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Home/ Questions/Q 8616645
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Editorial Team
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Editorial Team
Asked: June 12, 20262026-06-12T05:40:11+00:00 2026-06-12T05:40:11+00:00

I have read everywhere that for divide and conquer sorting algorithms like Merge-Sort and

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I have read everywhere that for divide and conquer sorting algorithms like Merge-Sort and Quicksort, instead of recursing until only a single element is left, it is better to shift to Insertion-Sort when a certain threshold, say 30 elements, is reached. That is fine, but why only Insertion-Sort? Why not Bubble-Sort or Selection-Sort, both of which has similar O(N^2) performance? Insertion-Sort should come handy only when many elements are pre-sorted (although that advantage should also come with Bubble-Sort), but otherwise, why should it be more efficient than the other two?

And secondly, at this link, in the 2nd answer and its accompanying comments, it says that O(N log N) performs poorly compared to O(N^2) upto a certain N. How come? N^2 should always perform worse than N log N, since N > log N for all N >= 2, right?

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  1. Editorial Team
    Editorial Team
    2026-06-12T05:40:13+00:00Added an answer on June 12, 2026 at 5:40 am
    1. Insertion sort is faster in practice, than bubblesort at least. Their asympotic running time is the same, but insertion sort has better constants (fewer/cheaper operations per iteration). Most notably, it requires only a linear number of swaps of pairs of elements, and in each inner loop it performs comparisons between each of n/2 elements and a “fixed” element that can be stores in a register (while bubble sort has to read values from memory). I.e. insertion sort does less work in its inner loop than bubble sort.
    2. The answer claims that 10000 n lg n > 10 n² for “reasonable” n. This is true up to about 14000 elements.
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