Roshan Mangal
BAN USERRoshan Mangal
; Email: roshanmangal@gmail.com
To excel with a progressive organization, which would utilize my technical skills, academic background and experience to achieve higher level assignments in the IT industry.
• 5+ years of experience in Software and Product Development.
• Presently Associated with Nokia India Pvt. Ltd., India
• In depth knowledge of C++; keen interest to learn new technologies and implement it.
• Strong Programming and debugging skills;
• Strong Knowledge of Algorithms and Data Structure;
• Strong ability in cracking complex problems;
• Efficiently work under pressure with an ability to deliver effectively on proposed time.
• Garnered appreciation from PDM; actively participate in presenting ideas and decision for the projects.
• Actively doing process improvements in Team, Made tools in current organization like Calvin (used for Task proposal and releasing activity), LXR (A well known source indexing tool), Jobbari (A tool to assign tasks within team).
• Received “Achieving Together“award in Nokia for being a Team Player.
• Three Months Onsite work experience in Nokia Oyj, Jyvaskyla, Finland.
Languages C++ and Qt(cute)
Operating system Windows, Unix
Database language SQL
Concepts Data Structures and Algorithms.
Version tool Synergy
Company Designation Stint
Nokia India Pvt. Ltd, Bangalore
R&D Software Engineer June 2007 – Present
HP-Invent (STSD), Bangalore
Intern software Engineer December 2006 – May 2007
Key Projects:
LKM Enabler
Company: Nokia India Pvt. ltd.
Duration: June 2010- till date
Team Size: 6-7
Job Responsibility: Development of License key management for content distribution on Nokia’s upcoming products/Platform.
E75, N73, N79, N81, N85, N86, N95, X6, 5800
Company: Nokia India Pvt. ltd.
Duration: June 2007-May 2010
Team Size: 5-8
Job Responsibility: Troubleshoots and debugs software programs for software enhancements and new products. Particular demands include the management of complexity, performance, user experience, efficient use of development tools and techniques, correctness of operation, robustness to failure and flexible adaptation to change.
This project includes timely enhancements of different software component of Device, their configuration and customizations in Product Development cycle.
MP Based Diagnostic Framework for HP Servers:
Company: HP-Invent (STSD).
Team Size: 4
Project Description:
This Project was to diagnose the High end server and give the UI interface to learn about Server Internals.
Secured 1st place and Received certificate of appreciation award for this project from Sitaram Dinkar, (Chief Technology Officer STSD) at HP-University student Conference 2007.
• Graphics Editor
Project Title: R-Paint
Semester: 6th
Team Size: 1
Place: BMSCE, Department of CSE, Bangalore.
Brief Overview of Project
This aim of this project is to develop a 2D graphics package which supports the basic operations which include creating objects like line, circles, polygon, etc and also transformation operations like translation, rotation, etc on such objects.
The project has been developed using procedural system design. It is written in Turbo C for the MS-DOS/windows 16 bit Environment.
• Linux shell
Project Title: Sea Shell
Semester: 5th
Team Size: 2
Place: BMSCE, Department of CSE, Bangalore.
Brief Overview of Project
This project is to design a simple SHELL for Linux for a set of commands. The project has been developed using ‘C’ language and is functional for Linux based platform. SHELL has been designed using standard “C’ library functions.
• Bachelors of Engineering with Computer Science specialization from B.M.S College of Engineering, Bangalore in 2007 with 74%.
• Intermediate Science with major in Mathematics from D. S. College, Katihar from Bihar Intermediate Education council, Bihar with 73%.
• Won second prize in the event ‘Coding and debugging’ held during Genesis 2006, Tecno-cultural fest of ISE Dept of BMSCE.
• Event coordinator of OSP Contest in UTSAV 2006, the national level fest of BMSCE.
• Event coordinator of Code Wars (OSP & Debugging Contest) during INIT7 2005, the intercollegiate technical fest of Dept. of Computer Science, BMSCE.
