Sort Colors Two Pointer Dutch Flag in DSA Python - 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 the two-pointer Dutch Flag method 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.

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

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

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: At most once per element, so it runs about n times where n is the number of elements.

How Execution Grows With Input

As the list gets longer, the number of steps grows roughly the same as the number of elements.

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

Pattern observation: The steps increase linearly with input size.

Final Time Complexity

Time Complexity: O(n)

This means the time to sort grows directly in 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 moves through the list only once, and swaps happen in constant time without extra loops, so the total steps grow linearly.

Interview Connect

Understanding this linear time sorting method shows you can handle problems efficiently with clever pointer use, a skill valued in many coding challenges.

Self-Check

"What if we used three separate loops to sort each color instead of one loop with pointers? How would the time complexity change?"

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