Sort Colors Two Pointer Dutch Flag in DSA C# - Cheat Sheet & Quick Revision

Recall & Review

beginner

What is the main idea behind the Dutch National Flag problem?

It is to sort an array containing three different values (like 0, 1, 2) in a single pass using three pointers to separate the values into three sections.

beginner

In the Dutch Flag algorithm, what do the three pointers represent?

The pointers represent the boundaries for the three colors: low for the end of 0s, mid for the current element, and high for the start of 2s.

intermediate

Why does the Dutch Flag algorithm only need one pass through the array?

Because it swaps elements in place and moves pointers accordingly, it sorts all elements without revisiting any index more than once.

intermediate

What happens when the mid pointer encounters a '2' in the Dutch Flag algorithm?

The element at mid is swapped with the element at high, and the high pointer is moved one step left. The mid pointer stays to check the swapped element.

beginner

Show the final sorted array after applying the Dutch Flag algorithm on [2, 0, 2, 1, 1, 0].

The sorted array is [0, 0, 1, 1, 2, 2].

What does the 'low' pointer represent in the Dutch Flag algorithm?

A The boundary for 2s

B The boundary for 1s

C The boundary for 0s

D The current element being checked

The 'low' pointer marks the end of the section containing 0s.

When the mid pointer finds a '1', what should be done?

A Stop the algorithm

B Swap with high and move high pointer

C Swap with low and move both pointers

D Just move mid pointer forward

When mid points to 1, just move mid forward because 1s are in the middle.

What is the time complexity of the Dutch Flag algorithm?

A O(n)

B O(n log n)

C O(1)

D O(n^2)

The algorithm runs in linear time, O(n), because it processes each element once.

Which pointer moves backward in the Dutch Flag algorithm?

A high

B mid

C low

D none

The 'high' pointer moves backward as 2s are placed at the end.

What is the main advantage of the Dutch Flag algorithm?

A Uses extra space for sorting

B Sorts in one pass with constant space

C Sorts in multiple passes

D Only works for two colors

It sorts the array in one pass using constant extra space.

Explain how the two-pointer Dutch Flag algorithm sorts an array with three colors.

Think about how you separate three groups in one go.

Describe the step-by-step process when the mid pointer encounters a '2' during sorting.

Remember why mid stays after swapping with high.

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