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Which is the best mvc training institute in noida?

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What is the difference between Interface and abstract classes? Give practical example from experience.

A group of friends are tracking the miles per gallon for each of their cars. Each time one of them fills up their gas tank, they record the following in a file: His or her name The type of car they drove How many miles driven since they last filled up How many gallons purchased at this fill up Date of the fill Their data is formatted as a comma separate value (csv) file with the following format for each row:(#person,carName,milesDriven,gallonsFilled,fillupDate) Miles are recorded as floating-point numbers and gallons as integers. Please create a program that allows members of this group to determine the miles per gallon (MPG) of each of their cars during a specific time range. Note: person may have more than one so a time range query might need to output data for one or more cars. A skeleton class will be provided; your job will be to complete the program. The principal function for querying MPG is of the form (the exact name, data types, etc., can be learned by inspecting the "solution" class in the skeleton code): GetRangeMPG(PersonName, StartDate, EndDate) Returns list of objects containing (CarName, MPG) MPG is calculated as (total miles traveled during time period)/ (total gallons filled during time period. The dates you receive in the query should be treated inclusively.

Given as list of movies in a csv file, extract the movie name and genre and be able to return a query based on the years and the genre.

Basically, Parsing a CSV input file and build an indexed data storage to allow search the data in time manner.

All Sim characters are loaded with the same copy of your software. Imagine two of these sims are positioned somewhere on a line that extends to infinity in both directions, left and right. Using the below interface, write a C# method called "Collide" that ensures that the two sims will collide at some point in the future. Remember, the code that you write will be deployed to each Sim. You cannot write code that will be deployed to one and not the other. The method should have no parameters. You do not need to implement the ISim interface.

public interface ISim
{
void MoveLeft(); // Moves the sim character one unit on the line to the left
void MoveRight(); // Moves the sim character one unit on the line to the right
void Relax(); // Commands the sim character to not do anything
void MarkPosition(); // Puts a mark on the line at the current position of the sim character
bool IsCurrentPositionMarked(); // Inspects the current position of the sim character for a mark and returns true if one is found, false if one is not found.
}

A number is special if it is possible to remove some digits from it to get a number having 3, 5 or 6 only.
For example, 38597 is special since it is possible to remove digits 8, 9, 7 to get 35. You cannot remove all the digits.
You can remove digits from left, right or middle.
Write a program in C which given a number N, calculate how many divisors of N are special

If two strings are identical Stringcmp() returns
A. -1 B. 1 C. 0 D. Yes

So recently to someone I know a question was ask how do you rotate a 2 dimensional array? He was asked to white board it. He could get diagram the problem but had trouble coding it.
This is not an easy answer because equal dimensional array are easy but require you know the trip that you divide the x and y dimension and then do four swapping rotation around the corners. Not equal X,Y are much harder. When I heard this question, I could get the equal dimensions done in an hour or two in a compiler. It took me two days to get the total program working. Here is the code:
{
#include <stdio.h>

int main()
{
#define xsize 4
#define ysize 8
int matrix[ysize*xsize] ;

typedef enum rotation
{
CW0,
CW90,
CW180,
CW270,
CCW0,
CCW90,
CCW180,
CCW270,
DONE0
} RotationTypes;

RotationTypes DesiredRotation = CW270;

#define min(a,b) ( (a) < (b) ? a : b)

int x,y,r,swap0, swap1,rotx,roty, rotidx;

//walk through every possibily
for (DesiredRotation = CW0; DesiredRotation <= DONE0 ; DesiredRotation++)
{

for (y=0;y<ysize;y++)
{
for (x=0;x<xsize;x++)
{

matrix[(y*xsize)+x] = y + x*10;

//print the base rotation for references
if ( DesiredRotation == CW0
|| DesiredRotation == CCW0
|| DesiredRotation == DONE0)
printf("%d ", matrix[(y*xsize)+x]);
}
if ( DesiredRotation == CW0
|| DesiredRotation == CCW0
|| DesiredRotation == DONE0)
printf("\n");

}

if ( DesiredRotation == CW0
|| DesiredRotation == CCW0
|| DesiredRotation == DONE0)
printf("\n");

//flips are generally more effeicient than rotation
//bu doing flips you only have to do one 90 degree
// type of rotation. flips work on uneven dimeions
if (DesiredRotation != CCW90 && DesiredRotation != CW270)
{
//horiztonal flip
for (y=0;y<ysize;y++)
{
for (x=0; x<xsize/2;x++)
{
swap0 = matrix[(y*xsize)+x];
matrix[(y*xsize)+x] = matrix[(y*xsize)+((xsize-1)-x)] ;
matrix[(y*xsize)+((xsize-1)-x)] = swap0;

