Quick Sort Partition Lomuto and Hoare in DSA C++ - 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 two partition methods: Lomuto and Hoare.

Scenario Under Consideration

Analyze the time complexity of these two partition functions 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++;
      std::swap(arr[i], arr[j]);
    }
  }
  std::swap(arr[i + 1], arr[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;
    std::swap(arr[i], arr[j]);
  }
}
    

These functions rearrange elements around a pivot to help Quick Sort divide the array.

Identify Repeating Operations

Both partitions scan parts of the array repeatedly:

• Primary operation: Comparing elements to the pivot and swapping.
• How many times: Each element in the current segment is checked once or twice.

How Execution Grows With Input

As the segment size n grows, the number of comparisons and swaps grows roughly in proportion.

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

Pattern observation: Operations grow linearly with input size for each partition call.

Final Time Complexity

Time Complexity: O(n) per partition call

This means the partition step looks at each element once or twice, so time grows directly with the number of elements.

Common Mistake

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

[OK] Correct: Partitioning must compare and possibly swap many elements, so it takes time proportional to the segment size, not a fixed small time.

Interview Connect

Understanding partition time helps you explain Quick Sort's efficiency and why pivot choice matters, a key skill in interviews.

Self-Check

"What if the pivot is always the smallest or largest element? How would that affect the time complexity of partition and overall Quick Sort?"

Practice

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

easy

A. Lomuto uses a single 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 chooses the last element as pivot, Hoare chooses the first element always

Solution

1. Step 1: Understand Lomuto partition

   Lomuto partition uses one 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 around the pivot.

3. Final Answer:

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

4. Quick Check:

   Lomuto single pointer, Hoare two pointers inward [OK]

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

Common Mistakes:

• Confusing pivot selection with partition method
• Thinking both use same pointer approach
• Assuming Hoare always picks first element as pivot

2. Which of the following is the correct Lomuto partition function header in C++?

easy

A. int partition(int arr[], int low, int high)

B. void partition(int arr[], int low, int high)

C. int partition(int arr[], int start, int end, int pivot)

D. bool partition(int arr[], int low, int high)

Solution

1. Step 1: Check return type

   Lomuto partition returns the pivot index as int after partitioning.

2. Step 2: Check parameters

   It takes array, low index, and high index as parameters to partition the subarray.

3. Final Answer:

   int partition(int arr[], int low, int high) -> Option A

4. Quick Check:

   Returns int pivot index, low high params [OK]

Hint: Partition returns pivot index as int with low and high params [OK]

Common Mistakes:

• Using void return type instead of int
• Adding extra pivot parameter unnecessarily
• Returning bool instead of pivot index

3. What is the output array after applying Lomuto partition on arr = [8, 3, 7, 6, 2] with low=0 and high=4?

medium

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

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

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

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

Solution

1. Step 1: Choose pivot and initialize pointers

   Pivot = arr[4] = 2, i = -1 initially.

2. Step 2: Iterate and swap smaller elements

   Check elements from index 0 to 3:
   - arr[0]=8 > 2 no swap
   - arr[1]=3 > 2 no swap
   - arr[2]=7 > 2 no swap
   - arr[3]=6 > 2 no swap
   After loop, swap arr[i+1]=arr[0] and pivot arr[4].

3. Final Answer:

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

4. Quick Check:

   Pivot swapped to front since no smaller elements [OK]

Hint: Pivot last element, swap smaller left, pivot final swap [OK]

Common Mistakes:

• Swapping elements incorrectly during iteration
• Confusing pivot position after partition
• Assuming pivot is first element

4. Identify the bug in this Hoare partition code snippet:

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;
    std::swap(arr[i], arr[j]);
  }
}

medium

A. Initialization of i and j is incorrect

B. No bug, code is correct

C. The loop condition should be i > j instead of i >= j

D. Pivot should be arr[high], not arr[low]

Solution

1. Step 1: Verify pointer initialization

   In standard Hoare partition, i starts at low-1 and j at high+1 to work with the do-while loops correctly.

2. Step 2: Check other parts

   Pivot selection arr[low], movement conditions arr[i] < pivot and arr[j] > pivot, i >= j return, and swap are all correct.

3. Final Answer:

   No bug, code is correct -> Option B

4. Quick Check:

   Standard Hoare uses i=low-1, j=high+1 [OK]

Hint: i starts at low, j at high in Hoare partition [OK]

Common Mistakes:

• Setting i and j outside array bounds
• Confusing pivot position with pointer initialization
• Misusing loop exit condition

5. Given the array [10, 80, 30, 90, 40, 50, 70], after applying Hoare partition with low=0 and high=6, what is the resulting array?

hard

A. [10, 30, 40, 50, 70, 90, 80]

B. [10, 30, 40, 50, 90, 80, 70]

C. [10, 80, 30, 90, 40, 50, 70]

D. [10, 30, 40, 50, 80, 70, 90]

Solution

1. Step 1: Set pivot and pointers

   Pivot = arr[0] = 10, i = low - 1 = -1, j = high + 1 = 7.

2. Step 2: Move pointers and check swaps

   do {i++;} while (arr[i] < 10): i=0, arr[0]=10 not <10, stop i=0.
   do {j--;} while (arr[j] > 10): j=6(70>10),5(50>10),4(40>10),3(90>10),2(30>10),1(80>10),0(10 not >10), stop j=0.
   i=0 < j=0 is false, so i >= j, return j=0, no swap.
   But since i == j, partition ends with pivot index 0.
   However, this means no swaps, so array remains same.

3. Step 3: Re-examine pointer movement

   Actually, the do-while loops stop at i=0 and j=0, so no swaps occur.
   Therefore, the array remains unchanged.

4. Final Answer:

   [10, 80, 30, 90, 40, 50, 70] -> Option C

5. Quick Check:

   All elements ≥ pivot, array unchanged [OK]

Hint: Track i and j moving inward, swap when i < j [OK]

Common Mistakes:

• Swapping after pointers cross
• Misplacing pivot after partition
• Confusing Hoare with Lomuto output