sahil.bathla1
BAN USER#include <iostream>
using namespace std;
struct list
{
char *value;
list * next;
}*first=NULL;
void JoinList(int A[],int size)
{
char *tempString = new char(15);
int i=0;list *currentNode;
while(i<size)
{
list *temp=new list;
temp>value=new char(15);
strcpy(temp>value,"");
temp>next=NULL;
do
{
itoa(A[i],tempString,10); //convert array element to string
strcat(temp>value,tempString); //concatenate to current node
i++;
if(i>=size)
break;
}while(A[i]==(A[i1]+1)); // till the next element is one greater than the current element
if(first==NULL)
{
first=temp;
currentNode=temp;
}
else
{
currentNode>next=temp;
currentNode=temp;
}
}
}
void display(list * node)
{
while(node!=NULL)
{cout<<node>value<<" ";
node=node>next;
}
}
int main()
{
int A[12]= {2,3,4,5,6,9,10,12,13};
JoinList(A,9);
display(first);
system("pause");
return 0;
}
Steps :
1) convert current array element to string & add to list
2) read & concatenate further elements provided the next element = previous element + 1
3) repeat 1 & 2 to form the linked list till you run out of array elements
Just see the comments in Join List Function. You will understand
wat do you mean by containing one sequence??
 sahil.bathla1 November 15, 2012your logic would work only if the tree is a perfect binary tree. it will fail otherwise... suppose root has only right subtree with one children i.e  tree=> a c
answer should be
c  2 0
a  1 1
but as per your solution it would be
c  1 0
a  0 1
iterative code
list * alternateSwap(list * node)
{
list *previous_joint,*new_joint,*temp;
previous_joint=new list;
while (node!=NULL&&node>next!=NULL)
{
temp=node>next>next;
node>next>next=node; //swap step 1 //break & update next nodes link
if(node==list_head)
list_head=node>next;
node=node>next; //swap step 2 update node
new_joint=node>next; // saving joining point
if(node!=list_head)
{
previous_joint>next=node; //joining previous joint with new swapped node
previous_joint=new_joint;
}
else
previous_joint=new_joint; //saving first joining point
node=temp;
} //end of while
previous_joint>next=node;
return list_head;
}

sahil.bathla1
November 13, 2012 list * alternateSwap(list * node)
{
if (node!=NULL&&node>next!=NULL)
{
list *temp= new list;
temp>next=node>next>next; //swap step 1 //save the next nodes link;
node>next>next=node; //swap step 2 //break & update next nodes link
node=node>next; //swap step 3 update node
node>next>next=alternateSwap(temp>next); //find next link of swapped node
delete temp;
}
return node;
}
example 1 > 2> 3
swap step 1 : save 2>next=3
swap step 2 : 2>next =1
swap step 3 : 2= node or head
step 4 : 2>next>next=alternateSwap(3) => 1>next=alternateSwap(3) = 3;
therefore final result = 2>1>3;
Using ashishanand's logic i made this code.Works fine :
#include <iostream>
using namespace std;
/*
Harcoded Today's Given Date 27/08/2012
Day Thursday
*/
class date
{
int dd,mm,yyyy;
public:
date(int d,int m,int y)
{dd=d;mm=m;yyyy=y;}
void compute_day();
};
void date :: compute_day()
{
int count_year=0,count_days=0;
int year=yyyy,month=mm,d=dd;
cout<<"Date is "<<d<<"/"<<month<<"/"<<year<<"\n";
while(year != 2012)
{
if(year>2012) // if year is greater compute odd days
{
if(year%400==0(year%4==0&&year%100!=0))
count_year+=2;
else
count_year++;
year=1;
}
else // if year is less compute odd days
{
if(year%400==0(year%4==0&&year%100!=0))
count_year=2;
else
count_year;
year+=1;
}
}
while(month != 8)
{
if(month>8)
{
if(month==1month==3month==5month==7month==8month==10month==12) // if month is greater compute odd days
count_days+=3;
else if(month==2)
;
else
count_days+=2;
month;
}
else
{
if(month==1month==3month==5month==7month==8month==10month==12)
count_days=3;
else if(month==2)
;
else
count_days=2;
month++;
}
}
if(d>27) // if days are greater compute odd days
{ count_days+=(d27);}
else
{ count_days=(27d);}
int gap=(count_days+count_year)%7;
if(gap<0) gap+=7; //mathematical modulo
if(gap==0)
cout<<"Thurday";
else if(gap==1)
cout<<"Friday";
else if(gap==2)
cout<<"Saturday";
else if(gap==3)
cout<<"Sunday";
else if(gap==4)
cout<<"Monday";
else if(gap==5)
cout<<"Tuesday";
else if(gap==6)
cout<<"Wednesday";
else ;
}
int main()
{
date toCheck(26,10,2016); // change it to any date you want to check
toCheck.compute_day();
system("pause");
return 0;
}

sahil.bathla1
November 13, 2012 Well the code does this :
Step 1 It keeps the track of the minimum element index all the time..
Step 2 Checks if the condition holds and keeps storing them in another array B till it holds
Step3 When condition fail the current B array is displayed and then step 1 and 2 are repeated starting from (minimum+1) index because all the elements before that index are irrelevant now
Step 4 if all the elements are traversed exit and display again
As you can see the logic is simple ... keep track of the minimum element. keep going till the condition holds and if it fails the minimum element and all the elements before it are irrelevant now because the condition would fail if we include them. so restart from minimum + 1 instead.
#include <iostream>
using namespace std;
void convert(int a)
{
if(a!=0)
{
int n=a%26;
if(n!=0)
{
convert(a/26);
cout<<(char)(96+n);
}
else
{convert((a/26)1);
cout<<"z";
}
}
}
int main()
{
int a;
cout<<"Enter number";
cin>>a;
convert(a);
system("pause");
return 0;
}

sahil.bathla1
November 12, 2012 exactly the solution i thought. Any way to improve upon it?
 sahil.bathla1 November 12, 2012i made a code to work it out .. complexity is surely less than O(n^2) . Please prove my code wrong because i haven't used either min heap or dequeue to solve it . :)
#include <iostream>
using namespace std;
void display(int B[],int n)
{
if(n>1)
{
for(int i=0;i<n;i++)
cout<<B[i]<<" ";
}
cout<<"\n";
}
int main()
{
int A[12]= {1,2,3,4,5,5,1,2,3,4,5,5};
int B[10];
int k,i=1;
int check=1;
int min=0,count=1,flag=0;
cout<<"Enter k\n";
cin>>k;
B[0]=A[0];
while(i<12) // 12 is harcoded .. take n instead
{
if(A[min]+A[i]>k) //to check if the condition satisfies
{
if(i>=check) flag=1; //minimum index to counter(to exclude any minimal subset)
if(A[i]<A[min]) //update min position
min=i;
B[count]=A[i];
count++; //total count in subarray
i++;}
else
{ if(flag==1) //if displayable ( new elements added)
{display(B,count);flag=0;check=i;} // it should not be displayed till the index i is rediscovered again
{min=min+1;B[0]=A[min];i=min+1;count=1;}
}
}
if(flag==1)
display(B,count);
system("pause");
return 0;
}
if(onParachute)
goto label execute_code
execute_code :
static int count;
count++;
if(count<3) // on landing each rover visit the label therefore count =2
{
while(1)
{
go left;
no operation; //to make the rover slow
}
}
else if(count ==3) // one rover has reached others parachute
{
while(1)
go left;
}
else ; //count can never be 4
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can you give the test case for which it failed
 sahil.bathla1 November 16, 2012