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

Practice - 5 Tasks

1fill in blank

easy

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

DSA Javascript

let left = [1];

Drag options to blanks, or click blank then click option'

Acolors.length

B1

C-1

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Starting left pointer at 1 or negative index.

The left pointer should start at index 0 to begin sorting from the start.

2fill in blank

medium

Complete the code to swap elements at indices i and left.

DSA Javascript

let temp = colors[i]; colors[i] = colors[[1]]; colors[[2]] = temp;

Drag options to blanks, or click blank then click option'

Aright

Bleft

Ci

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Swapping with right pointer or same index i.

We swap the element at i with the element at left pointer to move 0s to the front.

3fill in blank

hard

Fix the error in the while loop condition to continue while i is less than or equal to right.

DSA Javascript

while (i [1] right) {

Drag options to blanks, or click blank then click option'

A<=

B<

C>

D>=

Attempts:

3 left

💡 Hint

Common Mistakes

Using < instead of <= or wrong comparison operator.

The loop should run while i is less than or equal to right to cover all elements.

4fill in blank

hard

Fill both blanks to correctly update pointers after swapping a 0.

DSA Javascript

if (colors[i] === 0) { swap(colors, i, [1]); [2]++; i++; }

Drag options to blanks, or click blank then click option'

Aleft

Bright

Ci

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Incrementing right instead of left or wrong swap index.

After swapping with left, increment left and i to move forward.

5fill in blank

hard

Fill all three blanks to correctly update pointers after swapping a 2.

DSA Javascript

else if (colors[i] === 2) { swap(colors, i, [1]); [2]--; [3] = [3]; }

Drag options to blanks, or click blank then click option'

Aleft

Bright

Ci

Di + 1

Attempts:

3 left

💡 Hint

Common Mistakes

Incrementing i after swapping 2 or swapping with left pointer.

Swap with right pointer, decrement right, and keep i same to check swapped element.

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