Why Sort Colors Two Pointer Dutch Flag in DSA C?

The Big Idea

What if you could sort three colors perfectly in just one quick pass?

The Scenario

Imagine you have a box of colored balls mixed together: red, white, and blue. You want to arrange them so all reds come first, then whites, then blues. Doing this by picking each ball and checking where to put it manually is tiring and slow.

The Problem

Manually sorting each ball means checking every ball multiple times and moving them around again and again. This wastes time and can cause mistakes, like putting a red ball in the blue section by accident.

The Solution

The two-pointer Dutch Flag method uses just a few pointers to sort the balls in one pass. It smartly swaps balls to their right places without extra space or repeated checks, making sorting fast and simple.

Before vs After

✗ Before

for (int i = 0; i < n; i++) {
  for (int j = i + 1; j < n; j++) {
    if (colors[j] < colors[i]) {
      int temp = colors[i];
      colors[i] = colors[j];
      colors[j] = temp;
    }
  }
}

✓ After

int low = 0, mid = 0, high = n - 1;
while (mid <= high) {
  if (colors[mid] == 0) {
    int temp = colors[low];
    colors[low] = colors[mid];
    colors[mid] = temp;
    low++;
    mid++;
  } else if (colors[mid] == 1) {
    mid++;
  } else {
    int temp = colors[mid];
    colors[mid] = colors[high];
    colors[high] = temp;
    high--;
  }
}

What It Enables

This method lets you sort three groups quickly in one pass, saving time and effort in many real-world sorting tasks.

Real Life Example

Sorting laundry by color: whites, darks, and colors. Instead of sorting each piece multiple times, you can quickly separate them in one go using this method.

Key Takeaways

Manual sorting is slow and error-prone.

Two-pointer Dutch Flag sorts in one pass efficiently.

It uses swapping and pointers to organize colors fast.

Practice

1. What is the main goal of the Sort Colors Two Pointer Dutch Flag algorithm?

easy

A. To sort an array containing only 0s, 1s, and 2s in a single pass

B. To sort any array of integers in ascending order

C. To find the maximum element in an array

D. To reverse the elements of an array

Solution

1. Step 1: Understand the problem context

   The algorithm is designed specifically for arrays with only three distinct values: 0, 1, and 2.

2. Step 2: Identify the sorting goal

   The goal is to sort these values so that all 0s come first, then all 1s, and finally all 2s, in one pass without extra space.

3. Final Answer:

   To sort an array containing only 0s, 1s, and 2s in a single pass -> Option A

4. Quick Check:

   Sort Colors = Sort 0,1,2 in one pass [OK]

Hint: Remember: Dutch Flag sorts 0,1,2 in one pass [OK]

Common Mistakes:

• Thinking it sorts any integer array
• Confusing it with general sorting algorithms
• Assuming it reverses the array

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

easy

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

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

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

D. int low = 0, mid = 0, high = 0;

Solution

1. Step 1: Recall pointer roles

   Low starts at 0 to track the boundary of 0s, mid starts at 0 to scan elements, high starts at n-1 to track boundary of 2s.

2. Step 2: Match initialization

   Only int low = 0, mid = 0, high = n - 1; correctly initializes low and mid at 0 and high at n-1.

3. Final Answer:

   int low = 0, mid = 0, high = n - 1; -> Option A

4. Quick Check:

   Pointers start low=0, mid=0, high=end [OK]

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

Common Mistakes:

• Setting high to 0 instead of n-1
• Starting mid at n-1
• Confusing low and high initialization

3. What will be the output array after running the Dutch National Flag algorithm on arr = {2, 0, 2, 1, 1, 0}?

medium

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

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

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

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

Solution

1. Step 1: Trace the algorithm on input

   Initial array: {2, 0, 2, 1, 1, 0}. After sorting: all 0s first, then 1s, then 2s.

2. Step 2: Final sorted array

   Sorted array is [0, 0, 1, 1, 2, 2].

3. Final Answer:

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

4. Quick Check:

   Sorted colors = 0s, then 1s, then 2s [OK]

Hint: Sort 0s, then 1s, then 2s in one pass [OK]

Common Mistakes:

• Not sorting 0s and 2s correctly
• Mixing order of 1s and 0s
• Outputting original array

4. Identify the error in this C code snippet implementing the Dutch National Flag algorithm:

void sortColors(int* arr, int n) {
  int low = 0, mid = 0, high = n - 1;
  while (mid <= high) {
    if (arr[mid] == 0) {
      int temp = arr[low]; arr[low] = arr[mid]; arr[mid] = temp;
      low++;
      mid++;
    } else if (arr[mid] == 1) {
      mid++;
    } else {
      int temp = arr[mid]; arr[mid] = arr[high]; arr[high] = temp;
      high--;
    }
  }
}

medium

A. Missing decrement of high after swapping with low

B. Missing increment of mid after swapping with low

C. Missing increment of mid after swapping with high

D. Incorrect initialization of pointers

Solution

1. Step 1: Analyze pointer increments

   When swapping arr[mid] with arr[low] (for 0), low is incremented but mid is not incremented, causing infinite loop or wrong behavior.

2. Step 2: Correct pointer movement

   After swapping with low, both low and mid should be incremented to move forward.

3. Final Answer:

   Missing increment of mid after swapping with low -> Option B

4. Quick Check:

   Increment mid after swapping with low [OK]

Hint: Increment mid after swapping with low pointer [OK]

Common Mistakes:

• Not incrementing mid after swapping with low
• Incrementing mid after swapping with high
• Incorrect pointer initialization

5. Given an array arr = {1, 2, 0, 1, 2, 0, 1}, how many swaps will the Dutch National Flag algorithm perform to sort it?

hard

A. 3

B. 5

C. 6

D. 4

Solution

1. Step 1: Trace swaps during sorting

   Initial array: {1, 2, 0, 1, 2, 0, 1}
   - Swap 1: mid=1 (2) with high=6 (1) -> [1,1,0,1,2,0,2], high=5
   mid=1 now 1 -> mid=2
   - Swap 2: mid=2 (0) with low=0 (1) -> [0,1,1,1,2,0,2], low=1, mid=3
   mid=3 (1) -> mid=4
   - Swap 3: mid=4 (2) with high=5 (0) -> [0,1,1,1,0,2,2], high=4
   mid=4 now 0 ->
   - Swap 4: mid=4 (0) with low=1 (1) -> [0,0,1,1,1,2,2], low=2, mid=5 (stop)

2. Step 2: Count total swaps

   Total swaps counted are 4.

3. Final Answer:

   4 -> Option D

4. Quick Check:

   Count swaps carefully during pointer moves [OK]

Hint: Count swaps only when actual element exchange happens [OK]

Common Mistakes:

• Counting pointer increments as swaps
• Missing swaps when swapping with high
• Undercounting swaps for consecutive swaps on same mid