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DSA C#programming~10 mins

Quick Sort Partition Lomuto and Hoare in DSA C# - Interactive Practice

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Practice - 5 Tasks
Answer the questions below
1fill in blank
easy

Complete the code to select the pivot as the last element in Lomuto partition.

DSA C#
int pivot = arr[[1]];
Drag options to blanks, or click blank then click option'
Ahigh
Blow
C0
Dmid
Attempts:
3 left
💡 Hint
Common Mistakes
Choosing the first element as pivot instead of the last.
Using an incorrect index like 'mid' or 'low'.
2fill in blank
medium

Complete the code to swap elements in Lomuto partition when arr[j] is less than or equal to pivot.

DSA C#
if (arr[j] <= pivot) {
    i++;
    Swap(arr, i, [1]);
}
Drag options to blanks, or click blank then click option'
Ai
Bhigh
Cpivot
Dj
Attempts:
3 left
💡 Hint
Common Mistakes
Swapping with the pivot index instead of j.
Swapping with i itself, which does nothing.
3fill in blank
hard

Fix the error in the Hoare partition loop condition to avoid infinite loop.

DSA C#
while (arr[i] [1] pivot) i++;
while (arr[j] [2] pivot) j--;
Drag options to blanks, or click blank then click option'
A<=
B<
C>
D>=
Attempts:
3 left
💡 Hint
Common Mistakes
Using <= or >= causes infinite loops or wrong partitioning.
Swapping the comparison operators between i and j loops.
4fill in blank
hard

Fill both blanks to correctly swap and return the partition index in Lomuto partition.

DSA C#
Swap(arr, i + 1, [1]);
return [2] + 1;
Drag options to blanks, or click blank then click option'
Ahigh
Blow
Ci
Dj
Attempts:
3 left
💡 Hint
Common Mistakes
Returning low or j instead of i + 1.
Swapping with wrong index instead of high.
5fill in blank
hard

Fill all three blanks to correctly implement the Hoare partition swap and return index.

DSA C#
if (i >= j) return [1];
Swap(arr, [2], [3]);
Drag options to blanks, or click blank then click option'
Aj
Bi
Dhigh
Attempts:
3 left
💡 Hint
Common Mistakes
Returning i instead of j.
Swapping with wrong indices or pivot.

Practice

(1/5)
1. What is the main difference between Lomuto and Hoare partition schemes in Quick Sort?
easy
A. Lomuto uses recursion, Hoare uses iteration
B. Lomuto sorts the array, Hoare reverses the array
C. Lomuto uses a single pointer, Hoare uses two pointers moving inward
D. Lomuto always picks the last element as pivot, Hoare picks the first element

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 C
  4. Quick Check:

    Partition schemes differ by pointer usage = A [OK]
Hint: Remember: Lomuto one pointer, Hoare two pointers [OK]
Common Mistakes:
  • Confusing pivot selection with partition method
  • Thinking both use the same pointer approach
  • Assuming Hoare always picks first element as pivot
2. Which of the following is the correct C# syntax to swap two elements in an array arr at indices i and j?
easy
A. arr[i] = arr[j]; arr[j] = arr[i];
B. arr.Swap(i, j);
C. swap(arr[i], arr[j]);
D. int temp = arr[i]; arr[i] = arr[j]; arr[j] = temp;

Solution

  1. Step 1: Understand swapping in C# arrays

    Swapping requires a temporary variable to hold one value while assigning the other.
  2. Step 2: Check each option

    The option "arr[i] = arr[j]; arr[j] = arr[i];" overwrites arr[i] before saving its value. "int temp = arr[i]; arr[i] = arr[j]; arr[j] = temp;" correctly uses a temp variable. "swap(arr[i], arr[j]);" and "arr.Swap(i, j);" are invalid syntax in C#.
  3. Final Answer:

    int temp = arr[i]; arr[i] = arr[j]; arr[j] = temp; -> Option D
  4. Quick Check:

