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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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14y ago
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11y ago

T(n)=n+(n-1)+(n-2)+.............+2+1

make pair the first and last terms, the second and next to last terms, and so forth. it ends up looking like this:-

T(n)=[n+1]+[(n-1)+2]+[(n-2)+3]+.........

=(n+1)+(n+1)+(n+1)+............

In an algebra course, the sum of first N integers is given by the formula,

n(n+1)/2

therefore, T(n)=n(n+1)/2

=((n^2)+n)/2

=((n^2)/2)+(n/2)

So,Complexity of selection sort is O(n^2)

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12y ago

The worst case for merge sort is O(n*log2n)

The running time for merge sort is 2T(n/2)+O(n)

I hope this will help...

Jimmy Clinton Malusi (JKUAT-Main Campus)

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11y ago

calculate time complexity of heap sort

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12y ago

nlog2n

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Q: How do you calculate the time complexity of selection sort?
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