JSDUDE

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Reputation 2675

Page:

Comment hidden because of low score. Click to expand.

-1

of 1 vote

class Node {
    
    Node(String name) {
        this.nodeName = name;
        this.count = 0;
        this.children = new HasMap<>();
    }

    String nodeName;
    int count;
    Map<String Name, Node child> children;
}

class DataStructure {

    private Node parent = null;

    public DataStructure {
        parent = new Node("root");
    }
    
    public void populate(List<String> inputs) {
        if(null == inputs)
            return;
        		
		for(String input : inputs) {
            
			//input : /Electronics/Computers/Graphics Cards        
			
			List<String> categories = parseInput(input);  // categories : Electronics, Computers, Graphics Cards
            if(null == categories)
                continue;                
			
            Node current = parent;
            for(String category : categories) { // category : Electronics 
                if(!current.children.containsKey(category)) {
                    Node child = new Node(category);
                    current.children.add(category, child);
                }
                current.count++;
                current = current.children.get(category);                
            }
        }
    }
    
    private List<String> parseInput(String input) {
        if(input.isEmpty())
            return null;
            
		List<String> result = new ArrayList<String>(Arrays.asList(input.split("/")));
        return result;
    }
}

- JSDUDE April 19, 2017 | Flag Reply

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of 0 vote

I wrote almost the same code (I used switch case instead of if-else).
However your code misses the scenario "a(b"

Fix:

if(openParenCount > 0) {
	throw new IllegalArgumentException();
}

return maxOpenParen;

- JSDUDE June 04, 2015 | Flag Reply

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of 0 vote

Assume the dictionary is stored as a TRIE and has the following functions:
bool isPotentialWord(String str)
e.g. "qs" will return false since no word starts with qs
"ca" will return true as "cat", "category", "cast" etc.... are words
bool isWord(String str)
will return true if given word is a string

List<String> allPotentialStrings(String inputStr)
{
	if(String.IsEmpty(inputStr))
		return null;
	
	return allPotentialStrings(inputStr, 0);
}

List<String> allPotentialStrings(String inputStr, int startIndex)
{
	if(startIndex > inputStr.length - 1)
		return ".";
	
	List<String> outputStr = new List<String>();
	String currStr = "";
	
	for(int index = startIndex; index < inputStr.length; ++index)
	{
		currStr += ch;
		if(!isPotentialWord(currStr))
			break;
		else if(!isWord(currStr))
			continue;
		else
		{
			List<String> restStr = allPotentialStrings(inputStr, index + 1);
			if(null == restStr)
				continue;
			foreach(String str in restStr)
				outputStr.add(currStr + " " + str);
		}
	}
	
	return outputStr;
}

- JSDUDE January 20, 2015 | Flag Reply

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of 1 vote

Dang it ... I meant to write:
Design a Tiny URL system.

- JSDUDE January 19, 2015 | Flag Reply

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of 1 vote

void FindIntsWithSum(unsigned int[] array, unsigned int sum)
{
    if(0 == array.length)
        return;
        
   Dictionary<unsigned int, unsigned int> myDict = new Dict();
   
   for(int index = 0; index < array.length; ++index)
   {
       if(myDict.contains(sum - array[index]))   
       {
           Console.WriteLine("{0} & (1}", array[index], myDict.key(sum - array[index]));
           return;
       }
       else
       {
           myDict.add(array[index]);
       }
   }
   
   Console.WriteLine("This array does not contain values that sum upto: {0}", sum);
   return;
}

// Test cases
// Empty array
Single element array
Large array -- Blow the memory
Small array
Arrays with lot of duplicates
Array without sum of the integer ...
Array with multiple combination...
Performance, Memory: (Large and small array)
Simple tests: Small arrays with sum present : Basic P1
Check for buffer overflow

- JSDUDE November 22, 2014 | Flag Reply

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of 0 vote

@LV: psuedo code

class Product
{
	item#, prod#, category, title, author, etc...
}

void buildDS(File f, char delimiter)
{
	if(null == f)
		return;
	
