Merge Sort

Divide the list in half, sort each half, then merge — always O(n log n).

Intermediate

def merge_sort(a):
    if len(a) <= 1:
        return a
    mid = len(a) // 2
    left, right = merge_sort(a[:mid]), merge_sort(a[mid:])
    out, i, j = [], 0, 0
    while i < len(left) and j < len(right):
        if left[i] <= right[j]:
            out.append(left[i]); i += 1
        else:
            out.append(right[j]); j += 1
    return out + left[i:] + right[j:]

Guaranteed $O(n \log n)$ and stable, at the cost of $O(n)$ extra memory. Ideal for linked lists and for sorting data too big for RAM (external sorting).

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