Dutch National Flag Three Way Partition in DSA C++ - Time & Space Complexity

Time Complexity: Dutch National Flag Three Way Partition

O(n)

Understanding Time Complexity

We want to understand how the time needed to sort an array into three parts grows as the array gets bigger.

Specifically, how does the number of steps change when we separate values into less than, equal to, and greater than a pivot?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

void dutchFlagPartition(int pivot, std::vector<int>& arr) {
    int low = 0, mid = 0, high = arr.size() - 1;
    while (mid <= high) {
        if (arr[mid] < pivot) {
            std::swap(arr[low++], arr[mid++]);
        } else if (arr[mid] == pivot) {
            mid++;
        } else {
            std::swap(arr[mid], arr[high--]);
        }
    }
}
    

This code rearranges the array so that all elements less than the pivot come first, then elements equal to the pivot, then elements greater than the pivot.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: The while loop that moves through the array with the mid pointer.
• How many times: Each element is visited at most once by mid, so the loop runs roughly n times for an array of size n.

How Execution Grows With Input

As the array size grows, the number of steps grows roughly the same amount because each element is checked once.

Input Size (n)	Approx. Operations
10	About 10 checks and swaps
100	About 100 checks and swaps
1000	About 1000 checks and swaps

Pattern observation: The work grows linearly with the input size.

Final Time Complexity

Time Complexity: O(n)

This means the time to complete the partition grows directly in proportion to the number of elements.

Common Mistake

[X] Wrong: "The algorithm might check some elements multiple times, so it could be slower than linear."

[OK] Correct: Each element is moved or checked at most once, so the loop runs only about n times, keeping it efficient.

Interview Connect

Understanding this partitioning method shows you can handle in-place sorting and smart pointer movement, skills useful in many coding problems.

Self-Check

"What if we changed the pivot to be the median of the array instead of a fixed value? How would the time complexity change?"

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