Why does merge sort require o(n) steps to merge the two sorted halves?

Assuming this is the last merging step in the algorithm: 1 3 5 7 | 13 19 20 40

Doesn't this require only n/2 steps to compare two elements at a time?

1 Answer 1


First: O(n) is the same as O(n/2), as constant factors do not matter (O(n) means there's a factor so that runtime is always less than or equal that factor times n). But also, you're wrong with n/2 comparisons.

You start with two indices, pointing to the first elements in their half.

Then, for any of the n elements, as long as there are elements left in both half-arrays, you compare those two elements, take the smaller, and increment that index. So you need up to n-1 comparisons in n steps (n-1 if the two biggest elements are not in the same array, otherwise less). This makes the merge step O(n).

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