Dutch National Flag Three Way Partition in DSA Javascript - 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.

How does the number of steps change when we divide the array into three groups using the Dutch National Flag method?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

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

This code sorts the array into three parts: less than pivot, equal to pivot, and greater than pivot, all in one pass.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: The while loop that moves 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 amount.

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; doubling the array roughly doubles the work.

Final Time Complexity

Time Complexity: O(n)

This means the time to complete the partition grows directly in proportion to the size of the array.

Common Mistake

[X] Wrong: "Because there are swaps inside the loop, the time complexity must be quadratic O(n²)."

[OK] Correct: Each element is swapped at most once or twice, and the main loop moves forward without repeating elements, so the total steps stay linear.

Interview Connect

Understanding this linear time partition helps you handle problems that require grouping or sorting with limited passes, a valuable skill in many coding challenges.

Self-Check

"What if we used two pointers instead of three? 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 find the maximum element in an array

B. To sort an array in ascending order using quicksort

C. To reverse the elements of an array

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

Solution

1. Step 1: Understand the partition goal

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

2. Step 2: Identify the rearrangement

   It rearranges elements so that all smaller come first, then equal, then larger, without full sorting.

3. Final Answer:

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

4. Quick Check:

   Partitioning = Rearranging by pivot [OK]

Hint: Remember: three parts - 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 initialize pointers for the Dutch National Flag algorithm in JavaScript?

easy

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

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

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

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

Solution

1. Step 1: Identify pointer roles

   Low and mid start at 0, high at last index to cover full array.

2. Step 2: Match initialization to algorithm

   Correct initialization is low=0, mid=0, high=arr.length-1 to scan and partition properly.

3. Final Answer:

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

4. Quick Check:

   Pointers start at array ends and middle [OK]

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

Common Mistakes:

• Setting high to 0 instead of last index
• Starting mid at wrong position
• Using array length instead of length - 1

3. What will be the output of the following code snippet?

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

console.log(dutchFlag([2, 5, 1, 2, 3, 2, 4], 2));

medium

A. [1, 2, 2, 2, 5, 3, 4]

B. [2, 2, 2, 1, 3, 4, 5]

C. [1, 2, 2, 2, 3, 4, 5]

D. [1, 2, 2, 2, 4, 3, 5]

Solution

1. Step 1: Trace swaps for elements less than pivot

   Elements less than 2 (which is 1) move to front; 1 swaps to index 0.

2. Step 2: Place all 2s in middle and greater elements at end

   All 2s stay in middle; 3,4,5 move to the right side in any order preserving partition.

3. Final Answer:

   [1, 2, 2, 2, 3, 4, 5] -> Option C

4. Quick Check:

   Partitioned array with less, equal, greater [OK]

Hint: Swap less than pivot left, greater right [OK]

Common Mistakes:

• Not swapping correctly when element equals pivot
• Mixing order of greater elements
• Forgetting to increment mid after swap with low

4. Identify the bug in this Dutch National Flag partition code:

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

medium

A. The pointers low and mid are initialized incorrectly

B. The while loop condition should be mid <= high, not mid < high

C. The swap operation is missing for elements equal to pivot

D. The high pointer should be incremented, not decremented

Solution

1. Step 1: Check loop condition correctness

   The loop must run while mid is less than or equal to high to process all elements.

2. Step 2: Identify impact of incorrect condition

   Using mid < high skips processing when mid equals high, missing last element check.

3. Final Answer:

   The while loop condition should be mid <= high, not mid < high -> Option B

4. Quick Check:

   Loop must cover all elements [OK]

Hint: Use mid <= high to cover all elements [OK]

Common Mistakes:

• Using mid < high causes last element to be skipped
• Incorrect pointer initialization
• Misunderstanding swap necessity for equal elements

5. Given an array [4, 9, 4, 7, 4, 8, 7] and pivot value 7, what will be the final array after applying the Dutch National Flag partition algorithm?

hard

A. [4, 4, 4, 7, 7, 8, 9]

B. [4, 4, 4, 7, 8, 7, 9]

C. [4, 4, 4, 7, 8, 9, 7]

D. [7, 7, 4, 4, 4, 8, 9]

Solution

1. Step 1: Partition elements less than pivot (7)

   All 4s move to the front: [4, 4, 4, ...]

2. Step 2: Place all 7s in the middle and greater elements at the end

   7s stay in the middle; 8 and 9 move to the right side in sorted order.

3. Final Answer:

   [4, 4, 4, 7, 7, 8, 9] -> Option A

4. Quick Check:

   Less, equal, greater partitioned correctly [OK]

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

Common Mistakes:

• Mixing order of greater elements
• Placing pivot elements incorrectly
• Not moving all less elements to front