Answer / arefe

hello I'm not sure about what I'm going to say . please tell me if you think it's wrong.
at first let's write some steps of it .
1) i = j = n
2) i = n/2 , j = n/2 --> j = n
3) i = n/4 , j = n/4 --> j = n/2 --> j = n
4) i = n/8 , j = n/8 --> j = n/4 --> j = n/2 , j = n
.
.
.
now if we count ,
the number of n , which j is equal to is log(n) ,
the number of n/2 , which j is equal to is log(n)-1 ,
the number of n/4 , which j is equal to is log(n)-2 ,
the number of n/8 , which j is equal to is log(n)-3 ,
and so on ,
so the result is summation log(n) - i , which i is from 0 to log(n) ,
we write log(n) out of Sigma and multiple it with log(n) + 1,( if we write Sigma term we have log(n) +1 , log(n))
and summation i , which i is from 0 to log(n) , is equal to (log(n)+1)(log(n))/2

Answer / muhammad ijaz khan

The outer loop divides the working area in half in each
iteration. So the running time of this algorithm is
proportional to the number of times n can be divided by 2.
The inner loop will be executed unlimitted time. So
the time complexity of the outer loop will be the function
of ln and the total time is: T(n)= n*ln n

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