Problem:
There are 25 horses and only 5 tracks. So only 5 horses can run the race at a time. How many minimum no of races should be conducted to find the 3 best horses?
Sol: The answer is 7.
Round 1: Take 5 horses at a time. 5 winners of each race will go to round2.
Round 2: Winner of this will be top most horse.(call it W1)
Round 3: Final race comprises of following 5 horses.
h1: 2nd horse of round2
h2:3rd horse of round2
h3:2nd horse of W1’s group during round 1
h4:3rd horse of W1’s group during round1
h5:2nd horse of h1’s group during round1
There are 25 horses and only 5 tracks. No of matches to find top 3 horses.
Hat color problem–most asked question in interviews..
http://en.wikipedia.org/wiki/Prisoners_and_hats_puzzle
Two tribes one always tells true other false….
There were 2 tribes living on an island. The east tribal people always tell a lie. The west tribal people always tell the truth. Alan and Bryan came from the same tribe. However, we do not know which tribe they came from. When they were asked the marital status, Alan said: “Both of us are married.” But Bryan said: “I am not married”. Are they really married?
Ans:
Both are from same tribe. It means either both of them will lie or both will tell true. If we look at the statements both of them contradicts about Bryan’s marital status. It means one of them is not telling truth, as they are from same tribe it concludes that they are from east tribe. It means Bryan is married and Alan’s status is unclear.
Design a stack which have O(1) time complexity for finding the minimum element?
We can think of two different solutions -
Sol 1:
With additional memory where time complexity of findmin(),Push(),Pop() is O(1) but space complexity is O(n). Along with pushing the elements push the current minimum on to the stack. For pop operation pop 2 elements instead of 1 , it is still O(1).
For ex: Let us assume the following numbers which are pushed on to the stack.
4 2 9 3 6 1
Stack state will be something like below
4 4 2 2 9 2 3 2 6 2 1 1.
POP(x) = pop(), pop().
Where POP(x) is modified stack pop operation. pop() is standard pop() operation.
FINDMIN(x)=pop()
PUSH(x)=push() and push( MIN(pop(),elem)).
Where PUSH(x) is modified stack push operation. push() is standard push() operation.
Here storage needed for stack is 2n where n is number of elements in the stack.
Sol 2:
Difference Between Composition and Aggregation in Object Oriented Programming?
Both are one way of extending a class.
In a composition relationship, the whole has sole responsibility for the disposition of its parts, or as you put it above, the whole “controls the lifetime of” the part. In order for the whole
to have “sole disposition” or “control the lifetime” of its parts, the whole must be the only object that knows of the parts existence.
C++ Code Example:
class Whole {
Part* part;
public:
Whole(): part( 0 ) { }
~Whole() { delete part; }
};
Part is created inside Whole class constructor and it is destroyed when whole is destroyed.
Real Life Example:
University and Departments have a composition relationship. When University is created all the departments are created with it. When University is removed departments will not have the existance of their own.
In aggregation relationship, the part object reference can be re-used. Usually creation of part object is not the responsibility of the ‘whole ‘ object. It would have been created somewhere else, and passed to the ‘whole’ object as a method argument. Whole object will not have control on the lifetime of the part object.
class Whole {
Part* part;
public:
Whole() { }
~Whole() { }
};
Real Life Example:
University and Professors are in aggregation relationship. When University is formed professors will come and join the University. But, when University is removed Professors will still exists and they can work in other Universities.
Given a singly linked list, determine whether it contains a loop or not.
1. Start reversing the list. If you reach the head, then there is a loop.
But this changes the list. So, reverse the list again.
2. Second solution is called Hare and Tortoise approach. Basically have 2 ptrs both
pointing to the start of the list. Increment first pointer by one and
second pointer by 2. After each increment comapre if they are equal. They will meet if there is a loop.
p1 = p2 = head;
do {
p1 = p1->next;
p2 = p2->next->next;
} while (p1 != p2);
3. Hash all seen nodes and compare the next node with the nodes in the hash. If a node is already present in the hash then there is a loop.
Under what circumstances can one delete an element from a singly linked list in constant time?
If the list is circular and there are no references to the nodes in the list from anywhere else! Just copy the contents of the next node and delete the next node. If the list is not circular, we can delete any but the last node using this idea. In that case, mark the last node as dummy!
Remove duplicates from a linked list ?
NODE *DupRem(NODE *head)
{
NODE *p1 = head, *p2, *resultList = NULL, *temp = NULL;
while(p1 != NULL)
{
p2 = p1->link;
while(p2 != NULL)
{
if(p1->data == p2->data)
{
break;
}
else
{
p2 = p2->link;
}
}
if(p1->data==p2->data)
{
p1 = p1->link;
}else
{
if(resultList == NULL)
{
resultList = p1;
temp = p1;
p1 = p1->link;
}else
{
temp->link = p1;
temp = temp->link;
p1 = p1->link;
}
}
}
return resultList;
}
Given two sorted linked lists, write a function to merge them into one?
list* merge(LIST* list1, LIST* list2){
LIST *it1, *it2, *head;
if(!list1) return list2;
if(!list2) return list1;
head = (list1->value value) ? list1 : list2;
if(head == list1)
{ it1 = list1; it2 = list2;
}
else{
it1 = list2;
it2 = list1;
}
while(it1->next && it2){
if(it1->next->value >= it2->value){
list *temp = it1->next;
it1->next = it2;
it2 = it2->next;
it1->next->next = temp;
}
else{
it1 = it1->next;
}
}
if(it2)
it1->next = it2;
return head;
}
Given a single linked list write a function to swap each pair of nodes by manipulating with pointers (not values).?
Sol:
Original list: head->1->2->3->4->5->NIL should be transformed to head->2->1->4->3->5-> NIL
int reverse_pairs(LLIST ** head)
{
LLIST * temp = ULL;
LLIST* current_pair = *head;
LLIST** previouspair = head;
while((current_pair != NULL) && (current_pair->next != NULL))
{
temp = current_pair;
current_pair = current_pair->next;
temp->next = current_pair->next;
current_pair->next = temp;
*previouspair = current_pair;
current_pair = temp->next;
previouspair = &temp->next;
}
return 0;
}
