Consider the following piece of \’C\’ code fragment that removes duplicates from an ordered list of integers.
Node *remove-duplicates(Node *head, int *j)
{
Node *t1, *t2;
*j=0;
t1 = head;
if (t1! = NULL) t2 = t1 →next;
else return head;
*j = 1;
if(t2 == NULL)
return head;
while t2 != NULL)
{
if (t1.val != t2.val) --------------------------→ (S1)
{
(*j)++; t1 -> next = t2; t1 = t2: ----------→ (S2)
}
t2 = t2 →next;
}
t1 →next = NULL;
return head;
}Assume the list contains n elements (n≥2) in the following questions. a). How many times is the comparison in statement S1 made? b). What is the minimum and the maximum number of times statements marked S2 get executed? c). What is the significance of the value in the integer pointed to by j when the function completes?
(A)
- (a). n-1 times, since comparison is pairwise for n elements.
- (b). maximum : n-1 for all distinct elements, minimum: 0 for all same elements.
- (C). j keeps count of distinct nodes in the list.
(B)
- (a). n times, since comparison is pairwise for n elements.
- (b). maximum : n-1 for all distinct elements, minimum: 0 for all same elements.
- (C). j keeps count of distinct nodes in the list.
(C)
- (a). n-1 times, since comparison is pairwise for n elements.
- (b). maximum : n-1 for all distinct elements, minimum: 1 for all same elements.
- (C). j keeps count of distinct nodes in the list.
(D)
None of the above
Answer: (A)
Explanation:
- (a). n-1 times, since comparison is pairwise for n elements.
- (b). maximum : n-1 for all distinct elements, minimum: 0 for all same elements.
- (C). j keeps count of distinct nodes in the list.
Quiz of this Question
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