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 algorithm for sorting colors?

It uses three pointers to separate the array into three parts: all 0s to the left, all 2s to the right, and all 1s in the middle, sorting the colors in one pass.

beginner

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

The pointers represent: low - boundary for 0s, mid - current element being checked, high - boundary for 2s.

intermediate

Why does the Dutch Flag algorithm run in O(n) time?

Because it makes only one pass through the array, moving pointers and swapping elements without nested loops.

beginner

What happens when the mid pointer encounters a 0 in the Dutch Flag algorithm?

Swap the element at mid with the element at low, then increment both low and mid pointers.

intermediate

How does the algorithm handle encountering a 2 at the mid pointer?

Swap the element at mid with the element at high, then decrement high pointer without incrementing mid immediately.

What is the initial position of the 'high' pointer in the Dutch Flag algorithm?

A At the middle of the array

B At the first index of the array

C At the last index of the array

D At the second index of the array

The 'high' pointer starts at the last index to track where 2s should be placed.

When the mid pointer points to a 1, what does the algorithm do?

A Just increment mid pointer

B Swap with low pointer and increment both

C Swap with high pointer and decrement high

D Stop the algorithm

When mid points to 1, it is already in the correct middle section, so only mid is incremented.

Which of these is NOT a color represented in the Sort Colors problem?

A 0

B 1

C 2

D 3

The problem only involves colors 0, 1, and 2.

What is the main advantage of the Dutch Flag algorithm over sorting the array normally?

A It sorts in one pass with constant space

B It sorts in multiple passes

C It only works for sorted arrays

D It uses extra memory

The algorithm sorts the array in one pass and uses constant extra space.

What happens if you increment mid after swapping with high when mid points to 2?

A It will sort correctly anyway

B You might skip checking the swapped element

C The algorithm will be faster

D It will cause an infinite loop

Incrementing mid after swapping with high can skip checking the new element at mid, causing errors.

Explain the step-by-step process of the Dutch Flag algorithm for sorting colors.

Think about how the array is divided into three parts and how each color is placed.

Why is the Dutch Flag algorithm efficient for sorting an array with only three distinct values?

Consider time and space complexity compared to general sorting.

Practice

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

easy

A. To find the maximum element in an array

B. To sort any array of integers in ascending order

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

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 algorithm's purpose

   It sorts these values in one pass using two pointers, placing all 0s first, then 1s, then 2s.

3. Final Answer:

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

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 of pointers in the Dutch National Flag algorithm in C++?

easy

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

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

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

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

Solution

1. Step 1: Understand pointer roles

   Low and mid start at the beginning (index 0), high starts at the end (index n-1).

2. Step 2: Match initialization

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

3. Final Answer:

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

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 low or mid to n-1 incorrectly
• Initializing all pointers to zero
• Confusing high pointer position

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

std::vector<int> nums = {2, 0, 2, 1, 1, 0};
int low = 0, mid = 0, high = nums.size() - 1;
while (mid <= high) {
  if (nums[mid] == 0) {
    std::swap(nums[low], nums[mid]);
    low++; mid++;
  } else if (nums[mid] == 1) {
    mid++;
  } else {
    std::swap(nums[mid], nums[high]);
    high--;
  }
}
for (int num : nums) std::cout << num << " ";

medium

A. 0 0 1 1 2 2

B. 2 2 1 1 0 0

C. 0 1 0 1 2 2

D. 1 0 2 0 1 2

Solution

1. Step 1: Trace the sorting process

   Initially: {2,0,2,1,1,0} - Swap 2 and 0 at low and mid, low=1, mid=1 - mid points to 0, swap with low, low=2, mid=2 - mid points to 2, swap with high (0), high=4 - mid points to 0, swap with low, low=3, mid=3 - mid points to 1, mid++ - mid points to 1, mid++ Loop ends.

2. Step 2: Final array after sorting

   The array becomes: 0 0 1 1 2 2

3. Final Answer:

   0 0 1 1 2 2 -> Option A

4. Quick Check:

   Sorted colors output = 0 0 1 1 2 2 [OK]

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

Common Mistakes:

• Not swapping correctly when nums[mid] == 2
• Incrementing mid after swapping with high
• Confusing pointer updates

4. Identify the error in the following C++ code implementing the Dutch National Flag algorithm:

int low = 0, mid = 0, high = n - 1;
while (mid < high) {
  if (nums[mid] == 0) {
    std::swap(nums[low], nums[mid]);
    low++;
    mid++;
  } else if (nums[mid] == 1) {
    mid++;
  } else {
    std::swap(nums[mid], nums[high]);
    high--;
  }
}

medium

A. Pointers low and mid should start at n - 1

B. The high pointer should be incremented instead of decremented

C. Swapping should be done only when nums[mid] == 1

D. The loop condition should be 'mid <= high' instead of 'mid < high'

Solution

1. Step 1: Analyze loop condition

   The loop uses 'mid < high', which misses the case when mid equals high, leaving last element unsorted.

2. Step 2: Correct loop condition

   Changing to 'mid <= high' ensures all elements are processed including when mid equals high.

3. Final Answer:

   The loop condition should be 'mid <= high' instead of 'mid < high' -> Option D

4. Quick Check:

   Loop must cover mid == high for full sorting [OK]

Hint: Use 'mid <= high' to cover all elements [OK]

Common Mistakes:

• Using 'mid < high' misses last element
• Starting pointers incorrectly
• Wrong pointer increments or decrements

5. Given an array nums containing only 0s, 1s, and 2s, how would you modify the Dutch National Flag algorithm to count how many swaps are performed during sorting?

hard

A. Add a counter variable and increment it each time a swap is done inside the loop

B. Count the number of 0s, 1s, and 2s before sorting

C. Increment a counter every time mid pointer moves

D. Count swaps only when nums[mid] == 1

Solution

1. Step 1: Understand when swaps happen

   Swaps occur when nums[mid] is 0 or 2, swapping with low or high pointers respectively.

2. Step 2: Implement swap counting

   Introduce an integer counter initialized to zero, increment it inside the loop each time std::swap is called.

3. Final Answer:

   Add a counter variable and increment it each time a swap is done inside the loop -> Option A

4. Quick Check:

   Count swaps by incrementing on each swap call [OK]

Hint: Increment counter inside swap blocks only [OK]

Common Mistakes:

• Counting pointer moves instead of swaps
• Counting swaps for 1s which are not swapped
• Counting before sorting instead of during