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 Fortran 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.f90 
program sort_merge
    implicit none
    integer :: arr(5) = [5, 1, 4, 2, 8]
    call merge_sort(arr, 1, 5)
    print '(*(I0,1X))', arr
contains
    recursive subroutine merge_sort(arr, left, right)
        integer, intent(inout) :: arr(:)
        integer, intent(in) :: left, right
        integer :: mid
        if (left >= right) return
        mid = (left + right) / 2
        call merge_sort(arr, left, mid)
        call merge_sort(arr, mid + 1, right)
        call merge(arr, left, mid, right)
    end subroutine merge_sort

    subroutine merge(arr, left, mid, right)
        integer, intent(inout) :: arr(:)
        integer, intent(in) :: left, mid, right
        integer :: tmp(5), i, j, k, t
        i = left; j = mid + 1; k = 1
        do while (i <= mid .and. j <= right)
            if (arr(i) <= arr(j)) then
                tmp(k) = arr(i); i = i + 1
            else
                tmp(k) = arr(j); j = j + 1
            end if
            k = k + 1
        end do
        do while (i <= mid)
            tmp(k) = arr(i); i = i + 1; k = k + 1
        end do
        do while (j <= right)
            tmp(k) = arr(j); j = j + 1; k = k + 1
        end do
        do t = 1, k - 1
            arr(left + t - 1) = tmp(t)
        end do
    end subroutine merge
end program sort_merge

Complexity

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

Implementation notes

  • Keep the explicit algorithmic steps instead of calling a standard-library sort. The replay is meant to expose comparisons, movement, and recursion.
  • The implementation is intentionally compact for learning and replay, not a production sorting utility.