for loop - complexity - bigtheta 3 for cycle -


i resolve problem don't have solution of kindly ask if can confirm if solution correct or not

int h=1; int cont = 0;  (j = 2^n; j>1; j = j/2) {         h = h * 2;         (i =1; < j; = i*2)            (k=2; k<h; k++)                cont ++; }

i must find complexity of portion of code in bigtheta.

so, analyze third cycle grow in way

k -> linear until = h (h grow 2^w) - complexity log n.

about second, first cycles' limit 0 think complexity log n.

about first 2^n = 2^n-1 complexity n

the total complexity n * log n

you can proceed formally, step step, using sigma notation (i skipped steps, feel free ask more details if necessary):

enter image description here


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