Algorithms | Analysis of Algorithms | Question 2

What is the time complexity of fun()?

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 (n^2)
(C) Theta (n*Logn)
(D) Theta (nLognLogn)


Answer: (B)

Explanation: The time complexity can be calculated by counting 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(n^2)



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