}
}

//vertical flip
for (y=0;y<ysize/2;y++)
{
for (x=0; x<xsize;x++)
{
swap0 = matrix[(y*xsize)+x];
matrix[(y*xsize)+x] = matrix[(((ysize-1)-y)*xsize)+x] ;
matrix[(((ysize-1)-y)*xsize)+x] = swap0;

}
}
}



if (DesiredRotation != CW0
&& DesiredRotation != CW180
&& DesiredRotation != CCW0
&& DesiredRotation != CCW180
&& DesiredRotation != DONE0)
{

//do one CCW90 rotation which is the same as CW270
//if would be more effiecient to hard code every
//different combination but no way to make it elegant

int sqsize=min(ysize,xsize);

//first do the inplace square part because it is very efficient
for (y=0;y<sqsize/2;y++)
{
for (x=0;x<(sqsize+1)/2;x++)
{


rotx = x;
roty = y;
//store the top, left
int swap0 = matrix[(y*xsize)+x];

//rotate into top left from top right
matrix[(y*xsize)+x] = matrix[(x*xsize)+((sqsize-1)-y)];

//rotate into top right from bottom right
matrix[(x*xsize)+((sqsize-1)-y)] = matrix[(((sqsize-1)-y)*xsize)+((sqsize-1)-x)];

//rotate into bottom right from bottom left
matrix[(((sqsize-1)-y)*xsize)+((sqsize-1)-x)] = matrix[(((sqsize-1)-x)*xsize)+y];

//into into bottom left from original top left
matrix[(((sqsize-1)-x)*xsize)+y] = swap0;

}
}


//now the challenge is what to do with unequal array left overs
//we chose to hardcode inserting rotation and swapping until done

//we have to do two different sets of code if height is
//greater than width
if (ysize>xsize)
{
int rotinc=0;


int baseptr = sqsize*sqsize;
int rowsize = xsize;

for (x=xsize-1;x>=0;x--)
{
for (y=sqsize;y<ysize;y++)
{
//find where the extra non square starts after
//the sqaure part of the arrary
int rotptr = baseptr+((y-sqsize)*rowsize)+x;

//store it for later insertion
swap0 = matrix[rotptr];

//find where in the new dimensions
//the rotation from the end of the square should
//go
int swapptr = (((xsize-1)-x)*ysize)+y;

//swap upswards to make room for the
//insertion
while (rotptr>swapptr)
{
matrix[rotptr] = matrix[rotptr-1];

rotptr--;

}

//insert the rotation at the proper place
//which is now free
matrix[swapptr] = swap0;


}
//we have done one row so increase where the
//base pointer will be for the next time
baseptr+=ysize-sqsize;

//because we did one rotation the remaining rows
//are one rotation smaller
rowsize--;

}

}


//width is bigger than height
if (xsize>ysize)
{


int rotinc=0;


int rowsize = xsize;



for (x=sqsize;x<xsize;x++)
{

//insert happen at the begging of the array for
//this sizing
int swapptr = 0;

for (y=0;y<ysize;y++)
{
//find the rotation at the end of each row
int rotptr = rotinc+(y*rowsize)+sqsize;

//store it for latter use
swap0 = matrix[rotptr];



//swap upwards to made room for the insertion
while (rotptr>swapptr)
{
matrix[rotptr] = matrix[rotptr-1];

rotptr--;

}
//issert the rotation at the begining of the
//array
matrix[swapptr] = swap0;


//increasement the point
swapptr++;



}

//we have done one row so need to increase
//our start posiiton
rotinc+=ysize;

//we have done one row. the remain is one row smaller
rowsize--;

}

}


//print out the rotated and maybe flipped new array
for (x=0;x<xsize;x++)
{
for (y=0;y<ysize;y++)
{


printf("%d ", matrix[(x*ysize)+y]);
}

printf("\n");

}


} else {

//printout the vertical flipped only array
if (DesiredRotation != CW0
&& DesiredRotation != CCW0
&& DesiredRotation != DONE0)
{
for (y=0;y<ysize;y++)
{
for (x=0;x<xsize;x++)
{


printf("%d ", matrix[(y*xsize)+x]);
}

printf("\n");

}
}


}

printf("\n");

}




return 0;
}

}

How two compare two database using entity framework.

Description : Develop an application that transfer data from the test database to live database after showing comparison and user selection.

