Sanjay Kumar
BAN USEREvery thing is written in resume.
Sanjay Kumar. NIT Jamshedpur(CSE), C Developer
AMDOCS INDIA PRIVATE LIMITED, Pune.
Mail – Id : sanjay.2812@gmail.com
Contact No : 09049457250 / 9570592119
PROFESSIONAL SUMMARY
Around 3 year 0 months of professional IT experience.
Have experience in development of C integrated with Pro-C, UNIX, SQL, PL-SQL, Shell Scripting .
Good knowledge of Software Development Life Cycle
Ability to work in teams and handle wide range of tasks.
ACIEVEMENTS
Appreciated for quick learning, development and handling various responsibilities by the GL, PM, DI and VP.
Also appreciated by the client for in time deliveries with perfection.
TECHNICAL SKILLS
Languages/Packages C, UNIX, SQL, PL-SQL, Shell Scripting, Data Structure, Algorithmes, C++ (Basic)
Operating Systems Windows, Unix, Linux
Databases ORACLE 9
EMPLOYMENT DETAILS
Employer Designation Period
Amdocs Senior Subject Matter Expert Jul 2008 - till date
Academic Qualifications
Year Standard/Degree Board/University Percentage/CGPA
2004-2008 B.Tech. (CSE) N.I.T. Jamshedpur 7.83
2002 10+2 CBSE 80.2 (PCM->91.3)
2000 10th CBSE 74.8
Currently working in Amdocs DVCI, Pune
Project : Amdocs Ensemble AR
Domain : Telecom
Technology Used: C integrated with Pro-C, Cobol, UNIX, SQL, PL-SQL, Shell Scripting
Client : Sprint, Metro, Mobilicity.
JOB Descriptions:
Development and Production Support of AR & Collection (Account Receivable) module of Ensemble(product of Amdocs), for all the three clients (Sprint, Metro, Mobilicity) simultaneously. AR is the core module of product maintain all the details/financial activity on an account and make decision on which account will go for the collation.
Currently worked on Migration of AR module from Cobol to C, and change of platform from HP Unix to Linux. It all involves the effort in coding, debugging and testing of module having around 350 batch JOB's.
NIT Jamshedpur Project.
Underwent a Vocational Training of one month at B.S.N.L. Jamshedpur Telecom District in following units:-
GSM Mobile, Broadband , CDMA-WLL, OFC, Fixed Line Switching, PCM System.
PERSONAL INFORMATION
Date of Birth 15th Dec 1983
Gender Male
Marital Status Single
Passport No G5475322
Languages Known English, Hindi
Contact No. 09049457250
Email Id sanjay.2812@gmail.com
Present Address Cosmos, Block H, Flat no: 302 ,Magarpatta City, Pune, MH
Declaration: I hereby declare that all the relevant information mentioned here is true to best of my knowledge and belief.
Date:………….. Sanjay Kumar
Place:………….
- 0of 0 votes
AnswersIf A and B, two integers are given.
- Sanjay Kumar in India
compute A/B.
Ex. 2/5 --> Ans should be 0.4
225/1000 --> Ans should be 0.225
22/7 --> Ans Should be 3.(142857) where 142857 are repeating decimal| Report Duplicate | Flag | PURGE
Amazon Software Engineer / Developer Coding - 0of 0 votes
Answersthere is n node graph, each node having only one root. All nodes are labeled in a random order from int 0 to n-1
the graph is represented in a array format such that the value in the array at index equal to child node label is root node label.
For root assume the value is -1.
Ex.3 4 1 2 0 5
the array is
- Sanjay Kumar in India
value : 1, 3, 3, -1, 3, 1
Index : 0, 1, 2, 3, 4, 5
Find the hight of graph. I have given an answer in O(n^2)
follow up question, if you dont have any restiction on space reduced the time complexity.| Report Duplicate | Flag | PURGE
Amazon Systems Design Engineer Data Structures - 0of 0 votes
AnswersSuppose you have given a 10 digit mobile number (9903457235). Each digit in the number have 3 letters in mobile keypad. find/print all words which can be created by corresponding letter against digit given.. Total 3^10 words, I need complete recursive function.
