Print the desired pattern shown in the example

INPUT-1:

5

OUTPUT-1:

1

2 a

3 b 1

4 c

5

INPUT-2:

9

OUTPUT-2:

1

2 a

3 b 1

4 c 2 a

5 d 3 b 1

6 e 4 c

7 f 5

8 g

9

Java-Code

Compiler

import java.util.*;

public class Hello {

    public static void main(String[] args) {

   //Your Code Here

    Scanner sc=new Scanner(System.in);

    int N=sc.nextInt();

    int ctr=1,flag=1;

    //System.out.printf("%c",ch+3-1);

    for(int row=1;row<=N;row++)

    {   int temp=row;

        char ch=(char)('a'+row-2);

        if(ch>'z')

        ch='a';

        for(int col=1;col<=ctr;col++)

        {

            if(col%2==1)

            {

            System.out.print(temp+" ");

            temp-=2;

            }

            else if(col%2==0)

            {

            System.out.printf("%c ",ch);

            ch-=2;

            }

        }

        if(row-1==N/2)

        flag=0;

        else if(flag==1)

        ctr++;

        if(flag==0)

        ctr--;

        System.out.println();

    }

 }

}

Convert python code to c Example

Example INput and output:

INPUT:

5

1 1 1 1 0

1 1 1 0 0

1 1 1 0 0

1 0 1 1 1

1 1 1 0 1

OUTPUT:

9

INPUT:

2

0 0

0 0

OUTPUT:

0

Java-Code

Compiler

import java.util.*;

public class Hello {

    static int min(int num1,int num2){

        return num1<num2?num1:num2;

    }

    static int max(int num1,int num2){

        return num1>num2?num1:num2;

    }

    public static void main(String[] args) {

   Scanner sc=new Scanner(System.in);

  int n=sc.nextInt();

  int [][]arr=new int[n][n];

  int [][]dp=new int[n][n];

  int result=0;

  for(int i=0;i<n;i++){

      for(int j=0;j<n;j++){

          arr[i][j]=sc.nextInt();

          if(i==0||j==0)

          dp[i][j]=arr[i][j];

          else if(arr[i-1][j]==1&&arr[i][j-1]==1&&arr[i-1][j-1]==1&&arr[i][j]==1)

              dp[i][j]=min(min(dp[i-1][j],dp[i][j-1]),dp[i-1][j-1])+1;

          else if(arr[i][j]==1)

          dp[i][j]=1;

          else

          dp[i][j]=0;

          result=max(result,dp[i][j]);

      }

  }

  for(int i=0;i<n;i++){

      for(int j=0;j<n;j++){

          System.out.print(dp[i][j]+" ");

      }

      System.out.println();

  }

 System.out.print(result*result);

 }

}

Convert python code to c

N=int(input())

numlist=[int(val)for val in input().split()]

newlist=list(map(lambda x:x*2 if x%2==0 else x**2,numlist))

print(*newlist)

C-Code

Compiler

#include<stdio.h>

void main(){

    int n;

    scanf("%d",&n);

    for(int i=0;i<n;i++){

        int num;

        scanf("%d",&num);

        if(num%2)

        printf("%d ",num*num);

        else

        printf("%d ",num*2);

    }

}

The program must accept N integers and an integer k as the input.The program must print the sum of every K consecutive integers as the output.

Example input/output 1:

INPUT:

7

10 20 5 2 -5 66 -69

3

OUTPUT:

35 27 2 63 -8

Example input/output 2:

INPUT:

10

3 7 9 4 1 13 6 2 5 8

4

OUTPUT:

23 21 27 24 22 26 21

JAVA-Code

Compiler

import java.util.*;

public class Hello {

    public static void main(String[] args) {

   Scanner sc=new Scanner(System.in);

  String []str=new String[1000];

  str=sc.nextLine().split(" ");

  int n=sc.nextInt(),k=n-1;

  for(int i=0;i<str.length;i++)

  {

      if(i==k)

      {

         StringBuilder a=new StringBuilder(str[i]);

      System.out.print(a.reverse()+" ");

      k=k+n;

      }

      else

      System.out.print(str[i]+" ");

  }

 }

}

The program must accept N integers and an integer k as the input.The program must print the sum of every K consecutive integers as the output.