	List<Product> products = new List<Product>();
	Dictionary<string, List<int>> index_title = new Dictionary<>(string, List<int>);
	// Add one dictionary per queriable meta-data (title, category, author)
	
	while(!f.eof)
	{
		string line = f.readLine();
		
		// Handle bad line scenarios

		// Create product instance and populate
		string[] infos = string.tokenize(line, delimiter);
		products.add(new Product(infos[0], ... infos[infos.length - 1]));
		
		index_title.add(prod.title, products.length - 1);		
		// Update all the dictionaries created
	}
}

HTH :)

- JSDUDE October 24, 2014 | Flag Reply

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of 0 vote

void print(List<Node> listNodes)
{
	if(listNodes.isEmpty)
		return;
		
	Dictionary(int, List<Node>) dict = new ;
	int maxHeight = -1;
	int maxWidth = -1;
	
	// Update dictionary
	foreach(node in listNodes)
	{
		int height = getHeight(node);
		dict[index].List.AppendAtEnd(node);
		int width = dict[height].List.length;
		maxHeight = (height > maxHeight) ? height : maxHeight;
		maxWidth = (width > maxWidth) ? width : maxWidth;		
	}
	
	// Print
	for(int width = 0; width < maxWidth; ++width)
	{
		for(int index = 0; index < maxHeight; ++index)
		{
			Node temp = null;
			if(null != dict[index].List.getValueAt(width))
				temp = dict[index].List.getValueAt(width);
			string str = (null == temp) ? "\t" : temp.value.toString();
			print(str);
		}
		println();
	}
}

int getHeight(Node node)
{
	int height = 0;
	while(null != node.parent)
		node = node.parent;

	return height;
}

- JSDUDE October 24, 2014 | Flag Reply

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of 0 vote

@CT:
your implementation is correct.
Except for the initialization of diffs[0]. I think it should be as such:

diffs[0] = (bytes[0] == '#') ? -1 : 1;

- JSDUDE October 23, 2014 | Flag Reply

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of 0 vote

static int maxDivide(int num, int divider)
        {
            while(num % divider == 0)
                num = num / divider;

            return num;
        }

        static bool isNumberUgly(int num)
        {
            if (0 == num)
                return false;

            return maxDivide(maxDivide(maxDivide(num, 2), 3), 5) == 1;
        }

        static int findNthUglyNumber(int n)
        {
            int num = 1;
            int count = 1;

            while (count < n)
            {
                ++num;
                if (isNumberUgly(num))
                    ++count;
            }

            return num;
        }

- JSDUDE October 23, 2014 | Flag Reply

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of 0 vote

I have written code in C#

void printDiagonally(int max_int)
        {
            if (max_int < 0)
                return;

            int row_incrementer = 1;
            int row_start = 1;

            while (row_start <= max_int)
            {
                int col_incrementer = row_incrementer;
                int col_value = row_start;
                while (col_value <= max_int)
                {
                    Console.Write("{0} ", col_value);
                    col_value += col_incrementer;
                    col_incrementer++;
                }

                Console.WriteLine();
                row_incrementer++; 
                row_start += row_incrementer;                
            }

- JSDUDE October 23, 2014 | Flag Reply

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of 0 vote

I have not handled the trailing and ending zeros scenario in my code.

- JSDUDE October 22, 2014 | Flag Reply

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of 3 vote

static void Main(string[] args)
        {
            const int DIVIDER = 3;
            const int MAX_DIGITS = 3;
            const char BYTES = 'B';
            const char KILOBYTES = 'K';
            const char MEGABYTES = 'M';
            const char GIGABYTE = 'G';

            // Get the input from the console
            string stringInput = Console.ReadLine();
            
            // Extract the bytes from the input
            char lastChar = stringInput[stringInput.Length - 1];
            string stringBytes = stringInput.Substring(0, stringInput.Length - 1);

            // Get the length of the input bytes
            int lengthBytes = stringBytes.Length;

