Time ans space complexity Problem 1

#1

The answer of this should be log N instead of n

#2

Hey @Alfred
Please share the question

#3

nd the time complexity of the given code. int count = 0; for (int i = N; i > 0; i /= 2) for (int j = 0; j < i; j++) count++; O(log N) O(N) O(N*log N) O(N^2)

#4

This question I am confused little bit

#5

Hey @Alfred
in iteration 1 inner loop will run -> N time
in iteration 2 inner loop will run -> N/2 time
in iteration 3 inner loop will run -> N/2^2 time
in iteration 4 inner loop will run -> N/2^3 time

if add them u will get
N+ N/2+N/2^2 +N/2^3…+ N/2^K …+ 1

N( 1+1/2+1/4+1/8+1/16…) ==N(1/(1-1/2) ) [infinite gp sum] == 2N ==O(N)

#6

Okk so it’s approximately equal to n therefore bigOn

#7

Yeah …

#8

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