Sort Colors Two Pointer Dutch Flag in DSA Python - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of Dutch Flag Algorithm on Mixed Colors

What is the output of the following code after sorting the colors using the Dutch Flag algorithm?

DSA Python

def sortColors(nums):
    low, mid, high = 0, 0, len(nums) - 1
    while mid <= high:
        if nums[mid] == 0:
            nums[low], nums[mid] = nums[mid], nums[low]
            low += 1
            mid += 1
        elif nums[mid] == 1:
            mid += 1
        else:
            nums[mid], nums[high] = nums[high], nums[mid]
            high -= 1

nums = [2, 0, 2, 1, 1, 0]
sortColors(nums)
print(nums)

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

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

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

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

💡 Hint

Think about how the pointers low, mid, and high move and swap elements.

The Dutch Flag algorithm sorts the array in one pass by moving 0s to the front, 2s to the end, and 1s stay in the middle.

🧠 Conceptual

intermediate

Number of Passes in Dutch Flag Algorithm

How many passes over the array does the Dutch Flag algorithm make to sort colors?

A Depends on the input size

B One pass

C Three passes

D Two passes

💡 Hint

Consider how the pointers move through the array.

The algorithm uses three pointers and sorts the array in a single pass by swapping elements as needed.

🔧 Debug

advanced

Identify the Bug in Dutch Flag Implementation

What error will this code produce when run?

DSA Python

def sortColors(nums):
    low, mid, high = 0, 0, len(nums) - 1
    while mid < high:
        if nums[mid] == 0:
            nums[low], nums[mid] = nums[mid], nums[low]
            low += 1
            mid += 1
        elif nums[mid] == 1:
            mid += 1
        else:
            nums[mid], nums[high] = nums[high], nums[mid]
            high -= 1

nums = [2, 0, 1]
sortColors(nums)
print(nums)

A IndexError

B [0, 1, 2]

C SyntaxError

D Infinite loop

💡 Hint

Check the loop condition carefully.

The loop condition should be 'mid <= high' to ensure all elements are processed. Using 'mid < high' can cause an infinite loop.

❓ Predict Output

advanced

Final Array State After Sorting

What is the final state of the array after running this code?

DSA Python

def sortColors(nums):
    low, mid, high = 0, 0, len(nums) - 1
    while mid <= high:
        if nums[mid] == 0:
            nums[low], nums[mid] = nums[mid], nums[low]
            low += 1
            mid += 1
        elif nums[mid] == 1:
            mid += 1
        else:
            nums[mid], nums[high] = nums[high], nums[mid]
            high -= 1

nums = [1, 2, 0, 1, 2, 0, 1]
sortColors(nums)
print(nums)

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

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

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

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

💡 Hint

Remember the order: all 0s first, then 1s, then 2s.

The algorithm sorts the array so that all 0s come first, then 1s, then 2s.

🧠 Conceptual

expert

Why Dutch Flag Algorithm is Optimal for Sorting Colors?

Why is the Dutch Flag algorithm considered optimal for sorting an array containing only 0s, 1s, and 2s?

A It sorts the array by counting each color and then overwriting

B It uses recursion to divide and conquer the array

C It sorts the array in one pass with constant extra space

D It uses built-in sort which is optimized for small ranges

💡 Hint

Think about time complexity and space usage.

The Dutch Flag algorithm sorts the array in a single pass and uses only a few pointers, so it uses constant extra space and linear time.

Practice

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

easy

A. To reverse the elements of an array

B. To sort any array of integers in ascending order

C. To find the maximum element in an array

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

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 D

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 array, not just 0,1,2
• Confusing it with general sorting algorithms
• Assuming it requires multiple passes

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

easy

A. low, mid, high = 0, len(nums) - 1, 0

B. low, mid, high = 0, 0, len(nums) - 1

C. low, mid, high = len(nums) - 1, 0, 0

D. low, mid, high = len(nums) - 1, len(nums) - 1, 0

Solution

1. Step 1: Recall pointer roles

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

2. Step 2: Match initialization

   So, low = 0, mid = 0, high = len(nums) - 1 is correct.

3. Final Answer:

   low, mid, high = 0, 0, len(nums) - 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:

• Swapping high and mid initialization
• Starting pointers at wrong indices
• Using length instead of length-1 for high

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

nums = [2, 0, 2, 1, 1, 0]
low, mid, high = 0, 0, len(nums) - 1
while mid <= high:
    if nums[mid] == 0:
        nums[low], nums[mid] = nums[mid], nums[low]
        low += 1
        mid += 1
    elif nums[mid] == 1:
        mid += 1
    else:
        nums[mid], nums[high] = nums[high], nums[mid]
        high -= 1
print(nums)

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, 1, 0, 2, 2]

Solution

1. Step 1: Trace the algorithm on input

   Initially nums = [2,0,2,1,1,0]. The algorithm swaps 0s to front, 2s to end, 1s stay in middle.

2. Step 2: Final sorted array

   After the loop, nums becomes [0,0,1,1,2,2], sorted by color.

3. Final Answer:

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

4. Quick Check:

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

Hint: Remember: 0s front, 1s middle, 2s end [OK]

Common Mistakes:

• Not swapping correctly when nums[mid] == 2
• Incrementing mid after swapping with high
• Confusing order of pointers

4. Identify the bug in the following Dutch Flag algorithm code:

nums = [2, 0, 1]
low, mid, high = 0, 0, len(nums) - 1
while mid < high:
    if nums[mid] == 0:
        nums[low], nums[mid] = nums[mid], nums[low]
        low += 1
        mid += 1
    elif nums[mid] == 1:
        mid += 1
    else:
        nums[mid], nums[high] = nums[high], nums[mid]
        high -= 1
print(nums)

medium

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

B. Pointers low and mid are initialized incorrectly

C. Swapping logic for nums[mid] == 0 is wrong

D. The decrement of high should be after incrementing mid

Solution

1. Step 1: Analyze loop condition

   The loop runs while mid < high, so it stops when mid == high, missing last element check.

2. Step 2: Correct loop condition

   It should be mid <= high to process all elements including when mid equals high.

3. Final Answer:

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

4. Quick Check:

   Loop must cover mid == high [OK]

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

Common Mistakes:

• Using 'mid < high' misses last element
• Changing pointer initialization unnecessarily
• Incorrect swap logic for 0 or 2

5. Given an array nums = [1, 2, 0, 1, 2, 0, 1], how many swaps will the Dutch Flag algorithm perform to sort it?

hard

A. 4

B. 5

C. 3

D. 6

Solution

1. Step 1: Trace swaps during sorting

   Initial: [1,2,0,1,2,0,1] low=0, mid=0, high=6
   - mid=0 (1): mid=1 (no swap)
   - mid=1 (2): swap(1,6) -> [1,1,0,1,2,0,2] high=5 (swap 1)
   - mid=1 (1): mid=2
   - mid=2 (0): swap(0,2) -> [0,1,1,1,2,0,2] low=1 mid=3 (swap 2)
   - mid=3 (1): mid=4
   - mid=4 (2): swap(4,5) -> [0,1,1,1,0,2,2] high=4 (swap 3)
   - mid=4 (0): swap(1,4) -> [0,0,1,1,1,2,2] low=2 mid=5 (swap 4)

2. Step 2: Total swaps counted

   There are 4 swaps in total during the process.

3. Final Answer:

   4 -> Option A

4. Quick Check:

   Count swaps carefully during pointer moves [OK]

Hint: Count swaps when 0 or 2 moves happen [OK]

Common Mistakes:

• Counting pointer moves as swaps
• Missing swaps when swapping with same element
• Not tracing all pointer changes