            // If the length of input bytes is less than 3 digits, 
            // output is same as input
            // E.G. 34K
            if (lengthBytes <= MAX_DIGITS)
            {
                Console.WriteLine("The output is: {0}", stringInput);
                return;
            }

            // Length divided by 3 (10 to the power 3 is the multiple)
            // Decides if it is KB, MB or GB
            int quotient = (lengthBytes - 1)/ DIVIDER;
            
            // Length modulo 3 decides where the decimal is placed 
            int modulo = lengthBytes % DIVIDER;

            char unit = 'O';

            switch (quotient)
            {
                case 0:
                    unit = BYTES;
                    break;
                case 1:
                    unit = KILOBYTES;
                    break;
                case 2:
                    unit = MEGABYTES;
                    break;
                case 3:
                    unit = GIGABYTE;
                    break;
            }

            // Modulo 0 means no decimal required
            // It is a multiple of 3
            int max = (modulo == 0) ? MAX_DIGITS : modulo;

            string result_dec = stringBytes.Substring(0, max);

            if (result_dec.Length == MAX_DIGITS)
            {
                Console.WriteLine("The output is: {0}", string.Concat(result_dec, unit));
                return;
            }

            // Add the decimal after the digits
            string result = string.Concat(result_dec, '.', stringBytes.Substring(max , MAX_DIGITS - max), unit);
            Console.WriteLine("The output is: {0}", result);

            return;
        }

- JSDUDE October 22, 2014 | Flag Reply

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of 0 vote

Sonesh,
In your question you have mentioned Binary Search Tree. you might want to update the question.

- JSDUDE October 02, 2013 | Flag Reply

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of 0 vote

The proposed solution will work for BT. Solution is not utilizing the fact that it is a BST.

- JSDUDE October 02, 2013 | Flag Reply

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of 0 vote

What language is this?
Do you happen to know what's the time complexity of the 2nd line?

- JSDUDE July 13, 2013 | Flag Reply

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of 0 vote

struct Node
  {
      int data;
      Node* left;
      Node* right;
      Node* parent;
  };

Node* RightNeighbor(Node* node)
  {
      if(!node)
        return node;
      uint level  = 1;
      Node* currParent = node->parent;
      while(currParent)
      {
          Node* temp = GetLeftMostChild(currParent->right, level-1);
          if(temp)
            return temp;
          ++level;
          currParent = currParent->currParent->parent;
      }

      return null;
  }

Node* GetLeftMostChild(Node* node, uint level)
  {
      if(0 == level)
          return node;
      if(node->left)
          return GetLeftMostChild(node->left, level-1);
      if(node->right)
          return GetLeftMostChild(node->right, level-1);

       return null;
  }

- JSDUDE July 01, 2013 | Flag Reply

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of 0 vote

what does cur->children hold? Looks like an array. Can you please elaborate on the structure of your node ?

- JSDUDE July 01, 2013 | Flag Reply

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of 0 vote

Whatever data you want to send to the node with the id idTo.

- JSDUDE June 25, 2013 | Flag Reply

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of 0 vote

AAARGH ....
the last line should be
send(2*id + 2, sum);
Apologies for the mistake.

- JSDUDE June 14, 2013 | Flag Reply

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of 0 vote

Really stumped me. So I started with O(n^2) solution:

int GetSum()
{
    int sum = 0;
    // Get values from all nodes prior it self
    for(int i = 0; i < id; ++i)
    {
        sum + = recv(i);
    }
    
    // Send self value to all prior to self
    for(int i = 0; i < id; ++i)
    {
        send(i, val);
    }
    
    // Send self value to all after self
    for(int i = id+1; i < N; ++i)
    {
        send(i , val);
    }
    
    // Get values from all after self
    for(int i = id+1; i < N; ++i)
    {
        sum+= recv(i);
    }
    
     return sum+val;   
}

Then I realized it can be done in O(n):

