What is the time complexity of fun()?
C
int fun( int n)
{
int count = 0;
for ( int i = 0; i < n; i++)
for ( int j = i; j > 0; j--)
count = count + 1;
return count;
}
|
(A)
Theta (n)
(B)
Theta (n2)
(C)
Theta (n*log(n))
(D)
Theta (n*(log(n*log(n))))
Answer: (B)
Explanation:
The time complexity can be calculated by counting the number of times the expression “count = count + 1;” is executed. The expression is executed 0 + 1 + 2 + 3 + 4 + …. + (n-1) times.
Time complexity = Theta(0 + 1 + 2 + 3 + .. + n-1) = Theta (n*(n-1)/2) = Theta(n2)
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