Ravi Kant Pandey
BAN USERRavi Kant Pandey
Research Engineer,
Centre for Development of Telematics
New Delhi-110030
INDIA.
E-Mail:ravi_cdotd@yahoo.co.in
rpandey@cdotd.ernet.in
- » Resume
Work Experience:
September –2005 to Till date
Working as a Research Engineer in
Intelligent Network –Service Control Point(IN-SCP) Group,
Center for Development of Telematics(C-DOT),
Mandi Road, Mehrauli,
New Delhi -110030
Skill Sets:
Protocols: SS7, ANSI-TCAP
Tools Used: K-1297 Protocol Tester (G-20), SCENIC
Programming Language: C++, C, Pro C
Database: Oracle
Software Packages: Frame Maker, MS Office
Operating System: HP-UNIX, Windows-98/XP
Education :
B. Tech. (Computer Science & Engineering) 2001-2005
Aggregate percentage – 74.2 %
Kamla Nehru Institute of Technology, Sultanpur , U.P.
10+2th
Aggregate percentage – 71.2 %
Uttar Pradesh Board
Khair Inter College, Basti, U.P.
10th
Aggregate percentage – 72.8 %
Uttar Pradesh Board
Khair Inter College, Basti, U.P.
Project Undertaken:
1.Wireless Intelligent Network (WIN) - SCP for BSNL:
1.1 Pre-Paid Charging(PPC) for Wireless Intelligent Network
Overview:
PPC allows the subscriber to pay for voice telecommunication services prior to usage. A PPC subscriber establishes an account with the service provider to access voice telecommunication and SMS services. The PPC Subscriber may be notified about the account information at the beginning during or at the end of the voice telecommunications services. When the account balance is low, the subscriber use of voice telecommunication services may be de-authorized.
Core Features:
? Balance and Validity Enquiry
? Recharging of Account
? Language Preference
? Account Transaction (Balance Transfer, Balance to Validity Conversion and vice versa)
Subscribed Features:
? Call Forwarding on busy
? Call Forwarding on No answer
? Call Forwarding Default
? Do Not Disturb (DND)
? Selective Call Acceptance (SCA)
Team Size: 7
Project Status: Completed, Enhancement is going on.
Platform: HP-UX
Tools: K-1297 Protocol Tester
Language used: C, C++, PRO-C
Database: Oracle
Role: Software Developer
i. Design and coding of Charging for local, intra-circle, inter-circle calls.
ii. Design, Coding and Testing of User Distribution Algorithm.
iii. Design and coding for One India Plan charging.
iv. Design of new tables to support ONE INDIA PLAN.
v. Updating the schema of existing table to support the One India Plan.
vi. Design Coding and Testing of Balance Transfer.
vii. Design Coding and Testing of Balance to Validity Conversion and vice versa.
- 0of 0 votes
AnswersThe function (fopen() and fprintf()) returns EINTR when interrupted by a signal that was caught. We want to test this by sending a signal to interrupt fopen/fprintf when it is called. When should the signal be sent?
- Ravi Kant Pandey| Report Duplicate | Flag | PURGE
IBM Software Engineer / Developer Computer Architecture & Low Level - 0of 0 votes
AnswersThere are 3 societies a, b, and c. A lent tractors to B and C as many as they had. After some time B gave as many tractors to A and C as many as they have. After sometime c did the same thing. At the end of this transaction each one of them had 24.
- Ravi Kant Pandey
Find the tractors each originally had.| Report Duplicate | Flag | PURGE
Infosys Software Engineer / Developer Brain Teasers - 0of 0 votes
AnswersThree friends divided some bullets equally. After all of them shot 4 bullets the total no. of remaining bullets is equal to that of one has after division. Find the original number divided.
