Sorting

Merge Sort (Top-Down)

Split the array recursively, sort each half, then merge two sorted runs into one sorted result.

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.

divide and conquer Each recursive call solves a smaller sorted subproblem.
merge step Two sorted halves are combined by repeatedly taking the smaller front item.

Visual walkthrough

The pinned input is [5, 1, 4, 2, 8]. The diagrams show the split into recursive halves, the sorted subarrays, and the final merge choices.

Step 1 - Split the input

The first midpoint splits [5, 1, 4, 2, 8] into left [5, 1] and right [4, 2, 8].

Top-down split used by merge_sort.[5,1,4,2,8]mid = 2[5,1]left[4,2,8]right

Step 2 - Sorted halves return

Recursive calls return [1, 5] and [2, 4, 8] before the final merge begins.

Returned subarrays before the final merge.sidebefore sortafter sortleft[5, 1][1, 5]right[4, 2, 8][2, 4, 8]

Step 3 - Merge by taking smaller fronts

Take 1 from left, then 2 and 4 from right, then the remaining 5 and 8.

Final merge produces [1, 2, 4, 5, 8].choiceleft frontright frontmergedtake 112[1]take 252[1, 2]take 454[1, 2, 4]extend58[1, 2, 4, 5, 8]

Basic Implementation

basic.rb 
def merge_sort(values)
	return values if values.length <= 1
	mid = values.length / 2
	left = merge_sort(values[0...mid])
	right = merge_sort(values[mid..])
	merged = []
	i = 0
	j = 0
	while i < left.length && j < right.length
		if left[i] <= right[j]
			merged << left[i]
			i += 1
		else
			merged << right[j]
			j += 1
		end
	end
	merged + left[i..] + right[j..]
end

arr = [5, 1, 4, 2, 8]
puts merge_sort(arr).inspect

Complexity

  • Time: O(n log n)
  • Space: O(n)
  • Stable: yes

Implementation notes

  • merge_sort(values) is a Ruby method that returns an array; it does not mutate the original arr.
  • The base case is return values if values.length <= 1, so one-element slices are returned directly.
  • mid = values.length / 2 uses integer division, then values[0...mid] and values[mid..] create the left and right slices for recursive calls.
  • merged = [] collects the merged result, with i and j tracking current positions in the sorted left and right arrays.
  • The comparison uses left[i] <= right[j], so equal values would keep the left value first.
  • merged << left[i] and merged << right[j] append values; leftover tails are joined with merged + left[i..] + right[j..].
  • The trace shows the pinned input splitting into [5, 1] and [4, 2, 8], then returning sorted halves [1, 5] and [2, 4, 8].
  • The final merge returns [1, 2, 4, 5, 8], and puts merge_sort(arr).inspect prints that Ruby array representation.