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 Kotlin 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.kt 
fun mergeSort(values: List<Int>): List<Int> {
	if (values.size <= 1) return values
	val mid = values.size / 2
	val left = mergeSort(values.subList(0, mid))
	val right = mergeSort(values.subList(mid, values.size))
	val merged = mutableListOf<Int>()
	var i = 0
	var j = 0
	while (i < left.size && j < right.size) {
		if (left[i] <= right[j]) merged.add(left[i++]) else merged.add(right[j++])
	}
	merged.addAll(left.drop(i))
	merged.addAll(right.drop(j))
	return merged
}

fun main() {
	val arr = listOf(5, 1, 4, 2, 8)
	println(mergeSort(arr).joinToString(prefix = "[", postfix = "]"))
}

Complexity

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

Implementation notes

  • Kotlin stores the input in val arr = listOf(5, 1, 4, 2, 8), a val binding to a read-only List<Int>; the source never mutates arr.
  • The recursive signature is mergeSort(values: List<Int>): List<Int>. if (values.size <= 1) return values is the base case, while non-base calls build and return a new merged list.
  • Splitting uses values.subList(0, mid) and values.subList(mid, values.size); those list views are passed into recursive calls before merging.
  • The merge buffer is val merged = mutableListOf<Int>(). Cursor indexes i and j are mutable var Int values that advance with left[i++] and right[j++].
  • if (left[i] <= right[j]) uses Int comparison and takes the left value on ties, preserving stable ordering for equal values.
  • Remaining elements are appended with merged.addAll(left.drop(i)) and merged.addAll(right.drop(j)); drop creates suffix lists for the leftovers.
  • The trace shows [5, 1, 4, 2, 8] split into [5, 1] and [4, 2, 8], sorted to [1, 5] and [2, 4, 8], then merged into [1, 2, 4, 5, 8].
  • println(mergeSort(arr).joinToString(prefix = "[", postfix = "]")) formats the returned list as [1, 2, 4, 5, 8].