Sort Colors Two Pointer Dutch Flag in DSA Java - 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 two pointers 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.

public void sortColors(int[] nums) {
    int low = 0, mid = 0, high = nums.length - 1;
    while (mid <= high) {
        if (nums[mid] == 0) {
            int temp = nums[low];
            nums[low] = nums[mid];
            nums[mid] = temp;
            low++; mid++;
        } else if (nums[mid] == 1) {
            mid++;
        } else {
            int temp = nums[mid];
            nums[mid] = nums[high];
            nums[high] = temp;
            high--;
        }
    }
}
    

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

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: Each element is visited at most once by mid, so up to n times where n is array length.

How Execution Grows With Input

As the array size grows, 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 if-else checks, the time complexity must be quadratic O(n²)."

[OK] Correct: The loop runs only once through the array, and each element is handled in constant time, so the overall time is linear, not quadratic.

Interview Connect

Knowing this linear time sorting method shows you can handle problems efficiently by using pointers smartly, a skill valued in many coding challenges.

Self-Check

"What if we changed the array to contain more than three colors? 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 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 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 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 way to initialize the pointers for the Dutch Flag algorithm in Java?

easy

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

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

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

D. int low = 0, mid = 0, high = arr.length - 1;

Solution

1. Step 1: Identify pointer roles

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

2. Step 2: Match initialization with roles

   int low = 0, mid = 0, high = arr.length - 1; correctly sets low and mid to 0, and high to last index (arr.length - 1).

3. Final Answer:

   int low = 0, mid = 0, high = arr.length - 1; -> Option D

4. Quick Check:

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

Hint: Low and mid start at 0, high at array end [OK]

Common Mistakes:

• Setting pointers all to zero
• Starting high pointer at 0 instead of end
• Confusing mid and high initial positions

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

medium

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

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

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

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

Solution

1. Step 1: Apply Dutch Flag sorting logic

   Start with low=0, mid=0, high=5. Swap 2 with 0 at low, move pointers accordingly to group 0s, then 1s, then 2s.

2. Step 2: Final sorted array

   After sorting, the array becomes [0, 0, 1, 1, 2, 2].

3. Final Answer:

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

4. Quick Check:

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

Hint: 0s first, then 1s, then 2s after sorting [OK]

Common Mistakes:

• Not swapping correctly
• Mixing order of 1s and 2s
• Assuming original order stays same

4. Identify the error in this Java code snippet for the Dutch Flag algorithm:

int low = 0, mid = 0, high = arr.length - 1;
while (mid <= high) {
  if (arr[mid] == 0) {
    swap(arr, low, mid);
    low++;
  } else if (arr[mid] == 1) {
    mid++;
  } else {
    swap(arr, mid, high);
    high--;
  }
}

medium

A. Swap function parameters are wrong

B. Incorrect condition in while loop

C. Missing increment of mid pointer after swapping 0

D. No error, code is correct

Solution

1. Step 1: Analyze pointer increments after swaps

   When swapping 0, mid pointer must also increment to avoid infinite loop.

2. Step 2: Identify missing mid++ after swapping 0

   Code increments low but forgets to increment mid after swapping 0, causing repeated processing of same element.

3. Final Answer:

   Missing increment of mid pointer after swapping 0 -> Option C

4. Quick Check:

   mid++ needed after swapping 0 [OK]

Hint: Always increment mid after swapping 0 [OK]

Common Mistakes:

• Forgetting to increment mid after swap
• Changing while loop condition incorrectly
• Swapping wrong indices

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

hard

A. 6

B. 4

C. 5

D. 7

Solution

1. Step 1: Trace swaps during sorting

   Initial array: [1,2,0,2,1,0,0,2,1]. Swaps occur when 0 or 2 are moved. Counting swaps:
   - Swap 1: mid=1 (2) with high=8
   - Swap 2: mid=2 (0) with low=0
   - Swap 3: mid=3 (2) with high=7
   - Swap 4: mid=3 (2) with high=6
   - Swap 5: mid=3 (0) with low=1
   - Swap 6: mid=5 (0) with low=2

2. Step 2: Total swaps counted

   Total swaps performed are 6.

3. Final Answer:

   6 -> Option A

4. Quick Check:

   Count swaps carefully during pointer moves [OK]

Hint: Count swaps when 0 or 2 move positions [OK]

Common Mistakes:

• Counting swaps only for 0s
• Ignoring swaps when swapping with same value
• Miscounting pointer moves as swaps