#define SERVER = 0

int GetSum()
{
    int sum =0;
    if(id == SERVER)
    {
        for(int i = 1; i< N; ++i)
        {
            sum + = recv(i);
        }
        
        sum += val;

        for(int i = 1; i < N; ++i)
        {
            send(i, sum);
        }
        
        return sum;
    }
    
    send(SERVER, val);
    return recv(SERVER);
}

But he wanted better. After multiple clues, it was clear that since the IDs of the Nodes are sorted and in sequence from 0 to N-1, we can consider the nodes as tree, with 0 as root and the children of a Node will be (2*ID+1) parent would (ID-1)/2.
And we can take a recursive Binary approach.

int GetSum()
{
    // break condition
    if(2*id + 1 > N-1 )
        {
            send((id-1)/2, val);
            sum = recv((id-1)/2);
            return sum;
         }
    // recurse
    int sum = val + recv(2*id + 1) + recv(2*id + 2);
    
    // send to parent:
    if(id !=0)
    {
        // send partial sum to parent
        send((id-1)/2, sum);    
        
        // receive full sum from parent
        sum = recv((id-1)/2);      
    }    
    
    // Send final sum to children
    send(2*id + 1, sum);
    send(2*id + 1, sum);

}

- JSDUDE June 14, 2013 | Flag Reply

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of 0 vote

@Mystic
Only the function signature was provided. He didn't provide the function itself.
I had to code the solution without the code for send and receive.
I am curious as to how will it help to know the code for send and receive? Will help me improve my understanding.
Thanks.

- JSDUDE June 14, 2013 | Flag Reply

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of 0 vote

These are three solutions I implemented:

Solution 1 with HashTable:

//Global vairable
Hashtable<int, int> myHasTable = new Hashtable<int, int>();

void PopulateHashTable(int[] array1, int[] array2)
{
    foreach(int i in array1)
    {
        if(myHasTable(i).Value == 10)
            continue;
        myHasTable(i).Value = 10;
    }
    
    foreach(int i in array2)
    {
        if(myHasTable(i).Value==1 || myHasTable(i).Value==11)
            continue;
        myHasTable(i).Value++;
    }
    
    return;
}

int[] Union(int[] array1, int[] array2)
{
    int[] unionArray = new int[array1.length + array2.length];
    int index = 0;
    foreach(int i in array1)
    {
        if(myHasTable(i).Value > 1)
        {
            unionArray[index++] = i;
        }
    }
    
    foreach(int i in array2)
    {
        if(myHasTable(i).Value == 1)
        {
            unionArray[index++] = i;
        }
    }
    
    return unionArray;
}

int[] Intersection(int[] array1, int[] array2)
{
    int[] intersectionArray = new int[array1.length, array2.length];
    int index= 0;
    
    foreach(int i in array1)
    {
        if(myHasTable(i) == 11)
            intersectionArray[index++] = i;
    }
    
    return intersectionArray;
}

No extra memory O(n^2) Union only:

// No extra memory Union function
int[] Union(int[] array1, int[] array2)
{
    int unionArray = new int[array1.length, array2.length];
    int index = 0;
    int runner = 0;

    unionArray[index++] = array1[0];
    
    foreach(int i in array1)
    {
        runner = 0;
        while(runner < index)
        {
            if(uninoArray[runner] != i)
            {
                runner++;
            }
            else
            {
                break;
            }
        }        
        if(runner == index)
        {
            unionArray[index++] = i;
        }
    }
    
// Repeat for array2

// Sorted arrays union
// Sorting takes O(log n) time.
// Assume array1 and array2 are sorted
int[] Union (int[] array1, int[] array2)
{
    int unionArray;
    index1 = 0;
    int index2 = 0;
    //unionArray[index++] = array1[0]
    foreach(int i = 0; i < array1,length + array2.legnth; ++i)
    {
        if(array1[index1] == array2[index2])
        {
            if(index > 0 && array1[index1] != unionArray[index-1])
            {
            uninoArray[index++] = arra1[index1];
            index1++;
            index2++;
            }
        }
        else if(array1[index1] < array2[index2])
        {
            unionArray[index++] = array1[index1++]; 
        }
        else
        {
            unionArray[index++] = array2[index2++];
        }
        
