Q3 for int i0 ii j cout

Q3.
for (int i=0; i<n; i++)
for (int j=n; j>=i; j--)
cout << i << “,” << j <<endl;

Please use the table provided.

Solution

for (int i=0; i<n; i++)
for (int j=n; j>=i; j--)
cout << i << “,” << j <<endl;

Let the cost of each execution of line is 1

LineNo.   Time Taken To Run this Line of Code Once   Total number of times needed to run this code.
1                                                       1                                   n
2                                                       1                                   n+(n-1)+(n-2)+……1 = n*(n+1)/2
3                                                       1                                    n+(n-1)+(n-2)+……1 = n*(n+1)/2

Total Time needed To finish                                 

n*(n+1)/2 = O(n^2) because in asymptotic analysis higher order terms overpower lower order terms for very large value of n