Example input/output 1:

INPUT:

7

10 20 5 2 -5 66 -69

3

OUTPUT:

35 27 2 63 -8

Example input/output 2:

INPUT:

10

3 7 9 4 1 13 6 2 5 8

4

OUTPUT:

23 21 27 24 22 26 21

JAVA-Code

Compiler

import java.util.Scanner;

public class Hello {

    public static void main(String[] args) {

  Scanner sc=new Scanner(System.in);

 int n=sc.nextInt();

 int []arr=new int[n];

    for(int i=0;i<n;i++)

    arr[i]=sc.nextInt();

    int size=sc.nextInt();

    for(int i=0;i<=n-size;i++)

    {

        int sum=0;

        for(int j=i;j<i+size;j++)

        sum+=arr[j];

        System.out.print(sum+" ");

    }

 }

}

The program must accept an odd integer as the input and print the desired pattern as output as shown in the example section.

Note: Use _ for printing first line.

Example input/output 1:

INPUT:

3

OUTPUT:

____

\**/

*\/

Example input/output 2:

INPUT:

5

OUTPUT:

______

\****/

*\**/

**\/

JAVA-Code

Compiler

import java.util.*;

public class Hello {

    public static void main(String[] args) {

   Scanner sc=new Scanner(System.in);

  int n=sc.nextInt();

  String str="_";

  System.out.print(str.repeat(n+1)+"\n");

  for(int i=0;i<(n+1)/2;i++)

  {

      for(int j=0;j<n;j++)

      {

          if(j==i)

          System.out.print("\\");

          if(j==n-i-1)

          System.out.print("/");

          if(j>=i&&j<n-1-i)

          System.out.print("*");

          if(j<i)

          System.out.print("*");

      }

      System.out.println();

  }

 }

}

The program must accept the short code of a TV serial as the input.The format of the TV serial short code is "S followed by the season number E followed by the episode number". The program must expand and print the name of the TV serial as shown in the Example input/output section.

Note: The length of the TV serial short code is always 6.

Example input/output 1:

INPUT:

S02E05

OUTPUT:

Season 2, Episode 5

Example input/output 2:

INPUT:

S10E47

OUTPUT:

Season 10, Episode 47

JAVA-Code

Compiler

import java.util.*;

public class Hello {

    public static void main(String[] args) {

  Scanner sc=new Scanner(System.in);

 String s=sc.next();

 for(int i=1;i<s.length();i++)

 {

     if(s.charAt(i)=='E')

     {

 System.out.print("Season "+Integer.parseInt(s.substring(1,i))+", Episode "+Integer.parseInt(s.substring(i+1,s.length())));

     break;

     }

 }

}

}

The program for the flow chart

JAVA-Code

Compiler

import java.util.*;

public class Hello {

    public static void main(String[] args) {

   Scanner sc=new Scanner(System.in);

  int x,y,z,max,isprint=0;

  x=sc.nextInt();

  y=sc.nextInt();

  z=sc.nextInt();

  if(x>y)

  max=x;

  else

  max=y;

  int val=2;

  while(val<=max)

  {

      if(x%val==0&&y%val==0&&z%val!=0)

      {

          isprint=1;

          System.out.print(val+" ");

      }

      val++;

  }

  if(isprint==0)

  System.out.print("-1");

 }

}

The program must accept an R and C and accept the given input and print the desired output as shown int the Example Input/Output.

Example Input/Output 1:

Input:

5 5

1 2 3 4 5

1 2 3 4 5

1 2 3 4 5

1 2 3 4 5

1 2 3 4 5

Output:

1 2 + 4 5

1 2 + 4 5

+ + + + +

1 2 + 4 5

1 2 + 4 5

Example Input/Output 2:

Input:

3 5

1 2 3 4 5

6 7 8 9 10

11 12 13 14 15

Output:

1 2 + 4 5

+ + + + +

11 12 + 14 15

JAVA-Code

Compiler

import java.util.*;

public class Hello {

    public static void main(String[] args) {

  Scanner sc=new Scanner(System.in);

 int r,c;

 r=sc.nextInt();

 c=sc.nextInt();

 for(int row=0;row<r;row++)

 {

     for(int col=0;col<c;col++)

     {

         int num=sc.nextInt();

         System.out.print((row==r/2||col==c/2)?"+ ":num+" ");

     }

     System.out.println();

 }

 }

}

A table called laptop is created with the following DDL command:

CREATE TABLE laptop(id INT,name VARCHAR(100),price INT,processor VARCHAR(10),storage VARCHAR(10),ram VARCHAR(10));

Write the SQL query to select the id,name,price,processor,storage,ram from the table laptop where price is then 40000,the processor is equal to 'Core i5', the storage is equal to '1TB' and the ram is equal to '8GB'.

