Sorting

Quick Sort (Lomuto)

Choose the last item as a pivot, partition smaller values to its left, then recurse on the two sides.

Algorithm

The checked-in replay follows the same small input and final output across all 21 DSA books, so this Ruby DSA implementation can be compared directly with the other languages.

pivot The final element is moved to the boundary between smaller and larger values.
partition One scan rearranges the current range before the recursive calls.

Visual walkthrough

The pinned first partition uses [4, 1, 5, 2, 3] with pivot 3. The diagrams track the boundary, swaps, and recursive ranges.

Step 1 - Choose the last value as pivot

The pivot is arr[4] = 3, and i starts just before the current range.

Initial partition state for [4, 1, 5, 2, 3].i0i1i2i3i441523j startspivot

Step 2 - Swap small values left

1 and 2 are <= pivot, so they move into the left partition.

After scanning values before the pivot: [1, 2, 5, 4, 3].i0i1i2i3i412543<= 3<= 3> 3> 3pivot

Step 3 - Place pivot, then recurse

Swapping pivot 3 into index 2 gives [1, 2, 3, 4, 5]; recurse on [1, 2] and [4, 5].

Pivot lands at index 2 and splits the remaining work.left rangepivotright range[1, 2]3 at i2[4, 5]quick_sort(0,1)fixedquick_sort(3,4)

Basic Implementation

basic.rb 
def partition(arr, low, high)
	pivot = arr[high]
	i = low - 1
	(low...high).each do |j|
		if arr[j] <= pivot
			i += 1
			arr[i], arr[j] = arr[j], arr[i]
		end
	end
	arr[i + 1], arr[high] = arr[high], arr[i + 1]
	i + 1
end

def quick_sort(arr, low, high)
	if low < high
		pivot_index = partition(arr, low, high)
		quick_sort(arr, low, pivot_index - 1)
		quick_sort(arr, pivot_index + 1, high)
	end
end

arr = [4, 1, 5, 2, 3]
quick_sort(arr, 0, arr.length - 1)
puts arr.inspect

Complexity

  • Time: O(n^2) worst, O(n log n) average
  • Space: O(log n) average call stack
  • Stable: no

Implementation notes

  • quick_sort(arr, low, high) mutates the same Ruby Array; it does not return a new sorted array.
  • The recursion guard is if low < high, so empty and one-element ranges stop without calling partition.
  • partition(arr, low, high) chooses the last slot as the pivot with pivot = arr[high].
  • i = low - 1 marks the end of the <= pivot side, and (low...high).each scans j up to but not including the pivot slot.
  • When arr[j] <= pivot, Ruby parallel assignment arr[i], arr[j] = arr[j], arr[i] swaps two array slots in place.
  • After the scan, arr[i + 1], arr[high] = arr[high], arr[i + 1] places the pivot at its final index and returns i + 1.
  • The trace for [4, 1, 5, 2, 3] keeps 4 and 5 on the right of pivot 3, swaps 1 and 2 left, then places 3 at index 2.
  • The replay then summarizes recursive calls on [1, 2] and [4, 5]; it does not include a separate bad-pivot degradation case.
  • puts arr.inspect prints the mutated array as [1, 2, 3, 4, 5].