algorithm - Apply master theorem on T(n) = T(n/2) + n -

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I was just trying my hand at the master theorem and when I try to evaluate t (n) = t Using (N / 2) + An Master Theorem, the answer is O (N) evaluating.

But just go through the code given below: 

  Fun (n) {if (n == 1) Return; For (int i = 1; i & lt; = n; i ++) {printf ("*"); } Fun (N / 2); }

The recurring equation for the above code is T (n) = t (n / 2) + n, so the complexity of time for the above program should be o (n).

But if you think that logically, the number of times the program is running: n + n / 2 + n / 4 + n / 8 + ...... = nlogn Therefore, logical As per the above program, the complexity of time should be O (NLONP).

I am very confused now. Can someone help me, where is this happening to me?

No, evaluates section 2n

n + n / 2 + n / 4 + n / 8 + ...... = 2n

But if you have T (n) = 2 t (n / 2) + n, then It will be O (n log n)


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