Question

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Count the number of operations and the big-O time complexity in the worst-case and best-case for the following code

int small

for ( i n t i = 0 ; i < n ; i ++)

{

i f ( a [ i ] < a [ 0 ] )

{ small = a [ i ] ;

}

}

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Calculate the Big-O time complexity for the following code and explain your answer by showing all calculation.

for ( int i = 0 ; i < n ; i ++)

{

for ( int j = 0 ; j < n∗n ; j ++)

{ cout <<” t r i c k y ! ”<< endl ;

}

}

Answer 1

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- 1) int small for (int i=0; i<n; itt) ? if (a[i] caso]) {small = a[i]; o (nt2) 0(1) 0(1) = 0h42) [60674001) - 0 (nt2) big-o time. O(n). coorst-case time complexity O(n). in worst case complexity we have to check up to last element that's why it's worst case time complexity is ocn).

2) for (int i=0; i<n; itt) ~ o(at2) a for (int j = 0; je na n; j++) 0 (02) { count ch" tricky! " 22 endl; -3 0(1) - O(n+2) [O(12) +001)] = 0(m3 + 20%) = 0(m3) big-o time complexity is o (m3):