    }
}

Test cases:
1. Unsorted array with single value, all duplicates
2. Unsorted array some duplicates
3. Unsorted array no dupes
4. One array, 2nd empty
5. Sorted array, multiple dupes
6. Sorted array no dupes
7. Very large arrays
8. Both are empty arrays
9. 1 valued array(s)
10. Negative value and + values
11. both arrays with exactly same values
12. both arrays with unique values only
13. both large arrays containing only one value, duped

- JSDUDE June 14, 2013 | Flag Reply

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-1

of 1 vote

You are using one hash table for unions, and two for intersections.
Can you use one hash table for both union and intersection?

After giving it a thought I realized we can do it in O(m+n) time. i.e. one traversal of both arrays for both intersection and union.

- JSDUDE June 14, 2013 | Flag Reply

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of 0 vote

No doubt good explanation.
However they are recursively traversing the tree twice.
Same recursion, gathering different data. Should be able to do it in one recursion.

- JSDUDE June 11, 2013 | Flag Reply

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of 0 vote

You are not handling the return in the recursive call you are making:
f(n->left)
f(n->right)

- JSDUDE June 06, 2013 | Flag Reply

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of 0 vote

No specification about extra space so I propose using a stack.

Node* LowestCommonAncestor(Node* node1, Node* node2)
{
  if(!node1 && !node2)
    return null;
    
  if(!node1 || !node2)
    // Ask interviewer what needs to be returned here.
    
  Stack<Node*> s1 = new Stack<Node*>();
  Stack<Node*> s2 = new Stack<Node*>();
  
  Node* temp = node1;
  while(temp)
  {
    s1.push(temp);
    temp = temp->parent;
  }
  
  temp = node2;
  while(temp)
  {
    s2.push(temp);
    temp = temp->parent;
  }
  
  // Nodes from different trees
  if(s1.peek != s2.peek)
    return null;
  
  while(s1.peek == s2.peek)
  {
    temp = s1.pop;
    s2.pop;
  }
  
  return temp;
}

- JSDUDE June 06, 2013 | Flag Reply

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of 0 vote

Really awesome algorithm dude. Thanks for sharing :).
I spent some time: printing out one iteration and seeing how it can be manipulated to get the other two but this is thinking smart not hard.

- JSDUDE May 23, 2013 | Flag Reply

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of 0 vote

Queue SerializeBT(TreeNode* root)
{
  Queue<Node*> q = new Queue<Node*>();
  if(!root)
    q.enqueue(root);
  else
    SerializeBT(root, q);
  return q;
}

void SerializeBT(TreeNode* node, Queue q)
{
  if(!node)
  {
    q.enqueue(node);
    return;
  }
  
  q.enqueue(node);
  SerializeBT(node->left, q);
  SerializeBT(node->right, q);
  return;
}

- JSDUDE May 22, 2013 | Flag Reply

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of 0 vote

Onsite.
Not an useless Q. I did enter that this was an Onsite Q, but I don't see that information on this page.
If any one knows where it is shown, please enlighten us :).

- JSDUDE May 04, 2013 | Flag Reply

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of 0 vote

@DashDash,
They way i thought about it is, if the videosize is 2 hours long and your timestamp indicates that you have hit play more than Threshold times on this video within the timestamp span of 2 hours then it means you have issues.
Makes sense?
That is the way I am incrementing the count on the video in my list

But i guess i didnot consider re-viewing of the video.

- JSDUDE April 30, 2013 | Flag Reply

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of 0 vote

abc is followed by a '..' which in UNIX means that the parent of this directory. So the parent of abc, which in this case is the folder: home.

- JSDUDE April 30, 2013 | Flag Reply

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of 0 vote

Create a dictionary with Key as CustomerId (always unique) and value as a LinkList, whose each node represents the videoDetails (structure shown below):

We will have to update the LinkList Add function to do this:
1. Traverse the list and if videoId doesnt exist, create one at tail with count as 1
2. If videoId exists increment count

Traverse through each entry in the dictionary and if the number of play hit doesnt meet or exceeds the threshold, remove the node.

