nums[i] with nums[left], so both blanks use 'i' as the index for the first element.

5fill in blank

hard

Fill all three blanks to swap elements at indices i and right, then move right pointer and continue scanning.

DSA C#

int temp = nums[[1]]; nums[[2]] = nums[right]; nums[right] = temp; [3]--;

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

Aleft

Bi

Dright

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'left' instead of 'i' for the swap indices.

Not decrementing 'right' after swap.

We swap nums[i] with nums[right], so blanks 1 and 2 are 'i', and the last blank is 'right' to decrement.

Practice

1. What is the main idea behind the Sort Colors Two Pointer Dutch Flag algorithm?

easy

A. Swap elements randomly until sorted

B. Sort the array using built-in sort function

C. Count each color and then overwrite the array

D. Use three pointers to separate 0s, 1s, and 2s in one pass

Solution

1. Step 1: Understand the problem

   The algorithm sorts an array containing only 0s, 1s, and 2s representing colors.

2. Step 2: Use three pointers

   It uses three pointers: low, mid, and high to separate the colors in one pass efficiently.

3. Final Answer:

   Use three pointers to separate 0s, 1s, and 2s in one pass -> Option D

4. Quick Check:

   Three pointers = A [OK]

Hint: Remember three pointers separate colors in one pass [OK]

Common Mistakes:

• Using built-in sort instead of pointers
• Counting colors then rewriting array
• Swapping elements without pointers

2. Which of the following is the correct initialization of pointers in the Dutch Flag algorithm in C#?

easy

A. int low = 0, mid = 0, high = nums.Length - 1;

B. int low = nums.Length - 1, mid = 0, high = 0;

C. int low = 0, mid = nums.Length - 1, high = 0;

D. int low = 0, mid = 1, high = nums.Length;

Solution

1. Step 1: Identify pointer roles

   Low and mid start at 0, high starts at last index (length - 1).

2. Step 2: Check initialization correctness

   int low = 0, mid = 0, high = nums.Length - 1; correctly sets low and mid to 0, high to nums.Length - 1.

3. Final Answer:

   int low = 0, mid = 0, high = nums.Length - 1; -> Option A

4. Quick Check:

   Correct pointer init = C [OK]

Hint: Start low and mid at 0, high at last index [OK]

Common Mistakes:

• Setting high to 0 instead of last index
• Starting mid at length - 1
• Using invalid pointer values

3. What will be the output array after running the Dutch Flag algorithm on nums = new int[] {2, 0, 2, 1, 1, 0};?

medium

A. [0, 1, 0, 1, 2, 2]

B. [2, 2, 1, 1, 0, 0]

C. [0, 0, 1, 1, 2, 2]

D. [1, 0, 2, 0, 1, 2]

Solution

1. Step 1: Start with pointers low=0, mid=0, high=5

   Initial array: [2, 0, 2, 1, 1, 0]

2. Step 2: Process elements using Dutch Flag logic

   Swap 2 with 0 at high, move high left; swap 2 with 0 at low; move mid and low accordingly until sorted.

3. Final Answer:

   [0, 0, 1, 1, 2, 2] -> Option C

4. Quick Check:

   Sorted colors = A [OK]

Hint: Swap 0s left, 2s right, 1s stay middle [OK]

Common Mistakes:

• Not swapping correctly
• Moving pointers incorrectly
• Confusing order of colors

4. Identify the bug in the following C# code snippet for the Dutch Flag algorithm:

int low = 0, mid = 0, high = nums.Length - 1;
while (mid <= high) {
  if (nums[mid] == 0) {
    Swap(nums, low, mid);
    low++;
  } else if (nums[mid] == 1) {
    mid++;
  } else {
    Swap(nums, mid, high);
    high--;
  }
}

medium

A. Swap function parameters are incorrect

B. Missing increment of mid after swapping 0

C. high should be incremented instead of decremented

D. mid should not be incremented when nums[mid] == 1

Solution

1. Step 1: Check pointer increments after swaps

   When nums[mid] == 0, after swapping with low, mid must also increment to avoid infinite loop.

2. Step 2: Identify missing mid++

   The code increments low but misses mid++, causing mid to stay and loop endlessly.

3. Final Answer:

   Missing increment of mid after swapping 0 -> Option B

4. Quick Check:

   Increment mid after 0 swap = D [OK]

Hint: Always increment mid after swapping 0 [OK]

Common Mistakes:

• Forgetting mid++ after swapping 0
• Decrementing high incorrectly
• Incorrect swap parameters

5. Given an array nums containing only 0s, 1s, and 2s, how would you modify the Dutch Flag algorithm to sort an array that also contains 3s, placing all 3s at the end?

hard

A. Add a fourth pointer and extend the logic to handle 3s at the end

B. Use the same three pointers but treat 3s as 2s

C. Sort the array normally using built-in sort for 3s

D. Ignore 3s since they are already at the end

Solution

1. Step 1: Understand limitation of three pointers

   The original algorithm handles only three values (0,1,2). Adding 3 requires extra handling.

2. Step 2: Introduce a fourth pointer

   Use four pointers to separate 0s, 1s, 2s, and 3s, extending the logic to place 3s at the end.

3. Final Answer:

   Add a fourth pointer and extend the logic to handle 3s at the end -> Option A

4. Quick Check:

   Four pointers for four colors = B [OK]

Hint: Add pointer per unique value for correct sorting [OK]

Common Mistakes:

• Treating 3s as 2s causing wrong order
• Using built-in sort instead of pointers
• Ignoring 3s placement