Question

In the following code segement, how many times is STMT executed?

for (I=N; i >1 ; i = i/2)

for (j = 1; j <= N; j++)

STMT

I can do normal summation formula, what throws me off is the first for loop. Any help would be appreciated. Thank you .

Answer 1

(Please do rate the answer if you found useful - Assuming program is in C++ - Output is attached - Please read comments for explanation)

We can solve this problem by analyzing the time complexity. Total running time complexity of statement will be O(N) - (pronounced Big-Oh of N)

Let us breakdown:

Outer Loop Executes Log₂N times - Trace the value like first assume N=5 hence Outer loop executes Log₂5 = 2

next assume N=15 Outer Loop executes Log₂15 = 3 (Note - Example tracing is shown below as images)

Inner Loop Executes Count of outer loop times N - Trace value First assume N=5 hence Outer loop executes Log₂5 = 2 then Inner loop executes N*2 = 10, Now assume N=15 Outer Loop executes Log₂15 = 3 inner loop executes N*3=45

Hence lets call Outer loop executes Log₂N times and inner loop executes N*Log₂N times Hence complexity of total program is O(NLog₂N)

Simple Program for Clear Understanding

#include <iostream>

using namespace std;

int main()
{
int i,j,n;
cout<<"Enter N:";
cin>>n;
int firstloop=0,secondloop=0; //count for outerloop and innerloop
for (i=n; i >1 ; i = i/2)
{
firstloop++; //incrementing count
cout<<"First loop"<<i<<endl;
for (j = 1; j <= n; j++)
{
secondloop++;   //incrementing count
cout<<"Secondloop"<<j<<endl;
}
}
cout<<"First Loop Count :"<<firstloop<<endl<<"Second Loop Count :"<< secondloop<<endl;
return 0;
}

Output