Return the dictionary

Struct VideoDetails
{
  VideoID vID;
  VideoLength vLength;
  uint numberOfPlayHits;
}

- JSDUDE April 30, 2013 | Flag Reply

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of 1 vote

I also proposed stack to which he approved and then asked for a solution without using extra space.

- JSDUDE April 30, 2013 | Flag Reply

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of 0 vote

Sorry it becomes:
$/home/def/

- JSDUDE April 30, 2013 | Flag Reply

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of 1 vote

@Artist,
Good approach.I have a question though:
Worst case scenario (lets say the right most leaf and its parent form the sum), you will traverse the whole tree and search the whole tree for val correct? which would make it O(n^2) ?

My implementation of your algorithm ...

struct Node
{
  int data;
  Node* left;
  Node* right;
}

Driver Function

bool FindSum(Node* root, int sum, Node* node1, Node* node2)
{
  node1 = NULL;
  node2 = NULL;

  if(!root)
    return false;
    
  Node* current = root;
  
  while(current && current->data > sum)
    current = current->left;
    
  if(!current)
    return false;
    
  FindSum(current, current, sum, node1, node2);
  
  if(!node1)
    return false;
    
  return true;
}

Recursive Function to do a In-Order traversal

void FindSum(Node* root, Node* current, int sum, Node* node1, Node* node2)
{
  int value = sum - current->data;
  if(FindNode(root, value, node2))
  {
    node1 = current;
    return;
  }
  
  if(current->left)
    FindSum(root, current->left, sum, node1, node2);
    
  if(current->right)
    FindSum(root, current->right, sum, node1, node2);
}

Helper Function to find a node with given data

bool FindNode(Node* root, int data, Node* node)
{
  if(!root)
    return false;
    
  while(root)
  {
    if(root->data == data)
    {
      node = root;
      return true;
    }
    else if(root->data < data)
      root = root->right;
    else
      root = root->left;
  }
  
  return false;
}

- JSDUDE April 28, 2013 | Flag Reply

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of 0 vote

@DashDash,
I also thought the same during the interview. But the interviewer posed an example where the int variable wasn't sufficient. I can't remember the example right now. I am trying to remember\find it. Will udpate as soon as I do.

But yes you are correct int will work for most cases.

- JSDUDE April 27, 2013 | Flag Reply

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of 0 vote

int LevelWithMaxNodes(Node* root)
{
  if(!root)
    return -1;
  
  int maxNodes = 1;
  int runningMaxNodes = 1;
  int level = 1;
  int runningLevel = 1;
  Queue<Node*> myQ = new Queue();
  Node* DELIMITER = null;
  
  myQ.push(root);
  myQ.push(DELIMITER);
  
  while(!myQ.IsEmpty())
  {
    Node* current = myQ.dequeue();
    
    if(current)
    {
      ++runningMaxNodes;
      // The below condition returns the first Level with max nodes, >= will give the last level with maxNodes
      if(runningMaxNodes > maxNodes) 
      {
        maxNodes = runningMaxNodes;
        level = runningLevel;
      }
      
      if(current->left)
        myQ.enqueue(current->left);
      if(current->right)
        myQ.enqueue(current->right);
    }
    else if(current == DELIMITER)
    {
      ++runningLevel;
      myQ.enqueue(DELIMITER);
    }
  }
  
  return level;
}

- JSDUDE April 27, 2013 | Flag Reply

Comment hidden because of low score. Click to expand.

of 1 vote

Every time I dequeue a NULL (delimiter) I push a delimiter.
When 1 is queued in, I queued a NULL PTR as a delimiter. After which 2 & 3 got queued in followed by a NULL delemiter.

So when I dequeue 3, the front of the queue should be a null. So I will make the RightNeighbor of 3 as NULL. As expected.

This was my attempt, which the interviewer approved. Please let me know if i missed something.

Thanks.

