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DSA Pythonprogramming~10 mins

Sort Colors Two Pointer Dutch Flag in DSA Python - Interactive Practice

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Practice - 5 Tasks
Answer the questions below
1fill in blank
easy

Complete the code to initialize the left pointer at the start of the list.

DSA Python
left = [1]
Drag options to blanks, or click blank then click option'
A1
B0
C-1
Dlen(nums)
Attempts:
3 left
💡 Hint
Common Mistakes
Starting left pointer at 1 or -1 causes skipping the first element.
2fill in blank
medium

Complete the code to initialize the right pointer at the end of the list.

DSA Python
right = [1]
Drag options to blanks, or click blank then click option'
A0
B-1
Clen(nums) - 1
Dlen(nums)
Attempts:
3 left
💡 Hint
Common Mistakes
Using len(nums) causes index out of range errors.
3fill in blank
hard

Fix the error in the while loop condition to continue sorting while pointers do not cross.

DSA Python
while [1] <= right:
Drag options to blanks, or click blank then click option'
Anums
Bleft
Cright
Di
Attempts:
3 left
💡 Hint
Common Mistakes
Using 'left' instead of 'i' causes logic errors.
4fill in blank
hard

Fill both blanks to swap elements and move the left pointer forward.

DSA Python
nums[[1]], nums[[2]] = nums[[2]], nums[[1]]
left += 1
Drag options to blanks, or click blank then click option'
Ai
Bleft
Cright
D0
Attempts:
3 left
💡 Hint
Common Mistakes
Swapping with 'right' instead of 'left' causes wrong sorting.
5fill in blank
hard

Fill both blanks to swap elements, move the right pointer backward, and continue without incrementing i.

DSA Python
nums[[1]], nums[[2]] = nums[[2]], nums[[1]]
right -= 1
continue
Drag options to blanks, or click blank then click option'
Ai
Bleft
Cright
D0
Attempts:
3 left
💡 Hint
Common Mistakes
Incrementing 'i' after swapping with 'right' causes missed checks.

Practice

(1/5)
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
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