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How many people in party?

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At a party, everyone shook hands with everybody else. There were 66 handshakes. How many people were at the party?
See Solution: How many people in party?

lets say there are n persons
first person shakes hand with everyone else: n-1 times(n-1 persons)
second person shakes hand with everyone else(not with 1st as its already done): n-2 times
3rd person shakes hands with remaining persons: n-3

So total handshakes will be = (n-1) + (n-2) + (n-3) +…… 0;
= (n-1)*(n-1+1)/2 = (n-1)*n/2 = 66
= n^2 -n = 132
=(n-12)(n+11) = 0;
= n = 12 OR n =-11
-11 is ruled out so the answer is 12 persons.

Reverse a link list without using another

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How to reverse a link list without using another link list?

Note: Linked list is singly linked list (there is only next pointer no previous pointer).
See Solution: Reverse a link list without using another linked list

Take two pointers previous and current, increment both pointers by one till current reaches at the end of the list.

Reverse a Linked List

Reverse a Linked List

Node* reverse(Node* node)
{
	if(node == null)
		return null;
	Node * prev = node;
	Node* curr = node->next;
	
	while(curr != null){
		Node* tmp = curr->next;
		curr->next = prev;
		prev = curr;
		curr = temp;
	}
	return prev;
}

Detect if there is a cycle in a linked list

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How can you find if a given linked list has a cycle in it?

Note: Linked list is singly linked list (there is only next pointer no previous pointer).
See Solution: Detect if there is a cycle in the linked list

This can also be solved by using two pointers as we saw in Find nth element from end of linked list and Find middle element of linked list. Take two pointers pointer1 and pointer2, both pointing to head of the linked list initially, now increment pointer1 by one node and pointer2 by two nodes, If at any point of time pointer1 and pointer2 becomes equal then there is a cycle in the linked list, else pointer2 will reach end of the list and will become null eventually.

Code in C

bool isCyclePresent(Node* node){
	Node* pointer1 = node;
	Node* pointer2 = node;
	
	while(pointer2 != null && pointer2->next!=null){
		pointer1 = pointer1->next;
		pointer2 = pointer2->next->next;
		if(pointer1 == pointer2)
			return true;
	}
	return false;
}

How to measure 4 gallon of water

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How to measure exactly 4 gallon of water from 3 gallon and 5 gallon jars, Given, you have unlimited water supply from a running tap.

It was also solved by Bruce Willis in one of his Die Hard movie to stop a bomb.

See Solution: How to measure 4 gallon water from 3 gallon and 5 gallon jar
There are 2 possible solutions i can think of

Solution 1:

  1. Fill 3 gallon jar with water.
  2. Pour all its water into 5 gallon jar.
  3. Fill 3 gallon jar again.
  4. Pour its water into 5 gallon jar untill it is full. Now you will have exactly 1 gallon water remaining in 3 gallon jar.
  5. Empty 5 gallon jar, pour 1 gallon water from 3 gallon jar into it. Now 5 gallon jar has exactly 1 gallon of water.
  6. Fill 3 gallon jar again and pour all its water into 5 gallon jar, thus 5 gallon jar will have exactly 4 gallon of water.

Solution 2:

  1. Fill 5 gallon jar.
  2. Pour all its water to 3 gallon jar, thus leaving exactly 2 gallon water in 5 gallon jar.
  3. Empty 3 gallon jar and pour all of 5 gallon jar water into it, thus now 3 gallon jar will have exactly 2 gallon of water.
  4. Fill 5 gallon jar again and pour its water to 3 gallon jar until that is full. So exactly 4 gallon water will remain in 5 gallon jar.

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