Answer / rashid akhtar

We cant sort a singly list list in O(n) that is sure.
We can sort an array in O(k+n) if k is order of n,otherwise
counting sort is also not of O(n).
So moral of story,nobody can sort a singly list in O(n)...

Answer / alexht

Counting sort is a solution. Kormen 's book will help you.

Answer / sunil gupta

#include<stdio.h>
#include<conio.h>
#include<malloc.h>
typedef struct node
{
int info;
struct node *next;
}node;
node *temp,*ptr,*t=NULL,*end=NULL;
node *start=NULL;

void main()
{
int n,i,item;
clrscr();
i=0;
printf("enter num of node");
scanf("%d",&n);
while(n>0)
{
temp=(node*)malloc(sizeof(node));
temp->next=NULL;

printf("enter info");
scanf("%d",&item);
temp->info=item;
if(start==NULL)
{
start=temp;
}
else
{
ptr=start;
while(ptr->next!=NULL)
{

ptr=ptr->next;
}
ptr->next=temp;

} //else
n--;
}

ptr=start;
while(ptr!=NULL)
{
printf("%d",ptr->info);
ptr=ptr->next;
}
t=start;
while(t!=NULL)
{
ptr=t->next;
while(ptr!=NULL)
{
if(t->info>ptr->info)
{
temp->info=t->info;
t->info=ptr->info;
ptr->info=temp->info;
}
ptr=ptr->next;
}
t=t->next;
}

ptr=start;
while(ptr!=NULL)
{
printf("\n%d",ptr->info);
ptr=ptr->next;
}
getch();
}

Answer / hakr

Using Radix sort its the best solution it takes O(n).

typedef struct node *ptr;
struct node{
int element;
ptr next;
};
typedef ptr LIST;
typedef ptr POSITION;
int main()
{
.
.
.
.

for(i=0;i<10;i++)
{
p=A[i]->next;
while(A[i]->next!=NULL)
{
A[i]->next=p->next;
p->next=NULL;
digit=(p->element/pow(10,j))%10;
InsertLast(B[digit],p);//InsertLast is a function that
takes an int and position and then does as the name tells//
p=A[i]->next;
}
.
.
.
.
}

Answer / abc

singly linked lists can be sorted in minimum of O(n2) [n
square] time.

Answer / ajaay

struct NODE
{
int nodevalue;
struct NODE *next;
}
struct NODE *uslist, *slist;
struct NODE *cur_us, *cur_s; /*pointers to surf the lists*/
main()
{
..
/*uslist and slist have been initialised here*/
/*uslist holds the list to be sorted*/
/*slist is initially empty as nothing is sorted*/
..
..
LLsort();
}

LLsort ( )
{
int delval;
struct NODE *temp;

slist->nodevalue = 0; /*sorted list will have 0 as its first
element initially*/
slist->next = NULL;
cur_us = uslist;

while( cur_us != NULL)
{
cur_s = slist;
while(cur_s != NULL)
{
if(cur_us->nodevalue < cur_s->nodevalue) /*step-1 of
the algo.*/
{
target = cur_s; /*step-2: make cur_s as your TARGET*/
delval = delnode(cur_us); /*deletion has to be done to
the UNSORTED list*/
addnode(delval, target); /*remember addition has to be
done to the SORTED list*/
}
else
if(cur_s->next == NULL) /*step-3:if cur_s is the last
element in the sorted list*/
{
delval = delnode(cur_us);
temp_us = (struct NODE *)malloc(sizeof(NODE));
temp_us ->nodevalue = delval;
cur_s->next = temp_us; /*step-3:append the
unsorted element to the sorted list*/
temp_us->next = NULL;
}
cur_s = cur_s->next;
}
cur_us = cur_us->next;
}
}

In a gymnastic competition, scoring is based on the average of all scores given by the judges excluding the maximum and minimum scores. Let the user input the number of judges, after that, input the scores from the judges. Output the average score. Note: In case, more than two judges give the same score and it happens that score is the maximum or minimum then just eliminate two scores. For example, if the number of judges is 5 and all of them give 10 points each. Then the maximum and minimum score is 10. So the computation would be 10+10+10, this time. The output should be 10 because 30/3 is 10.

0 Answers   TCS,

void main () { int x = 10; printf ("x = %d, y = %d", x,--x++); } a. 10, 10 b. 10, 9 c. 10, 11 d. none of the above

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void main() { if(~0 == (unsigned int)-1) printf(“You can answer this if you know how values are represented in memory”); }

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1 Answers   Zoho,

struct aaa{ struct aaa *prev; int i; struct aaa *next; }; main() { struct aaa abc,def,ghi,jkl; int x=100; abc.i=0;abc.prev=&jkl; abc.next=&def; def.i=1;def.prev=&abc;def.next=&ghi; ghi.i=2;ghi.prev=&def; ghi.next=&jkl; jkl.i=3;jkl.prev=&ghi;jkl.next=&abc; x=abc.next->next->prev->next->i; printf("%d",x); }

1 Answers  

4. Main() { Int i=3,j=2,c=0,m; m=i&&j||c&I; printf(“%d%d%d%d”,I,j,c,m); }

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What is the main difference between STRUCTURE and UNION?

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what will be the output of this program? void main() { int a[]={5,10,15}; int i=0,num; num=a[++i] + ++i +(++i); printf("%d",num); }

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Given a list of numbers ( fixed list) Now given any other list, how can you efficiently find out if there is any element in the second list that is an element of the first list (fixed list)

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