Quick Sort Partition Lomuto and Hoare in DSA Java - Time & Space Complexity

Time Complexity: Quick Sort Partition Lomuto and Hoare

O(n)

Understanding Time Complexity

We want to understand how the time taken by Quick Sort's partition steps grows as the input size increases.

Specifically, we compare the Lomuto and Hoare partition methods to see how their operations scale.

Scenario Under Consideration

Analyze the time complexity of these two partition methods used in Quick Sort.

// Lomuto Partition
int lomutoPartition(int[] arr, int low, int high) {
  int pivot = arr[high];
  int i = low - 1;
  for (int j = low; j < high; j++) {
    if (arr[j] <= pivot) {
      i++;
      swap(arr, i, j);
    }
  }
  swap(arr, i + 1, high);
  return i + 1;
}

// Hoare Partition
int hoarePartition(int[] arr, int low, int high) {
  int pivot = arr[low];
  int i = low - 1, j = high + 1;
  while (true) {
    do { i++; } while (arr[i] < pivot);
    do { j--; } while (arr[j] > pivot);
    if (i >= j) return j;
    swap(arr, i, j);
  }
}
    

Both methods rearrange elements around a pivot but use different approaches.

Identify Repeating Operations

Both methods repeatedly compare and swap elements to partition the array.

• Primary operation: Comparing elements to the pivot and swapping if needed.
• How many times: Each element between low and high is checked at least once.

How Execution Grows With Input

As the number of elements (n) increases, the number of comparisons and swaps grows roughly in proportion to n.

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

Pattern observation: The operations grow linearly with input size.

Final Time Complexity

Time Complexity: O(n)

This means the partition step takes time proportional to the number of elements it processes.

Common Mistake

[X] Wrong: "Partitioning takes constant time because it just picks a pivot."

[OK] Correct: Partitioning must check and possibly swap many elements, so its time grows with input size.

Interview Connect

Understanding partition time helps you explain Quick Sort's efficiency and shows you can analyze key algorithm steps clearly.

Self-Check

"What if the pivot is always the smallest element? How would that affect the partition's time complexity?"

Practice

1. What is the main difference between Lomuto and Hoare partition schemes in Quick Sort?

easy

A. Lomuto uses one pointer, Hoare uses two pointers moving inward

B. Lomuto sorts in descending order, Hoare sorts in ascending order

C. Lomuto uses recursion, Hoare uses iteration

D. Lomuto swaps only once, Hoare swaps multiple times

Solution

1. Step 1: Understand Lomuto partition

   Lomuto partition uses a single pointer to track the boundary of smaller elements and swaps elements accordingly.

2. Step 2: Understand Hoare partition

   Hoare partition uses two pointers starting from both ends moving inward to swap elements that are out of place.

3. Final Answer:

   Lomuto uses one pointer, Hoare uses two pointers moving inward -> Option A

4. Quick Check:

   Pointer count difference [OK]

Hint: Remember: Lomuto one pointer, Hoare two pointers [OK]

Common Mistakes:

• Confusing sorting order difference
• Thinking both use same pointer strategy
• Assuming Hoare uses recursion only

2. Which of the following is the correct syntax to swap elements at indices i and j in Java array arr?

easy

A. int temp = arr[i]; arr[i] = arr[j]; arr[j] = temp;

B. arr[i], arr[j] = arr[j], arr[i];

C. swap(arr[i], arr[j]);

D. arr.swap(i, j);

Solution

1. Step 1: Identify Java swap syntax

   Java does not support tuple unpacking or built-in swap method for arrays, so manual swap using a temporary variable is needed.

2. Step 2: Check each option

   int temp = arr[i]; arr[i] = arr[j]; arr[j] = temp; uses a temporary variable to swap correctly; others are invalid Java syntax.

