Algorithm Interview questions Answers ( Microsoft, Google and Amazon )

1. Reverse a singly linked list


//
// iterative version
//
Node* ReverseList( Node ** List )
{

Node *temp1 = *List;
Node * temp2 = NULL;
Node * temp3 = NULL;

while ( temp1 )
{
*List = temp1; //set the head to last node
temp2= temp1->pNext; // save the next ptr in temp2
temp1->pNext = temp3; // change next to privous
temp3 = temp1;
temp1 = temp2;
}

return *List;
}


2. Delete a node in double linked list


void deleteNode(node *n)
{
node *np = n->prev;
node *nn = n->next;
np->next = n->next;
nn->prev = n->prev;
delete n;
}


3. Sort a linked list

//sorting in descending order
struct node
{
int value;
node* NEXT;
}
//Assume HEAD pointer denotes the first element in the //linked list
// only change the values…don’t have to change the //pointers

Sort( Node *Head)
{
node* first,second,temp;
first= Head;
while(first!=null)
{
second=first->NEXT;
while(second!=null)
{
if(first->value < second->value)
{
temp = new node();
temp->value=first->value;
first->value=second->value;
second->value=temp->value;
delete temp;
}
second=second->NEXT;
}

first=first->NEXT;
}
}


4. Reverse a string

void ReverseString (char *String)
{
char *Begin = String;
char *End = String + strlen(String) - 1;
char TempChar = '\0';

while (Begin < End)
{
TempChar = *Begin;
*Begin = *End;
*End = TempChar;
Begin++;
End--;
}
}


5. Insert a node a sorted linked list

void sortedInsert(Node * head, Node* newNode)
{
Node *current = head;

// traverse the list until you find item bigger the // new node value
//
while (current!= NULL && current->data < newNode->data)
{
current = current->next);
}
//
// insert the new node before the big item
//
newNode->next = current->next;
current = newNode;
}



6. Covert a string to upper case

void ToUpper(char * S)
{
while (*S!=0)
{
*S=(*S >= 'a' && *S <= 'z')?(*S-'a'+'A'):*S;
S++;
}
}


7. Multiple a number by 7 without using * and + operator.

NewNum = Num << 3; // mulitplied by 2 ^ 3 = 8

NewNum = NewNum - Num; // 8 – 1 = 7


8. Write a function that takes in a string parameter and checks to see whether or not it is an integer, and if it is then return the integer value.

#include

int strtoint(char *s)
{
int index = 0, flag = 0;

while( *(s+index) != '\0')
{
if( (*(s + index) >= '0') &&
*(s + index) <= '9')
{
flag = 1;
index++;
}
else
{
flag = 0;
break;
}
}

if( flag == 1 )
return atoi(s);
else
return 0;
}

main()
{
printf("%d",strtoint("0123"));
printf("\n%d",strtoint("0123ii"));
}



9. Print a data from a binary tree – In-order(ascending)

//
// recursive version
//

Void PrintTree ( struct * node node )
{
if ( node == NULL )
return;

PrintTree(node->left );
Printf(“%d”, node->data);
PrintTree(node->right );
}

10. print integer using only putchar

//
// recursive version
//
void PrintNum ( int Num )
{
if ( Num == 0 )
return;
PrintNum ( Num / 10 );
Puthcar ( ‘0’+ Num % 10 );
}

11. Find the factorial of number

//
// recursive version
//

int Factorial( int Num )
{

If ( num > 0 )
return Num * Factorial ( Num –1 );
else
return 1;
}


//
// iterative version
//

int Factorial( int Num )
{
int I
int result = 1;

for ( I= Num; I > 0; I-- )
{
result = result * I;
}

return result;
}


12. Generate Fib numbers:

int fib( n ) // recursive version
{

if ( n < 2 )
return 1;
else
return fib ( n –1 ) + fib ( n –2 );

}

int fib( n ) //iterative version
{
int f1 =1, f2 = 1;

if ( n < 2 )
return 1;
for ( i = 1; i < N; i++)
{
f = f1 + f2;
f1= f2;
f = f1;
}

return f;
}







13. Write a function that finds the last instance of a character in a string

char *lastchar(char *String, char ch)
{
char *pStr = NULL;

// traverse the entire string

while( * String ++ != NULL )
{
if( *String == ch )
pStr = String;
}

return pStr;
}

14. Return Nth the node from the end of the linked list in one pass.

Node * GetNthNode ( Node* Head , int NthNode )
{
Node * pNthNode = NULL;
Node * pTempNode = NULL;
int nCurrentElement = 0;

for ( pTempNode = Head; pTempNode != NULL; pTempNode = pTempNode->pNext )
{
nCurrentElement++;
if ( nCurrentElement - NthNode == 0 )
{
pNthNode = Head;
}
else
if ( nCurrentElement - NthNode > 0)
{
pNthNode = pNthNode ->pNext;
}
}
if (pNthNode )
{
return pNthNode;
}
else
return NULL;
}



15. Counting set bits in a number.

First version:

int CoutSetBits(int Num)
{

for(int count=0; Num; Num >>= 1)
{
if (Num & 1)
count++;
}
return count;
}


Optimized version:

int CoutSetBits(int Num)
{

for(int count =0; Num; count++)
{
Num &= Num -1;
}
}