Sort Colors Two Pointer Dutch Flag in DSA C# - Time & Space Complexity

Time Complexity: Sort Colors Two Pointer Dutch Flag

O(n)

Understanding Time Complexity

We want to understand how the time needed to sort colors using two pointers changes as the list grows.

How does the number of steps grow when we have more colors to sort?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

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

void Swap(int[] arr, int i, int j) {
    int temp = arr[i];
    arr[i] = arr[j];
    arr[j] = temp;
}
    

This code sorts an array containing 0s, 1s, and 2s by moving pointers and swapping elements.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: The while loop that moves the mid pointer through the array.
• How many times: Each element is visited at most once by mid, so up to n times where n is array length.

How Execution Grows With Input

As the array size grows, the number of steps grows roughly the same amount.

Input Size (n)	Approx. Operations
10	About 10 steps
100	About 100 steps
1000	About 1000 steps

Pattern observation: The steps grow linearly with input size, doubling input roughly doubles steps.

Final Time Complexity

Time Complexity: O(n)

This means the time to sort grows in direct proportion to the number of elements.

Common Mistake

[X] Wrong: "Because there are nested ifs and swaps, the time must be quadratic O(n²)."

[OK] Correct: The loop runs once per element, and swaps happen at most once per element, so total steps grow linearly, not squared.

Interview Connect

Understanding this linear time sorting method shows you can handle problems efficiently by using pointers smartly, a skill valued in many coding challenges.

Self-Check

"What if we used three separate loops to sort instead of two pointers? How would the time complexity change?"

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