- Sanjay Kumar in India| Report Duplicate | Flag | PURGE
Microsoft Software Engineer / Developer Arrays - 0of 0 votes
AnswersFind the number of nodes in a singly link list, it can be both cyclic and acyclic
- Sanjay Kumar in India| Report Duplicate | Flag | PURGE
Amazon Algorithm
what about 1, 3, 5 whose sum is 9 and can be divided in 3 and 1,5 :) check with example man.
- Sanjay Kumar January 18, 2012can you please check correctly ..
My algo will give output as
19 5 3 1
15 4 2
which is also correct... :) there may be more then one solution and mine algo is giving one of them...
okay one question why you took P+4 why not P+3 that you need to explain right.. you need to take the combination of P and P+2 to explain this, only P+2 not signify that P+1 is divisible by 3.
- Sanjay Kumar January 17, 2012I have manually tested it.. Might be not full proof ..
Sort the array in decreasing order..
1. take the first number and add to one diff array.. and second number add to 2nd array.. avg of 1st will 1st number avg of 2nd will be 2nd number.
2. take the next element from original array and add to array having greater average and keep track of new avg and number of element added to that array for further calculation.
3. Do the step 2 untill the last element of original array..
4. If possibility exist you will find the arrays having equal average.
This will fail with this array... 10 5 7 6 2 9 3
the sets are possible like 5 7 6 and 10 2 9 3 avg to 6
But I don't think your code is handling this...
I don't understand... If we will not have such array then..any algo can't grantee the division of array as expected. for example : 1, 3, 4, 15. how can you divide this array.
- Sanjay Kumar January 17, 2012If I am understanding the question correctly
The main point is find out the time t1 having average value as MAX(Sell Point) and t2 having average value as MIN(Buy Point).
P is prime that mean P is not divisible by 2. that gives P+1 is divisible by 2.
P and P+2 is prime that mean P+3 is not divisible by 3 which will makes P divisible by 3 but P is prime. that's why P+4 divisible by 3 as P+2, P+3 are not divisible by 3. that means P+1 divisible by 3.
That mean P+1 divisible by both 2 and 3 that is 6
which proof P+1+6 is divisible by 6 i.e. P+7 divisible by 6.
this is more complicated then this... you need to get profiling to the user. you can guess to complete the search sentence using trie and profile + use the type of keyword you were guessing like if software then add download or something like that, which has maximum search with that keyword... many thing also need to be consider like location and more..
- Sanjay Kumar January 15, 2012The above solution is wrong
Check out this..
6
2 9
1 7 3 11
The above tree will satisfy all your check but is not a BST.
Inorder will shorted array..
edition in first line
"use a queue and inqueue the node with level information... transverse the tree using the node which we dqueue from queue(not tree)...
thanks nmc for explaining and justifying my answer..
- Sanjay Kumar January 02, 2012Fine this is a program which replace all the fragment from a string.
so this can be used as function also which accept the original string
and fragment string and return the string without the fragment .
Rest you can find if you execute the code.
Please don't mind the example
- Sanjay Kumar January 02, 2012I think this is one function can be used..
# include <conio.h>
# include <stdio.h>
#include <string.h>
int main ()
{
char s1[]="sfufuckckanfuckjayfufuckck kumarffuckuck fuck", s2[]="fuck";
char *s3; int i,len;
s3=strstr(s1,s2);
while (1)
{
if(s3==NULL)
{
break;
}
strncpy(s3,s3+strlen(s2),strlen(s3+strlen(s2)));
len=strlen(s1)-strlen(s2);
for(i=strlen(s1);i>=len;i--)
s1[i]='\0';
s3=strstr(s1,s2);
}
printf("%s",s1);
getch();
return 0;
}
try o do it without reversing the linklist
- Sanjay Kumar December 29, 2011#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <conio.h>
char * test(int v)
{
char *buf;
buf=(char *)malloc(7);
memset(buf,' ',6);
switch(v)
{
case 1:
strncpy(buf, "Case 1", strlen(buf));
break;
case 2:
strncpy(buf,"Case 2", strlen(buf));
break;
case 3:
strncpy(buf, "Case 3", strlen(buf));
break;
default:
strncpy(buf, "defult", strlen(buf));
}
return buf;
}
int main()
{
char *p = test(2);
printf("%s",p);
getch();
}
use malloc
- Sanjay Kumar December 29, 2011tell me what will be the ans of 195+15 i think it will be 210
- Sanjay Kumar December 28, 2011you can use extern or define the array allocated at runtime as pointer and return the address....
