Data Structure online quiz on “Largest and Smallest Number in a Linked List using Recursion”.
1. Which of the following methods can be used to find the largest and smallest number in a linked list?
a) Recursion
b) Iteration
c) Both Recursion and iteration
d) Impossible to find the largest and smallest numbers
View Answer
Answer: c
Clarification: Both recursion and iteration can be used to find the largest and smallest number in a linked list.
2. Consider the following code snippet to find the largest element in a linked list:
struct Node{
int val;
struct Node *next;
}*head;
int get_max()
{
struct Node* temp = head->next;
int max_num = temp->val;
while(______)
{
if(temp->val > max_num)
max_num = temp->val;
temp = temp->next;
}
return max_num;
}
Which of the following lines should be inserted to complete the above code?
a) temp->next != 0
b) temp != 0
c) head->next != 0
d) head != 0
Answer: b
Clarification: The line “temp != 0” should be inserted to complete the above code.
3. Consider the following code snippet to find the smallest element in a linked list:
struct Node
{
int val;
struct Node* next;
}*head;
int get_min()
{
struct Node* temp = head->next;
int min_num = temp->val;
while(temp != 0)
{
if(_________)
min_num = temp->val;
temp = temp->next;
}
return min_num;
}
Which of the following lines should be inserted to complete the above code?
a) temp > min_num
b) val > min_min
c) temp->val < min_num
d) temp->val > min_num
Answer: c
Clarification: The line “temp->val = min_num” should be inserted to complete the above code.
4. What is the output of the following code:
#include
#include
struct Node
{
int val;
struct Node* next;
}*head;
int get_max()
{
struct Node* temp = head->next;
int max_num = temp->val;
while(temp != 0)
{
if(temp->val > max_num)
max_num = temp->val;
temp = head->next;
}
return max_num;
}
int main()
{
int n = 9, arr[9] ={5,1,3,4,5,2,3,3,1},i;
struct Node *temp, *newNode;
head = (struct Node*)malloc(sizeof(struct Node));
head -> next =0;
temp = head;
for(i=0;i<n;i++)
{
newNode =(struct Node*)malloc(sizeof(struct Node));
newNode->next = 0;
newNode->val = arr[i];
temp->next =newNode;
temp = temp->next;
}
int max_num = get_max();
printf("%d %d",max_num);
return 0;
}
a) 5
b) 1
c) runtime error
d) garbage value
Answer: c
Clarification: The variable temp will always point to the first element in the linked list due to the line “temp = head->next” in the while loop. So, it will be an infinite while loop and the program will produce a runtime error.
5. What is the output of the following code?
#include
#include
struct Node
{
int val;
struct Node* next;
}*head;
int get_max()
{
struct Node* temp = head->next;
int max_num = temp->val;
while(temp != 0)
{
if(temp->val > max_num)
max_num = temp->val;
temp = temp->next;
}
return max_num;
}
int get_min()
{
struct Node* temp = head->next;
int min_num = temp->val;
while(temp != 0)
{
if(temp->val < min_num)
min_num = temp->val;
temp = temp->next;
}
return min_num;
}
int main()
{
int i, n = 9, arr[9] ={8,3,3,4,5,2,5,6,7};
struct Node *temp, *newNode;
head = (struct Node*)malloc(sizeof(struct Node));
head -> next =0;
temp = head;
for(i=0;i<n;i++)
{
newNode =(struct Node*)malloc(sizeof(struct Node));
newNode->next = 0;
newNode->val = arr[i];
temp->next =newNode;
temp = temp->next;
}
int max_num = get_max();
int min_num = get_min();
printf("%d %d",max_num,min_num);
return 0;
}
a) 2 2
b) 8 8
c) 2 8
d) 8 2
View Answer
Answer: d
Clarification: The program prints the largest and smallest elements in the linked list, which are 8 and 2 respectively.
6. What is the time complexity of the following iterative code used to find the smallest and largest element in a linked list?
#include
#include
struct Node
{
int val;
struct Node* next;
}*head;
int get_max()
{
struct Node* temp = head->next;
int max_num = temp->val;
while(temp != 0)
{
if(temp->val > max_num)
max_num = temp->val;
temp = temp->next;
}
return max_num;
}
int get_min()
{
struct Node* temp = head->next;
int min_num = temp->val;
while(temp != 0)
{
if(temp->val < min_num)
min_num = temp->val;
temp = temp->next;
}
return min_num;
}
int main()
{
int i, n = 9, arr[9] ={8,3,3,4,5,2,5,6,7};
struct Node *temp, *newNode;
head = (struct Node*)malloc(sizeof(struct Node));
head -> next =0;
temp = head;
for(i=0;i<n;i++)
{
newNode =(struct Node*)malloc(sizeof(struct Node));
newNode->next = 0;
newNode->val = arr[i];
temp->next =newNode;
temp = temp->next;
}
int max_num = get_max();
int min_num = get_min();
printf("%d %d",max_num,min_num);
return 0;
}
a) O(1)
b) O(n)
c) O(n2)
d) O(n3)
Answer: b
Clarification: The time complexity of the above iterative code used to find the largest and smallest element in a linked list is O(n).