Precondition: database and schema information provided and list of tables.

Condition Of Acceptance:
1. Application should select the source and destination databases and be capable of showing comparisons.
2. It should be capable transferring data from source to destination database/deleting the selected data from destination.
Technologies: C#, Sql server, entity framework, object oriented design principle.

Dear Friends,
please help to get the logic for the below , basically it is atoi implimentation question.

input : string as 0x1234
output : in int formate 0x1234

if input: 0x12Ab43
output should be 0x12Ab43

if input : 0xdY6
outpust : should give an error , since Y is out of scope.

Testing Strings
Mr X and Mr Y, his friend are programmers and testers respectively working in the same company. So, once they faced the following scenario :

Mr X wrote an application that took as input some user data. The data the application took as input was a string in some strange language. That language consisted of
K
K distinct letters. However, due to some issues, the application got corrupted and one particular String among many was lost.

However, Mr X had seen that String once before it got lost. He remembers some info about it. Particu-larly, he remembers the lost String had length equal to
M
M.

Mr Y, being the chief QA person in his company needs to try and figure out the number of possible different possible candidate Strings for the lost String.

Mr X remembers N pieces of info. For each one, he gives you 2 numbers
L
L and
R
R and a number
Z
Z. He remembers for sure that the
Z
t
h
Zth letter of the language of the string did not occur between positions
L
L and
R
R inclusive of the lost String.

Input Format :

The first line contains
3
3 space separated integers
N
N ,
M
M and
K
K.

Each of the next
N
N contains
3
3 space separated integers, denoting
L
L ,
R
R and
Z
Z respectively.

Output Format :

You need to find and print the number of different possible candidate Strings for the lost String based on the info Mr X remembers. As the number of such Strings can be large, print it Modulo
10
6
+
3
106+3
Constraints:

1
≤
N
,
M
,
K
≤
10
5
1≤N,M,K≤105

1
≤
L
≤
R
≤
M
1≤L≤R≤M
1
≤
Z
≤
K
1≤Z≤K
Sample Input
1 2 26
1 2 1
Sample Output
625

Today Jin has given a task to Shino, Shino has to travel from cell
(
1
,
1
)
(1,1) to cell
(
N
,
M
)
(N,M) in a grid of size
N
∗
M
N∗M. But in order make this task interesting for Shino, Jin has decided to keep some special candies in some
K
K special cells of the grid, where each candy has an amount of happiness associated with it.

Shino can travel only in right & down direction in the grid, as he is too careful & does not want to fall out of grid. Now, we call the value of a path the happiness of all cells lying on the path. All non-special cells have happiness equal to
0
0.

Now, you need to find and print the sum of the values of all paths from
(
1
,
1
)
(1,1) to
(
N
,
M
)
(N,M), traveling only right and down to an adjacent cell.

As Shino is not good at counting help him find the answer.

Input Format

The first Line contains a single integer
T
T denoting number of test cases

The next line contains
3
3 space separated integers,
N
N,
M
M and
K
K where N * M is the size of grid & K denoting number of special cells

The next
K
K lines contains three integers
X
i
,
Y
i
,
H
i
Xi,Yi,Hi where
(
X
i
,
Y
i
)
(Xi,Yi) is cell coordinate &
H
i
Hi is the amount of happiness Shino will get from a candy at cell
(
X
i
,
Y
i
)
(Xi,Yi)

Constraints

1
≤
T
≤
3
1≤T≤3
1
≤
N
,
M
,
K
≤
10
5
1≤N,M,K≤105

1
≤
X
i
≤
N
,
1
≤
Y
i
≤
M
1≤Xi≤N,1≤Yi≤M

1
≤
H
i
≤
10
5
1≤Hi≤105

Output Format

For each test case you will output a single integer denoting the total amount of happiness Shino will get. As the answer can be quiet large you can output answer modulo
10
9
+
7
109+7
Sample Input
1
2 2 2
1 2 4
2 1 7
Sample Output
11

The input consists of N cases(equal to about 10000). The first line of the input contains only positive integer N. Then follow the cases. Each case consists of exactly one line with two positive integers separated by space. These are the numbers whose reverse you have to add

The input consists of N cases(equal to about 10000). The first line of the input contains only positive integer N. Then follow the cases. Each case consists of exactly one line with two positive integers separated by space. These are the numbers whose reverse you have to add

What are two possible ways of passing by reference in C# language ?

Follow up : (Answer to above is "ref" and "out").
Which one would you use and why ?