• Member of Protocol, Computer science club of BMSCE.
• Ranked AIR-380 in K-CET 2003(Non Karnataka Quota).
• Ranked AIR-4400 in IIT 2003.
Hobbies : Listening to music, Solving puzzles like Sudoku, Movies.
Address : 102, CKB Layout, Marathahalli, Bengaluru, Karnataka, India
main(){
printf("10"); //or printf("01");
main();
}
void func()
{
static int a =1;
cout<<a++<<" ";
if(a<=500) func();
}
rand_0_or_1() {
while( (( a=rand5() ) ==5 ) ) {} // if rand5 returns 5 call again
return a & 0x1 ; // return last bit of number so there is equal chance of getting 0 or 1
}
rand7() {
while( (a = 4*rand_0_or_1() + 2*rand_0_or_1() + rand_0_or_1())==0) {} //if it is zero call again
return a;
}
int DivideArray(int *a,int length)
{
int l=0;
int r=length-1;
if(length<2) return -1;
int leftSum = a[l];
int rightSum =a[r];
while (l<r)
{
if(l==r-1)
{
if(leftSum ==rightSum )
return 1;//true
return -1;//false
}
if(abs(leftSum ) < abs(rightSum)) //abs=absolute value
{
l++;
leftSum += a[l];
}
else
{
r--;
rightSum += a[r];
}
}
return -1
}
Always 3 groups,coz you can eliminate 66% in one weigh.
Guys,reply if you have different idea?
Check your code for two lists,which are not intersecting.
Please tell me,for which case,my code will give wrong result?
1)Sort both array (nlogn)
2)now we have A[n] and B[m] two sorted array.
take iA and iB two index variable
lets create a minimum heap minHeap(max size=k);
n+
for each iA+iB=0 Add A[iA]+A[iB](only one elem in heap till now)
for each iA+iB=1 Add A[iA]+A[iB](only 2 more elem in heap added)
.....
do the same thing till you are able to add elemment in k space heap
suppose
for each iA+iB=x Add A[iA]+A[iB](no element found space in k spaced heap)
after this point every element formed will be lesser than found in last iteration
here minimum value of heap will be the kth element.
ps: the idea is to create virtual young's table and iterate thru each diagonal,till one diagonal miss entry in minheap
Complexity of step 2 will be about K(logK).
Total complexity = nlogn +nlogn + klogk =O(nlogn)
Nice solution:
some optimization suggested in your code
int p = i - 1;
int q = i + 1;
replaced by
int p = i - 2;
int q = i + 2;
No! for example {1,3,1,1,3}.
answer will be 2nd and 5th house.3+3 = 6.
I hope this logic might be simple.
Dice has 6 faces,hence total 36 pairs are possible.
A/c to question, 1-12 all should be having equal probability.
hence each sum should come 3 times(36/12).
so we need 12 three times, hence three faces should have 6 in blank dice.
and similarly,we need 1 also three times, so rest 3 faces should have 0.
void printFib(int pCount, int i=0,int j=1)
{
if(pCount){ cout<<i; pcount--;}
if(pCount) { printFib(pCount,j,i+j); }
}
void printFib(int pCount, int i=0,int j=1)
{
if(pCount){ cout<<i; pcount--;}
if(pCount) { printFib(pCount,j,i+j); }
}
int check( struct node * list1,struct node * list2)
{
int count1 = countLL(list1);
int count2 = countLL(list2);
while(count1> count2)
{
list1 = list1->next;
count1--;
}
while(count2>count1)
{
list2 = list2->next;
count2--;
}
while(list1 ! = list2)
{
list1 = list1->next;
list2 = list2->next;
}
if(list1== null)
{
return 0;/ do not intersect
}
else
return 1; //it intersects
}
result = 0;
for each element
{
result ^= element[i];
}
return result;
- Roshan Mangal June 23, 2014