    Use temp variable for swap = B [OK]
Hint: Always use a temp variable to swap array elements [OK]
Common Mistakes:
  • Overwriting one element before saving it
  • Using non-existent swap functions
  • Confusing syntax with other languages
3. What is the output array after applying Lomuto partition on arr = [8, 3, 7, 6, 2] with pivot as last element?
medium
A. [2, 3, 7, 6, 8]
B. [3, 2, 6, 7, 8]
C. [2, 3, 6, 7, 8]
D. [8, 3, 7, 6, 2]

Solution

  1. Step 1: Identify pivot and initialize pointers

    Pivot is 2 (last element). 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. Finally swap pivot with arr[i+1] = arr[0].
  3. Step 3: Final array after swap

    Swap arr[0] and arr[4]: [2, 3, 7, 6, 8]
  4. Final Answer:

    [2, 3, 7, 6, 8] -> Option A
  5. Quick Check:

    Lomuto partitions around pivot = D [OK]
Hint: Pivot ends at correct place after Lomuto partition [OK]
Common Mistakes:
  • Swapping incorrectly during iteration
  • Confusing pivot position after partition
  • Mixing Hoare and Lomuto steps
4. Identify the error in this Hoare partition code snippet in C#:
int HoarePartition(int[] arr, int low, int high) {
    int pivot = arr[low];
    int i = low + 1;
    int j = high;
    while (true) {
        while (arr[i] < pivot) i++;
        while (arr[j] > pivot) j--;
        if (i >= j) return j;
        int temp = arr[i]; arr[i] = arr[j]; arr[j] = temp;
    }
}
medium
A. Pivot should be last element, not first
B. Pointers i and j can go out of array bounds causing runtime error
C. Return statement should be i, not j
D. Swap should happen before checking i >= j

Solution

  1. Step 1: Analyze pointer increments

    i increments while arr[i] < pivot without boundary check; similarly j decrements without boundary check.
  2. Step 2: Identify risk of out-of-bounds

    If all elements are less or greater than pivot, i or j can move beyond array limits causing runtime error.
  3. Final Answer:

    Pointers i and j can go out of array bounds causing runtime error -> Option B
  4. Quick Check:

    Missing boundary checks cause errors = A [OK]
Hint: Always check array bounds when moving pointers [OK]
Common Mistakes:
  • Ignoring boundary checks in loops
  • Assuming pivot must be last element
  • Misplacing swap and return statements
5. Given the array arr = [10, 80, 30, 90, 40, 50, 70], what is the array state after applying Hoare partition with pivot as arr[low] (first element)?
hard
A. [70, 30, 40, 50, 10, 80, 90]
B. [10, 80, 30, 90, 40, 50, 70]
C. [40, 30, 10, 50, 70, 80, 90]
D. [50, 30, 40, 10, 70, 80, 90]

Solution

  1. Step 1: Initialize pointers and pivot

    Pivot = 10 (arr[0]), i = low + 1 = 1, j = high = 6 (following the Hoare code style).
  2. Step 2: Move pointers inward and swap

    i moves forward while arr[i] < pivot: arr[1]=80 not <10, i stays 1. j moves backward while arr[j] > pivot: arr[6]=70 >10, j=5; arr[5]=50 >10, j=4; arr[4]=40 >10, j=3; arr[3]=90 >10, j=2; arr[2]=30 >10, j=1; arr[1]=80 >10, j=0. Now i=1, j=0, i >= j, so partition ends. But before that, swaps happen when i < j. The first swap occurs between arr[i]=80 and arr[j]=70 (positions 1 and 6), then pointers move inward. After completing swaps, the array becomes [70, 30, 40, 50, 10, 80, 90].
  3. Step 3: Resulting array after partition

    The array after Hoare partition is [70, 30, 40, 50, 10, 80, 90].
  4. Final Answer:

    [70, 30, 40, 50, 10, 80, 90] -> Option A
  5. Quick Check:

    Hoare partitions around first element pivot = B [OK]
Hint: Hoare swaps inward around first element pivot [OK]
Common Mistakes:
  • Assuming pivot moves to final position immediately
  • Confusing Lomuto and Hoare partition results
  • Not tracing pointer movements carefully
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