CareerCup

1> p= XOR 1 to n

for i=1 to n-1
p=p XOR a(i)

p is the ans .

2 > p= XOR 1 to N-1

for i=1 to N
p= p XOR a(i)

p is the ans .

3 >

p=0;
for i=1 to N
p=p XOR a(i)

p= ans .

4 >
bool CheckBST ( node* T , int min , int max )
{
if ( (T-> data < min ) || (T-> data >= max ) )
return false;

bool p=true,q=true;

if( T-> left !=NULL )
p=CheckBST ( T->left , min, T->data) ;

if( T-> right !=NULL )
q=CheckBST ( T->right , T->data , max ) ;

return p&q ;

}

6 > ATOI()

int atoi( char* str )
{
if( str==NULL )return 0;
int res=0;
bool neg=false;
int i=0;

if( str[0]=='-' )
{
neg=true;
i=1;
}

for( ;str[i]!='\0' ;i++)
res = res*10+ (str[i]-'0');

if( neg )
res=-res ;


return res;

}

bool WildCardMatch ( char* pat , char* text )
{

while ( *text )
{

switch ( * pat )
{
case '?' :
if( *text=='.' )return false;
break;


case '*' :
{
while( * pat== '*' )pat++;

if( !*pat )return true ;

while ( *text )
if( WildCardMatch( pat,text++) )return true ;

return false;

}

default :
if( *pat != *text )return false ;


}

pat++;
text++;

}

while( *pat=='*' )
pat++;

return !(*pat) ;

}

1> Implement the Dictionary using TST .

2> Using BFS we can determine the shortest path between one word to another if we consider all words to be nodes .

12 > Thrashing !

15 >

void Delnode( node* ptr )
{
if( ptr==NULL )return ;

if( ptr->next==NULL ) return ; /* We cannot do anything :-( */

node* t=ptr->next;

copy( ptr,t ); /* Copy data of t into ptr */

ptr->next=t->next;

t->next=NULL;

free(t);

}

15>

void DelNode(node **head, node *del_ptr)
{
   node *temp;
   if (*head == NULL || del_ptr == NULL) return;
   if (*head->next == del_ptr) {
       *head->next = del_ptr->next;
       free(del_ptr);
   }
   temp = *head;
   while(temp->next) {
      if (temp->next == del_ptr) {
         temp->next = del_ptr->next;
         free(del_ptr);
          return;
      }
      temp = temp->next;
   }  
}