7. Consider the following recursive implementation to find the largest element in a linked list:
struct Node
{
int val;
struct Node* next;
}*head;
int max_of_two(int a, int b)
{
if(a > b)
return a;
return b;
}
int recursive_get_max(struct Node* temp)
{
if(temp->next == 0)
return temp->val;
return max_of_two(______, _______);
}
Which of the following arguments should be passed to the function max_of two() to complete the above code?
a) temp->val,recursive_get_max(temp->next)
b) temp, temp->next
c) temp->val, temp->next->val
d) temp->next->val, temp
Answer: a
Clarification: The arguments {temp->val,recursive_get_max(temp->next)} should be passed to the function max_of_two() to complete the above code.
8. What is the output of the following code?
#include
#include
struct Node
{
int val;
struct Node* next;
}*head;
int max_of_two(int a, int b)
{
if(a > b)
return a;
return b;
}
int recursive_get_max(struct Node* temp)
{
if(temp->next == 0)
return temp->val;
return max_of_two(temp->val,recursive_get_max(temp->next));
}
int min_of_two(int a, int b)
{
if(a < b)
return a;
return b;
}
int recursive_get_min(struct Node* temp)
{
if(temp->next == 0)
return temp->val;
return min_of_two(temp->val,recursive_get_min(temp->next));
}
int main()
{
int n = 9, arr[9] ={1,3,2,4,5,0,5,6,7},i;
struct Node *temp, *newNode;
head = (struct Node*)malloc(sizeof(struct Node));
head -> next =0;
temp = head;
for(i=0;i<n;i++)
{
newNode =(struct Node*)malloc(sizeof(struct Node));
newNode->next = 0;
newNode->val = arr[i];
temp->next =newNode;
temp = temp->next;
}
int max_num = recursive_get_max(head->next);
int min_num = recursive_get_min(head->next);
printf("%d %d",max_num,min_num);
return 0;
}
a) 7 1
b) 0 7
c) 7 0
d) 1 1
Answer: c
Clarification: The program prints the largest and the smallest elements in the linked list, which are 7 and 0 respectively.
9. What is the time complexity of the recursive implementation used to find the largest and smallest element in a linked list?
#include
#include
struct Node
{
int val;
struct Node* next;
}*head;
int max_of_two(int a, int b)
{
if(a > b)
return a;
return b;
}
int recursive_get_max(struct Node* temp)
{
if(temp->next == 0)
return temp->val;
return max_of_two(temp->val,recursive_get_max(temp->next));
}
int min_of_two(int a, int b)
{
if(a < b)
return a;
return b;
}
int recursive_get_min(struct Node* temp)
{
if(temp->next == 0)
return temp->val;
return min_of_two(temp->val,recursive_get_min(temp->next));
}
int main()
{
int n = 9, arr[9] ={1,3,2,4,5,0,5,6,7},i;
struct Node *temp, *newNode;
head = (struct Node*)malloc(sizeof(struct Node));
head -> next =0;
temp = head;
for(i=0;i<n;i++)
{
newNode =(struct Node*)malloc(sizeof(struct Node));
newNode->next = 0;
newNode->val = arr[i];
temp->next =newNode;
temp = temp->next;
}
int max_num = recursive_get_max(head->next);
int min_num = recursive_get_min(head->next);
printf("%d %d",max_num,min_num);
return 0;
}
a) O(1)
b) O(n)
c) O(n2)
d) O(n3)
Answer: b
Clarification: The time complexity of the above recursive implementation used to find the largest and smallest element in linked list is O(n).
10. What is the output of the following code?
#include
#include
struct Node
{
int val;
struct Node* next;
}*head;
int min_of_two(int a, int b)
{
if(a < b)
return a;
return b;
}
int recursive_get_min(struct Node* temp)
{
if(temp->next == 0)
return temp->val;
return min_of_two(temp->val,recursive_get_min(temp->next));
}
int main()
{
int n = 5, arr[5] ={1,1,1,1,1},i;
struct Node *temp, *newNode;
head = (struct Node*)malloc(sizeof(struct Node));
head -> next =0;
temp = head;
for(i=0;i<n;i++)
{
newNode =(struct Node*)malloc(sizeof(struct Node));
newNode->next = 0;
newNode->val = arr[i];
temp->next =newNode;
temp = temp->next;
}
int min_num = recursive_get_min(head->next);
printf("%d",min_num);
return 0;
}
a) 1
b) 0
c) compile time error
d) runtime error
Answer: a
Clarification: The program prints the smallest element in the linked list, which is 1.
11. How many times will the function recursive_get_min() be called when the following code is executed?
#include
#include
struct Node
{
int val;
struct Node* next;
}*head;
int min_of_two(int a, int b)
{
if(a < b)
return a;
return b;
}
int recursive_get_min(struct Node* temp)
{
if(temp->next == 0)
return temp->val;
return min_of_two(temp->val,recursive_get_min(temp->next));
}
int main()
{
int n = 5, arr[5] ={1,1,1,1,1},i;
struct Node *temp, *newNode;
head = (struct Node*)malloc(sizeof(struct Node));
head -> next =0;
temp = head;
for(i=0;i<n;i++)
{
newNode =(struct Node*)malloc(sizeof(struct Node));
newNode->next = 0;
newNode->val = arr[i];
temp->next =newNode;
temp = temp->next;
}
int min_num = recursive_get_min(head->next);
printf("%d",min_num);
return 0;
}
a) 4
b) 5
c) 6
d) 7
Answer: b
Clarification: The function recursive_get_min() will be called 5 times when the above code is executed.