Consider the method below, which is intended to return true if there is at least one duplicate in the vector, and false if there are no duplicates.

bool hasdupicate($($vectornums)$)$

{$\{$

for($($int k=0$=0$; k <$<$ nums.size()−1$()-1$; k++)$++)$

{$\{$

if($($nums[$[$k]==$]==$nums[$[$k+1])$+1])$

return true;

}$\}$

return false;

}$\}$

Under which condition will the method not necessarily produce the desired result?

Question 28$28$ options:

A)$)$

When the vector is sorted in decreasing order.

B)$)$

When the vector has at most 2$2$ elements.

C)$)$

When the values in the vector are all positive.

D)$)$

When the vector is sorted in increasing order.

E)$)$

When the values in the vector are all the same.

Answer rating: 100% (QA)

The method will not necessarily produce the desired result under option B When the vector has at most 22 elements This is because the loop condition k numssize 1 will not be appropriate for vectors wi... View the full answer