Question

/* * Find the first element in A that is * not also in B. */

01: int firstNotIn (int* A, int* B, int nA, int nB)

02: {

03: int pos = 0;

04: bool found = false;

05: while (pos < nA && !found)

06: {

07: if (find(B, B+nB, A[pos]) != B+nB)

08: found = true;

09: else

10: ++pos;

11: }

12: if (!found)

13: {

14: pos = -1;

15: }

16: return pos;

17: }

According to the copy-and-paste technique, how would you annotate line 5 of this function by the time you had determined the complexity of the entire loop?

Answer 1

Answer:

Currently, given function is returing the position of first element in A that is also in B.

But now, question is asking to fund the position of first element in A that is not in B.

Please find my modification in given function:

/*
* Find the first element in A that is
* not also in B.
*/
int firstNotIn (int A, int B, int nA, int nB)
{
int pos = 0;
bool found = false;

// this will loop until we find a element in A that is not in B or till end of A, So it runs O(N) times
// N is the size of A
while (pos < nA && !found)
{
if (seqSearch(B, nB, A[pos]) == -1) //this takes O(M) time, M is the size of B
found = true;
else
++pos;
}
if (!found)
{
pos = -1;
}
return pos;
}

Time complexity: (M*N)

Please upvote, as i have given the exact answer as asked in question. Still in case of any concerns in code, let me know in comments. Thanks!