Sort Colors Two Pointer Dutch Flag in DSA Javascript - 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.

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

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: Single while loop that moves pointers and swaps elements.
• How many times: The loop runs at most once through the array, so up to n times where n is the array length.

How Execution Grows With Input

As the array size grows, the number of steps grows roughly the same amount.

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

Pattern observation: The steps grow linearly with input size; doubling input roughly doubles the work.

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 swaps, the time must be quadratic O(n²)."

[OK] Correct: The swaps happen inside a single loop that moves pointers forward or backward, so each element is handled only once or twice, keeping time linear.

Interview Connect

Understanding this linear time sorting method shows you can solve problems efficiently by clever pointer use, 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 find the maximum value in an array

B. To sort any array of integers in ascending order

C. To reverse the elements of 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 sorting goal

   The goal is to sort these values so that all 0s come first, then all 1s, and finally all 2s, in one pass without extra space.

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 integer array
• Confusing it with general sorting algorithms
• Assuming it reverses the array

2. Which of the following is the correct initialization for the two pointers in the Dutch Flag algorithm in JavaScript?

easy

A. let low = 0, mid = 0, high = nums.length - 1;

B. let low = 0, mid = nums.length - 1, high = 0;

C. let low = nums.length - 1, mid = 0, high = 0;

D. let low = nums.length - 1, mid = nums.length - 1, high = 0;

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 to roles

   Correct initialization is let low = 0, mid = 0, high = nums.length - 1;.

3. Final Answer:

   let low = 0, mid = 0, high = nums.length - 1; -> Option A

4. Quick Check:

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

Hint: Start low and mid at 0, high at array end [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 array after running the Dutch Flag algorithm on nums = [2, 0, 2, 1, 1, 0]?

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: Apply Dutch Flag sorting logic

   Sort 0s to front, 1s in middle, 2s at end in one pass.

2. Step 2: Result after sorting

   The sorted array is [0, 0, 1, 1, 2, 2].

3. Final Answer:

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

4. Quick Check:

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

Hint: Sort 0s front, 1s middle, 2s end in one pass [OK]

Common Mistakes:

• Not sorting all 0s to front
• Mixing order of 1s and 2s
• Returning original unsorted array

4. Identify the bug in this snippet of the Dutch Flag algorithm:

let low = 0, mid = 0, high = nums.length - 1;
while (mid <= high) {
  if (nums[mid] === 0) {
    [nums[low], nums[mid]] = [nums[mid], nums[low]];
    low++;
  } else if (nums[mid] === 2) {
    [nums[mid], nums[high]] = [nums[high], nums[mid]];
    high--;
  } else {
    mid++;
  }
}

medium

A. Incorrect comparison operator in while loop

B. Missing increment of mid after swapping 0s

C. Swapping 2s with low instead of high

D. No bug, code is correct

Solution

1. Step 1: Analyze pointer increments after swaps

   When swapping 0s, mid pointer must also move forward to avoid infinite loop.

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

   The code increments low but forgets to increment mid after swapping 0s.

3. Final Answer:

   Missing increment of mid after swapping 0s -> Option B

4. Quick Check:

   mid++ missing after 0 swap causes infinite loop [OK]

Hint: Increment mid after swapping 0s to avoid infinite loop [OK]

Common Mistakes:

• Forgetting mid++ after swapping 0s
• Swapping 2s with wrong pointer
• Incorrect loop condition

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

hard

A. Use a separate array to track swaps

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

C. Add a counter variable and increment it each time a swap occurs

D. Increment a counter only when mid pointer moves

Solution

1. Step 1: Understand swap operations in algorithm

   Swaps happen when 0 or 2 is encountered and elements are exchanged.

2. Step 2: Track swaps with a counter

   Introduce a variable initialized to zero and increment it every time a swap line executes.

3. Final Answer:

   Add a counter variable and increment it each time a swap occurs -> Option C

4. Quick Check:

   Count swaps by incrementing counter on each swap [OK]

Hint: Increment counter inside swap blocks to count swaps [OK]

Common Mistakes:

• Counting pointer moves instead of swaps
• Counting elements instead of swaps
• Using extra arrays unnecessarily