- Sanjay Kumar December 28, 2011I don't know who is this.. but I completely agree with this solution..
You are returning the max sum and node address.
Kundan I need second idea on my algo also.. Please comment.
- Sanjay Kumar December 27, 2011this is fine did you handled 8888889 and 111
- Sanjay Kumar December 27, 2011let me try this...
start parsing both the link list one node at a time.. to find out which is the largest one and what is the diff in the length..
suppose list1 1->2->3->4
and list2 5->3->4->6->9->2->1
given example L1=4 L2=7 and diff=3
create a function like below..
it can be passed to function
Node* add_node(Node *N1,Node *N2,size)
{
static int carry;
int sum,remainder;
Node *temp_node;
if(N1==NULL || N2==NULL)
return NULL;
if(size<=diff)
{
size++;
temp_node=(Node*)malloc(sideof(Node*));
temp_node->next=add_node(N1,N2->next,size);
sum=N2->value+carry;
}
else
{
size++;
temp_node=(Node*)malloc(sideof(Node*));
temp_node->next=add_node(N1->next,N2->next,size);
sum=N1->value+N2->value+carry;
}
carry=sum/10;
remainder=sum%10;
temp_node->value=remainder;
return temp_node;
}
tere might more exception handling and we need to take care.. but this is my over all idea..
Above one replace only the first occurance.. this one replace all..
# include <conio.h>
# include <stdio.h>
#include <string.h>
int main ()
{
char s1[]="sanjay kumar", s2[]="a";
char *s3; int i,len;
s3=strstr(s1,s2);
while (1)
{
if(s3==NULL)
{
break;
}
strncpy(s3,s3+strlen(s2),strlen(s3+strlen(s2)));
len=strlen(s1)-strlen(s2);
for(i=strlen(s1);i>=len;i--)
s1[i]='\0';
s3=strstr(s1,s2);
}
printf("%s",s1);
getch();
return 0;
}
what about this... ?
# include <conio.h>
# include <stdio.h>
#include <string.h>
int main ()
{
char s1[]="sanjay kumar", s2[]="y k";
char *s3; int i,len;
s3=strstr(s1,s2);
strncpy(s3,s3+strlen(s2),strlen(s3+strlen(s2)));
len=strlen(s1)-strlen(s2);
for(i=strlen(s1);i>=len;i--)
s1[i]='\0';
printf("%s",s1);
getch();
}
BFS...
Push all the element in a queue with level details..
This is my logic..
#include <stdio.h>
#include <conio.h>
int main()
{
long long number=7654932810;
long long i, k, j=10000, MAX=0, NUM=0, temp, temp1;
for(i=0;i<7;i++)
{
NUM=(number>>i);
for(k=0;k<i;k++)
{
NUM=NUM/5;
}
temp=(NUM/j);
temp1= NUM - temp*j;
if(MAX<temp1)
MAX=temp1;
}
printf("MAX=%lld",MAX);
getch();
}
And in the question its given its a number not a string.
- Sanjay Kumar December 24, 2011Use Bit wise shifting and division at the same time to get the 4 digit number in a loop and find the MAX one..
10 digit shift 0 bit divided by 1000000 --> First for digit number
10 Digit shift 1 bit divided by 100000 --> 2nd to 5th digit number
10 Digit shift 2 bit divided by 10000 --> 3nd to 6th digit number
and so on ..
Compare every number to MAX and if greater then MAX assign it to MAX.
I have one more Doubt.. you are getting the input from command prompt that mean it's string. If it's a numaric variable, suppose getting output from some other function then it will not work.
- Sanjay Kumar December 24, 2011I had given the same answer in MS interview.
- Sanjay Kumar December 24, 2011that's my solution... :)
- Sanjay Kumar October 15, 2011
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@Hruday.. I agree..
- Sanjay Kumar January 19, 2012@eugene.yarovoi my algo will work fine with your example...