Question

Consider the following algorithm:

M: matrix of integer numbers
N: number of rows (columns) of M
x: integer number
function F2(M,N,X)
if(N==0):
return -1
for 0 <= i < N
for 0 <= j < N
if(i==j):
if(M[i,j]==x):
return i
return -1
end function
Choose ONE option:
a. The worst-case time complexity of F2 is Q(N2)
b. The worst- and best-case time complexities of F2 are the same
c. The best-case time complexity of F2 is Q(N)
d. The worst-case time complexity of F2 is Q(N)

Answer 1

The algorithm F2's worst-case time complexity is O(N) because it may need to check every element on the diagonal of the matrix. The best-case time complexity is O(1) if x is found at the first position. Therefore, the best-case time complexity is indeed O(N), not O(N^2).

Step-by-step explanation:

The algorithm F2 is designed to find the index of the first occurrence of the integer x on the diagonal of a square matrix M of size N. In the worst-case scenario, x is not on the diagonal or is in the last position, so the algorithm must check every element on the diagonal, resulting in O(N) time complexity. However, the best-case scenario occurs when x is in the first position on the diagonal, resulting in O(1) time complexity. Therefore, the correct answer is that the best-case time complexity is O(N) and the worst-case scenario is O(N), not O(N^2) since the inner loop does not iterate N times for each outer loop iteration but only checks when i equals j, which happens N times.