- JSDUDE April 27, 2013 | Flag Reply

Comment hidden because of low score. Click to expand.

of 3 vote

struct Node
{
  int data;
  Node* left;
  Node* right;
}

Use Iteration

Node* FindCeiling(Node* head, int key)
{
  if(!head)
    return head;
  
  Stack<Node*> myStack = new Stack();
  Node* current = head;
  
  while(current)
  {
    if(current->data > key)
    {
      myStack.push(current);
      current = current->left;
    }
    if(current->data <= key)
      current = current->right;
  }
  
  return myStack.Pop();
}

Use Recursion

Node* FindCeiling(Node* head, int key)
{
  if(!head)
    return head;
  
  if(head->data <= key)
    return FindCeiling(head->right, key);
    
  Node* temp = FindCeiling(head->left, int key);
  if(!temp)
    return head;
  
  return temp;
}

- JSDUDE April 27, 2013 | Flag Reply

Comment hidden because of low score. Click to expand.

of 2 vote

struct Node
{
  int data;
  Node* right;
  Node* left;
  Node* RightNeighbor;
}

void UptateTreeToPointToRightNeighbor(Node* head)
{
  if(!head)
    return null;
    
  Queue<Node*> myQ = new Queue();
  myQ.push(head);
  myQ.push(NULL);  
  
  while(!myQ,IsEmpty)
  {
    Node* curr = myQ.Dequeue();
    
    if(curr)
    {
      curr->RightNeighbor = myQ.Front;
      if(curr->left)
        myQ.push(curr->left);
      if(curr->right)
        myQ.push(curr->right);
    }
    else
        myQ.push(NULL);
  }
}

- JSDUDE April 27, 2013 | Flag Reply

Comment hidden because of low score. Click to expand.

of 2 vote

Can you please provide an Example?
If you are taking x and y as input then there is only one value for (x^3 + y^3).
May be I am missing something here.

- JSDUDE April 22, 2013 | Flag Reply

Comment hidden because of low score. Click to expand.

of 0 vote

Nope.

- JSDUDE April 15, 2013 | Flag Reply

Comment hidden because of low score. Click to expand.

of 1 vote

Sorry I meant 27th Feb and 27th March are exactly one month apart.
This resolves the Leap year issue, 30/31 day month issue.
Apologies for the earlier post.

- JSDUDE April 08, 2013 | Flag Reply

Comment hidden because of low score. Click to expand.

of 0 vote

So i assumed one month as:
Date of One month to the Same Date of the next month, irrespective of the month.

So 27th Feb and 28th Feb are exactly one month apart.

- JSDUDE April 08, 2013 | Flag Reply

Comment hidden because of low score. Click to expand.

of 0 vote

I coded with both dates inclusive.

- JSDUDE April 08, 2013 | Flag Reply

Comment hidden because of low score. Click to expand.

of 1 vote

Class Player
{
    public:
        void play(string bandName, string songName);
        string topSong(string bandName);
        
    private:
        Hash<string bandName, Class Band> bands = new Hash<string, Class Band>();
}

Class Band
{
    public:
        string Name;
        void play(string songName);
        string topSong();
        
     private:
        Hash songs = new <string songName, Song songs>();      
        // Will ensure that the topSong variable is updated by one thread at a time
        volatile song topSong;
}

Class Song
{
    public:
        Song song;
        void IncrementRequest();
        int GetNumberofRequests();
        
        private:
        int numberOfRequests;    // initialized to 0 in the constructor
}

string topSong()
{
    return this.topSong;
}

void play(string SongName)
{
    // Assuming that in case of same number of plays, the latest song wins
    if (topSpong.GetNumberOfRequests <= (song.GetNumberOfRequests + 1))
     {
        topSong = song;
     }
        
     song.incrementRequest();   
    
    song.play();
}

void IncrementRequest()
{
    this.numberOfRequests++;
}

int GetNumberOfRequests()
{
    return numberOfRequests;
}

- JSDUDE April 06, 2013 | Flag Reply

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