3. Final Answer:

   int temp = arr[i]; arr[i] = arr[j]; arr[j] = temp; -> Option A

4. Quick Check:

   Manual swap with temp variable [OK]

Hint: Use temp variable to swap array elements in Java [OK]

Common Mistakes:

• Using Python-style tuple swap
• Calling non-existent swap method
• Trying to swap without temp variable

3. What is the output array after applying Lomuto partition on arr = [8, 3, 7, 6, 2] with pivot as last element?

medium

A. [8, 3, 7, 6, 2]

B. [3, 2, 7, 6, 8]

C. [2, 3, 7, 6, 8]

D. [2, 3, 6, 7, 8]

Solution

1. Step 1: Identify pivot and initialize pointers

   Pivot is last element 2. Initialize i = -1.

2. Step 2: Iterate and swap elements smaller than pivot

   Check elements: 8 > 2 no swap; 3 > 2 no swap; 7 > 2 no swap; 6 > 2 no swap. Then swap pivot with element at i+1 (index 0): swap arr[0] and arr[4].

3. Final Answer:

   [2, 3, 7, 6, 8] -> Option C

4. Quick Check:

   Lomuto partition output [OK]

Hint: Pivot last element, swap smaller left, pivot to i+1 [OK]

Common Mistakes:

• Not swapping pivot at end
• Confusing pivot position
• Swapping elements incorrectly

4. Find the error in this Hoare partition code snippet:

int hoarePartition(int[] arr, int low, int high) {
  int pivot = arr[low];
  int i = low + 1, j = high;
  while (true) {
    while (i <= high && arr[i] < pivot) i++;
    while (j >= low && arr[j] > pivot) j--;
    if (i >= j) break;
    int temp = arr[i]; arr[i] = arr[j]; arr[j] = temp;
  }
  int temp = arr[low]; arr[low] = arr[j]; arr[j] = temp;
  return j;
}

medium

A. Pointers i and j initialization is incorrect

B. Pivot should be last element, not first

C. The inner while conditions should use <= and >= incorrectly

D. No error, code is correct

Solution

1. Step 1: Check pointer initialization

   In Hoare partition, i starts at low + 1 and j at high, which is a valid approach.

2. Step 2: Analyze pointer movement and swapping

   The code correctly moves pointers inward and swaps elements until i >= j, then swaps pivot with arr[j].

3. Final Answer:

   No error, code is correct -> Option D

4. Quick Check:

   Pointer start positions and loop logic [OK]

Hint: Hoare pointers start inside range: i=low+1, j=high [OK]

Common Mistakes:

• Starting pointers outside array incorrectly
• Using wrong pivot position
• Misplacing swap after loop

5. Given array arr = [5, 3, 8, 4, 2, 7, 1, 10], after applying Hoare partition with pivot as first element, what is the correct partitioned array?

hard

A. [1, 3, 2, 4, 5, 7, 8, 10]

B. [2, 3, 1, 4, 5, 7, 8, 10]

C. [1, 2, 3, 4, 5, 7, 8, 10]

D. [5, 3, 8, 4, 2, 7, 1, 10]

Solution

1. Step 1: Initialize pointers and pivot

   Pivot = 5 (first element), i = low - 1 = -1, j = high + 1 = 8.

2. Step 2: Move pointers and swap

   i moves right skipping elements < 5: i=0(5), stop; j moves left skipping elements > 5: j=7(10),6(1 stop at 1). Swap arr[i]=5 and arr[j]=1. Array becomes [1,3,8,4,2,7,5,10]. Continue until i >= j.

3. Step 3: Final partitioned array

   After all swaps, array is [2, 3, 1, 4, 5, 7, 8, 10] with pivot index at 4.

4. Final Answer:

   [2, 3, 1, 4, 5, 7, 8, 10] -> Option B

5. Quick Check:

   Hoare partition swaps around pivot [OK]

Hint: Hoare swaps elements crossing pointers around pivot [OK]

Common Mistakes:

• Confusing pivot position after partition
• Not swapping correctly when pointers cross
• Assuming sorted order after one partition