- Ravi Kant Pandey| Report Duplicate | Flag | PURGE
Infosys Software Engineer / Developer Math & Computation - 0of 0 votes
Answerstwo link lists starts from head1 and head2 but their some nodes are common as shown below so they end at the same node
- Ravi Kant Pandey
like
a-b-c-d-e-f-
g-h-i
x-y-z-w-
write an algo to find the first commom element of the list ie g.| Report Duplicate | Flag | PURGE
Adobe Software Engineer / Developer Algorithm
Hi,
if u have O(n) space you can do it in O(n) run time
traverse preorder the addresses of the nodes and store it.
then traverse inorder the addresses of the nodes and store it.
eg tree
a
b c
d e f
have PreOrder:a b d e c f
Inorder: d b e a c f
for two node ptrs eg d and c
mark the positions in inorder array and find first preorder ptr which seprates it into two
different parts of Inorder array
for above example first preorder ptr is a which seprates inorder into dbe and cf
so a is first commom ancester of d and c
In the previous post i put whole program
APPEND & JOIN are helping functins.
1.APPEND function append a node to a list.
and sets ptr to next node.
2. JOIN joins two lists.
3.MergeSort function devides the list into two lists recursively and sorts it using Function
merge
NOTE: in linked list we can add any node to a list
without using extra space. so there is no need of
extra spce
so merge sort of linked list can be done in O(nlogn) time without using extra space
1.void linklist :: APPEND(node ** ptr,node **head,node **tail){
if(*head == NULL){
*head = *ptr;
}
else{
(*tail)->next=*ptr;
}
*tail = *ptr;
*ptr=(*ptr)->next;
(*tail)->next = NULL;
}
2.void linklist :: JOIN(node **head_1,node **tail_1,node **head_2,node **tail_2){
if(*head_2 == NULL)
{
*tail_2=*tail_1;
(*tail_2)->next=NULL;
*tail_1=NULL;
}
else if(*head_1 == NULL)
{
*head_1=*head_2;
(*tail_2)->next = NULL;
*head_2=NULL;
}
else
{
(*tail_1)->next =*head_2;
*tail_1=NULL;
(*tail_2)->next = NULL;
*head_2=NULL;
}
}
3.void linklist :: MergeSort(){
if(p == NULL || p->next == NULL)
return;
p=MSort(p);
}
linklist :: node* linklist :: MSort(node* frst){
node* list1Head = NULL;
node* list1Tail = NULL;
node* list2Head = NULL;
node* list2Tail = NULL;
if((frst == NULL) || (frst->next == NULL))
return frst;
while(frst != NULL){
APPEND(&frst,&list1Head,&list1Tail);
if(frst != NULL)
APPEND(&frst,&list2Head,&list2Tail);
}
list1Head = MSort(list1Head);
list2Head = MSort(list2Head);
return merge(list1Head,list2Head);
}
4.
linklist :: node* linklist :: merge(node* list1, node* list2){
node *temp = NULL;
node *t =NULL;
node *pos = NULL;
if(list2==NULL)
return list1;
if(list1->data < list2->data){
temp=list1;
list1=list1->next;
}
else{
temp=list2;
list2=list2->next;
}
pos=temp;
while(list1!=NULL && list2 != NULL){
if(list1->data < list2->data){
t=list1;
list1=list1->next;
}
else{
t=list2;
list2=list2->next;
}
temp->next=t;
temp=t;
}
if(list1==NULL)
temp->next=list2;
else
temp->next=list1;
return pos;
}
It takes only O(n) time.........
- Ravi Kant Pandey March 26, 2007
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to do it in one pass i think the no of elements
- Ravi Kant Pandey March 26, 2007shall be known.
suppose no of elements are n.
then following works well.
#define MAX n
#include<stdio.h>
int main(){
int a[MAX]={0,1,1,1,0,1,0,0,0,1.......};
int i,j,temp;
i=0;
j=9;
while(i<j){
if(a[i]==0)
i++;
else{
while(a[j]==1 && j>i)
j--;
if(j>i){
temp=a[i];
a[i]=a[j];
a[j]=temp;
i++;
j--;
}
}
}
printf("\n\nSorted Data: ");
for(i=0;i<MAX;i++)
printf(" %d",a[i]);
}