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Suppose (for simplicity) that n = 2k for some entire k. Let T(n) the time used to sort n elements. As we can perform separation and merging in linear time, it takes cn time to perform these two steps, for some constant c. So, T(n) = 2T(n/2) + cn.

In the same way:

T(n/2) = 2T(n/4) + cn/2, so

T(n) = 4T(n/4) + 2cn. Going in this way ... T(n) = 2mT(n/2m) + mcn, and

T(n) = 2kT(n/2k) + kcn = nT(1) + cnlog2n = O(n log n). Remember, as n=2k k = log2n! The general case requires a bit more work, but it takes O(n log n) time anyway.

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โˆ™ 2009-08-09 13:25:02
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Q: How do you calculate time complexity for merge sort?
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