Dutch National Flag Three Way Partition in DSA C++ - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to initialize the low pointer to start of the array.

DSA C++

int low = [1];

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

An

B1

C-1

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Starting low at 1 instead of 0.

Starting low at -1 which is invalid index.

The low pointer should start at index 0 to begin partitioning from the start.

2fill in blank

medium

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

DSA C++

std::swap(arr[[1]], arr[low]);

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Ahigh

Bi

Cpivot

Dlow

Attempts:

3 left

💡 Hint

Common Mistakes

Swapping with high instead of low.

Swapping with pivot index which is not used here.

We swap the current element at i with the element at low to move smaller elements to the front.

3fill in blank

hard

Fix the error in the loop condition to process all elements.

DSA C++

while (i [1] high) {

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A<=

B<

C>

D>=

Attempts:

3 left

💡 Hint

Common Mistakes

Using < causes last element to be skipped.

Using > or >= reverses the loop logic.

The loop should run while i is less than or equal to high to process all elements.

4fill in blank

hard

Fill both blanks to correctly update pointers after swapping smaller element.

DSA C++

std::swap(arr[i], arr[[1]]);
[2]++;
i++;

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

Alow

Bhigh

Ci

Dpivot

Attempts:

3 left

💡 Hint

Common Mistakes

Incrementing high instead of low.

Not incrementing i causing infinite loop.

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

5fill in blank

hard

Fill both blanks to correctly swap and update pointers for larger elements.

DSA C++

std::swap(arr[i], arr[[1]]);
[2]--;
// do not increment i here

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

Alow

Bhigh

Ci

Dpivot

Attempts:

3 left

💡 Hint

Common Mistakes

Incrementing i after swapping with high causes skipping elements.

Swapping with low instead of high.

Swap with high, decrement high pointer, and do not increment i to recheck swapped element.

Practice

1. What is the main goal of the Dutch National Flag Three Way Partition algorithm?

easy

A. To rearrange an array so that all elements less than a pivot come first, then equal, then greater

B. To sort an array in ascending order using quicksort

C. To find the maximum element in an array

D. To reverse the elements of an array

Solution

1. Step 1: Understand the partitioning goal

   The algorithm divides the array into three parts: less than pivot, equal to pivot, and greater than pivot.

2. Step 2: Recognize the rearrangement purpose

   This helps in sorting or grouping elements efficiently without full sorting.

3. Final Answer:

   To rearrange an array so that all elements less than a pivot come first, then equal, then greater -> Option A

4. Quick Check:

   Partitioning = Rearranging by pivot [OK]

Hint: Remember: three groups - less, equal, greater [OK]

Common Mistakes:

• Confusing partition with full sorting
• Thinking it only finds max or min
• Assuming it reverses the array

2. Which of the following is the correct way to declare the pointers used in the Dutch National Flag algorithm in C++?

easy

A. int low = n - 1, mid = 0, high = 0;

B. int low, mid, high = 0;

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

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

Solution

1. Step 1: Identify pointer initialization

   In the algorithm, low and mid start at 0, high starts at n-1 (last index).

2. Step 2: Match correct initialization

   int low = 0, mid = 0, high = n - 1; correctly initializes low and mid to 0, high to n-1.

3. Final Answer:

   int low = 0, mid = 0, high = n - 1; -> Option C

4. Quick Check:

   Pointers start: low=0, mid=0, high=n-1 [OK]

Hint: Start low and mid at 0, high at last index [OK]

Common Mistakes:

• Initializing pointers incorrectly
• Setting high to 0 instead of n-1
• Not initializing mid at 0

3. What will be the output array after applying the Dutch National Flag algorithm on arr = [2, 0, 2, 1, 1, 0] with pivot = 1?

medium

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

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

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

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

Solution

1. Step 1: Partition array by pivot 1

   Elements less than 1: 0,0; equal to 1: 1,1; greater than 1: 2,2.

2. Step 2: Rearrange array accordingly

   Place all 0s first, then 1s, then 2s resulting in [0, 0, 1, 1, 2, 2].

3. Final Answer:

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

4. Quick Check:

   Less, equal, greater order = [0,0,1,1,2,2] [OK]

Hint: Group by less, equal, greater around pivot [OK]

Common Mistakes:

• Not grouping equal elements in the middle
• Mixing order of groups
• Assuming full sort instead of partition

4. Given the following code snippet for Dutch National Flag partition, what is the error?

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

medium

A. Pivot value is not defined

B. Incorrect swap function usage

C. Loop condition should be mid < high

D. Missing increment of mid when arr[mid] < pivot

Solution

1. Step 1: Analyze pointer increments

   When arr[mid] < pivot, low is incremented but mid is not incremented, causing infinite loop.

2. Step 2: Identify missing mid increment

   Mid must also increment after swapping to move forward.

3. Final Answer:

   Missing increment of mid when arr[mid] < pivot -> Option D

4. Quick Check:

   Increment mid after swap for less than pivot [OK]

Hint: Always increment mid except when swapping with high [OK]

Common Mistakes:

• Forgetting to increment mid after swap with low
• Changing loop condition incorrectly
• Ignoring pivot initialization

5. You have an array arr = [3, 5, 2, 3, 8, 3, 1] and want to partition it using the Dutch National Flag algorithm with pivot = 3. Which of the following is the correct final partitioned array?

hard

A. [1, 2, 3, 3, 3, 5, 8]

B. [3, 3, 3, 1, 2, 5, 8]

C. [1, 2, 3, 3, 3, 8, 5]

D. [5, 8, 3, 3, 3, 1, 2]

Solution

1. Step 1: Group elements by pivot 3

   Less than 3: 1, 2; equal to 3: 3, 3, 3; greater than 3: 5, 8.

2. Step 2: Arrange groups in order

   Place less first, then equal, then greater: [1, 2, 3, 3, 3, 5, 8].

3. Step 3: Check options for exact order

   [1, 2, 3, 3, 3, 5, 8] matches the correct partition with greater elements 5 and 8 in any order, which is valid partitioning.

4. Final Answer:

   [1, 2, 3, 3, 3, 5, 8] -> Option A

5. Quick Check:

   Partition order less, equal, greater = [1,2,3,3,3,5,8] [OK]

Hint: Partition keeps groups, order inside groups can vary [OK]

Common Mistakes:

• Assuming full sort instead of partition
• Mixing order of greater elements
• Placing equal elements incorrectly