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Magento Plugin for Twitter Card

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What is Twitter Card?

I recently came to know about twitter cards from my brother, As such twitter allows only 140 characters to tweet, but to compete with Facebook they have launched a feature called twitter cards. Basically after adding the specific twitter meta tags to your page, twitter will give an option to expand the post, after expanding the post, user will see photo, title, description with link to your site, for example visit this funnychutkule’s twitter page. This is a great feature from twitter to promote your website and in my point of view it is a must.

To know more about types of twitter cards please visits official twitter page

I have written a small Twitter card plugin to add the twitter meta tags to your website’s product page, Click here to visit official magento connect page of this twitter card plugin.

How to install Twitter card Magento plugin from puzzlersworld?

  1. Login to your magento admin panel.
  2. Go to System->Magento Connect->Magento Connect Manage.
  3. You need to login again with the same credentials as in first step.
  4. Paste this extension key http://connect20.magentocommerce.com/community/Pw_Twittercard under Install New Extensions section and click on install.
  5. Confirm the installation once again, after installation you have to set your twitter username.

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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.

Rotate an array by k elements

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Given an array of n elements, write an algorithm to rotate it right by k element without using any other array. In other words rotate an array in place.
you can have temp variable to swap two elements.
Ex:-
Initial Array: A = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
Rotating array A right by 3 elements should result in
A = {8, 9, 10, 1, 2, 3, 4, 5, 6, 7}
Hint: There is a trick, think before doing much of the programming 🙂
See Solution: Rotate an array by k elements

So the trick is to reverse the whole array first then reverse the array from 0 to k-1 and k to n-1.
For example
Array A after reversing whole array: {10, 9, 8, 7, 6, 5, 4, 3, 2, 1}
Array A after reversing 0 to 2 elements: {8, 9, 10, 7, 6, 5, 4, 3, 2, 1}
Array A after reversing 3 to 9 elements: {8, 9, 10, 1, 2, 3, 4, 5, 6, 7}
Code:

void rotateRight(int[] a, int k){
	int length = a.length();//in C/C++ it will be given as input
	reverse(a, 0, length -1);//reverse whole array
	reverse(a, 0, k-1);//assuming 0 < k < length
	reverse(1, k, n-1);
}
void reverse(int[] a, int start, int end){
	while(start < end){
		swap(a, start, end);
		start++;
		end--;
	}
}
void swap(int[] a, int pos1, int pos2){
	int temp = a[pos1];
	a[pos1]= a[pos2];
	a[pos2] = temp;
}
//Note: in interviews you might be asked to handle negative scenarios
// like array out of bound etc.

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;
}

Find nth element from end of linked list

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How to find nth element from the end of the linked list?

Or in other words Given an integer ‘n’, return a pointer to the n-th node from the end of the list.
Note: Linked list is singly linked list (there is only next pointer no previous pointer).

See Solution: find nth element from end of linked list

Keep two pointers pointer1 and pointer2, both pointing to head of the linked list initially. Increment pointer1 by one node till it reaches (n+1)th node. Now increment both pointers by one node till pointer1 becomes null. At this point the node which pointer2 is pointing to will be the nth node from the last.

Code in C/C++

Node * getNthNodeFromLast(Node * node, int n){
	Node* pointer1 = node;
	Node* pointer2 = node;

	while(pointer1 != null && n > 0){
		pointer1 = pointer1->next;
		n--;
	}
	//List does not have n nodes, so return null
	if(n > 0)
		return null;
	//traverse till pointer1 becomes null
	while(pointer1 != null){
		pointer1 = pointer1->next;
		pointer2 = pointer2->next;
	}
	return pointer2;
}

Find middle element of linked list

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How to Find the middle element of a linked list without traversing twice ?

Linked list is singly linked list(there is only next pointer no previous pointer).

See Solution: Find middle element of a linked list

Take two pointers both pointing to head of the linked list( pointer1 and pointer2 ). Now increment pointer1 by one node and pointer2 by two nodes at a time in a while loop till pointer 2 reaches end of the list or becomes null, at this point the node which pointer1 is pointing to will be the middle element/node of the linked list.

Code in C/C++

Node* getMiddleElement(Node * node){
	Node* pointer1 = node;
	Node* pointer2 = node;

	//traverse till pointer2 reaches end of the list or becomes null
	while(pointer2 != null && pointer2->next != null){
		pointer1 = pointer1->next;
		pointer2 = pointer2->next->next;
	}
	return pointer1;
}

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