Programming Interview Questions

hari@hbiz.in

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Comment hidden because of low score. Click to expand.

of 0 vote

knock outs

- hari@hbiz.in September 26, 2012 | Flag Reply

Comment hidden because of low score. Click to expand.

of 0 vote

Not sufficient in this case. Amazon expects us to check for subtrees.

- hari@hbiz.in August 29, 2012 | Flag Reply

Comment hidden because of low score. Click to expand.

of 0 vote

Very good point. But preorder testing is required only if you are given the tree in form of list or array. If the tree itself is given, then any one traversal is sufficient with node by node comparison.

- hari@hbiz.in August 27, 2012 | Flag Reply

Comment hidden because of low score. Click to expand.

of 0 vote

What is the logic behind checking preorder too?

- hari@hbiz.in August 26, 2012 | Flag Reply

Comment hidden because of low score. Click to expand.

of 0 vote

Sorry. It is a binary tree but not a BST

- hari@hbiz.in August 26, 2012 | Flag Reply

Comment hidden because of low score. Click to expand.

of 0 vote

typedef struct TNode
{
	int value;
	TNode *left;
	TNode *right;
} TNode;

bool ifExist(const TNode* treeA, const TNode* treeB);

bool isSubtree(const TNode* treeA, const TNode* treeB)
{
     if (treeB == NULL) 
     {
         // if treeB is null no point in searching.
         return true;
     }
     bool a_has_b = ifExist(treeA,treeB);
     bool b_has_a = false;	
     // Check only if treeB is not found in treeA.
     if (!a_has_b)      
        b_has_a= ifExist(treeB,treeA);
      
     return a_has_b || b_has_a;
}


bool ifExist(const TNode* treeA, const TNode* treeB)
{
    if ( treeA == NULL && treeB != NULL)  
    {
	   // We reached the end of path. No subtree found.
	   return false;
    }
    if ( treeA == treeB)
    {
	   return CompareTheSubtree(treeA,treeB);	   
    }
    return ifExist(treeA->left,treeB)|| ifExist(treeA->right,treeB);
}

Note : I have not written CompareTheSubtree() method as it is trivial.

- hari@hbiz.in August 26, 2012 | Flag Reply

Comment hidden because of low score. Click to expand.

of 4 vote

Travel through tree A, and search for node B if not found, travel through Tree B and search for node A. If you found in either case, compare the elements.

- hari@hbiz.in August 26, 2012 | Flag Reply

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of 0 vote

Why recursion when the following code can do the same.

K=A;

for ( i=1;i<n;i++)
{
k= K*A;
}

- hari@hbiz.in August 16, 2012 | Flag Reply

Comment hidden because of low score. Click to expand.

of 0 vote

Can somebody throw some explanation here.

- hari@hbiz.in August 16, 2012 | Flag Reply

Comment hidden because of low score. Click to expand.

of 0 vote

This is what it doing

Input

                           4
			2    5
	              1        7

After completing swapping at 2

                            4
                      2         5
                           1        7

Swapping at 4 

                           4
                    5              2
                        7              1

Swapping at  2
   
                             4
                 5                       2
                       7            1

  This is mirror image of structure but not data.

- hari@hbiz.in August 16, 2012 | Flag Reply

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of 0 vote

Say the set N =3 { 1,2,3}
Combinations are {1,2} {2,3} {1,3}

Say the set N= 4 {1,2,3,4}
Combinations are {1,2,3 }, {1,2,4}, {1,3,4}, {2,3,4}

- hari@hbiz.in August 16, 2012 | Flag Reply

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of 0 vote

Slightly modified

void verticalSum( TNode* node, int *arr, int Level)
{
  if (node==NULL) return;  

  verticalSum(node->Left,arr, Level-1);
  arr[Level]=arr[Level]+node->value;
  verticalSum(node->Right,arr,Level+1);  
}

- hari@hbiz.in August 14, 2012 | Flag Reply

Comment hidden because of low score. Click to expand.

of 2 vote

Solution using a queue and stack. No hack. o(n2)

Convert2DLL(Queue queue,int level, list** head)
{
 Queue nextqueue;
 Stack stack;

  if (queue.empty())
     return;

 while ( !queue.emtpy() )
 {
    TNode *node= queue.pop();

    if (level%2 == 1)
    {    
        stack.push(node->left);
       stack.push(node->right);
    }
    else
    {
        nextqueue.push(node->left);
        nextqueue.push(node->right);       
    }

    *head->right=node;
     node->left=*head;   
     *head=node;
     *head->right=null;        
 }

 if ( level%2 == 1)
 {
   while ( !stack.empty)
   {
      node *temp = stack.pop();
      nextqueue.push(temp);
   }
 }

 queue = nextqueue;
 Convert2DLL(queue,level++, head);
}

- hari@hbiz.in August 14, 2012 | Flag Reply

Comment hidden because of low score. Click to expand.

of 0 vote

No need of 2 digits if we can XOR the a[num] with 1 (assume all the a[elements] initialized with 0) . If at all an XOR with 1 returns a 0, we can assume it is a duplicate and return it. Only thing is we don't know whether a number occurred or not, which is not required in the current case.

- hari@hbiz.in August 12, 2012 | Flag Reply

Comment hidden because of low score. Click to expand.

of 0 vote

Can you please help me to understand why do we require 2 bits? A 1 bit array of size 4294967295 is sufficient right?

- hari@hbiz.in August 12, 2012 | Flag Reply

Comment hidden because of low score. Click to expand.

of 0 vote

I believe your algorithm won't work or I misunderstood it, can you prove how your algorithm will generate a value 15?

- hari@hbiz.in August 11, 2012 | Flag Reply

Comment hidden because of low score. Click to expand.

of 0 vote

Good approach. But when large number of elements checked out, a BST, would give a better performance than linked list.

- hari@hbiz.in August 11, 2012 | Flag Reply

Comment hidden because of low score. Click to expand.

of 0 vote

For min-heap, we can use a BST.

- hari@hbiz.in August 11, 2012 | Flag Reply

Comment hidden because of low score. Click to expand.

of 0 vote

No. Complexity is O(n+m+ (n-m) ). Worst case is o(3n.)

- hari@hbiz.in August 10, 2012 | Flag Reply

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