Note:The id of the laptops must be sorted in ascending order.

SQL-Query

select * from mobile where imei like '%545%' order by id;

The program must accept an even number as the input and print the desired pattern as shown in the example input/output.

Example Input/Output 1:

Input:

6

Output:

1 2 3 6 5 4

* 1 2 4 3

* * 1 2

Example Input/Output 2:

Input:

8

Output:

1 2 3 4 8 7 6 5

* 1 2 3 6 5 4

* * 1 2 4 3

* * * 1 2

C-Code

Compiler

#include<stdio.h>

#include <stdlib.h>

int main()

{

int size;

scanf("%d",&size);

int ctr=size,temp=size,secondhalf=size;

for(int i=0;i<size/2;i++)

{   int first=1;

    for(int j=0;j<temp;j++)

    {

        if(j>=i)

        {   if(j<=(size/2)-1)

            {

            printf("%d ",first);

            first++;

            }

            else

            {

                printf("%d ",secondhalf);

                secondhalf--;

            }

        }

        else

            printf("* ");

    }

    printf("\n");

    ctr-=2;

    secondhalf=ctr;

    temp--;

}

}

complete the getRepeatedString to run successfully.

Example Input/Output 1:

Input:

code

Output:

ccooddee

Example Input/Output 2:

Input:

gmail

Output:

ggmmaaiill

C-Code

Compiler

char* getRepeatedString(char str[])

{

    long int i=0,len=strlen(str);

char *temp=malloc(sizeof(char)*(len*3));

for(long int index=0;index<len;index++)

{

    temp[i++]=str[index];

    temp[i++]=str[index];

}

temp[i]=NULL;

return temp;

}

A table called laptop is created with the following DDL command:

CREATE TABLE laptop(id INT,name VARCHAR(100),price INT,processor VARCHAR(10),storage VARCHAR(10),ram VARCHAR(10));

Write the SQL query to select the id,name,price,processor,storage,ram from the table laptop where price is then 40000,the processor is equal to 'Core i5', the storage is equal to '1TB' and the ram is equal to '8GB'.

Note:The id of the laptops must be sorted in ascending order.

SQL-Query

select * from laptop where price<40000 and processor ='Core i5' and storage ='1TB' and ram='8GB' order by id;

Flow chart

C-Code

Compiler

#include<stdio.h>

#include <stdlib.h>

int main()

{

int x,y;

scanf("%d%d",&x,&y);

while(x<y)

{

    printf("%d%d",x%10,y%10);

    x/=10;

    y/=10;

}

}

The program must accept a string S containing only lower case alphabets as the input.The program must print all the palindromic sub string values of length 3 in the string S as the output.If there is no such palindrome sub string the program must print -1 as the output.

Example Input/Output 1:

Input:

eyelens

Output:

eye

ele

Example Input/Output 2:

Input:

skillrack

Output:

-1

C-Code

Compiler

#include<stdio.h>

#include <stdlib.h>

void print(char str[],int start,int end)

{

    for(;start<=end;start++)

    printf("%c",str[start]);

    printf("\n");

}

int main()

{

char str[100];

int len,ctr=0;

scanf("%s%n",str,&len);

for(int i=0;i<len-2;i++)

{

    if(str[i]==str[i+2])

    {

    print(str,i,i+2);

    ctr=1;

    }

}

(ctr==0)&&(printf("-1"));

}

A table called event is created with the following DDL command:

CREATE TABLE event(id INT,title VARCHAR(100),type VARCHAR(10),attendeescount INT,active BOOLEAN);

Write the SQL query to select the id,title and attendeescount from the table event where active is equal to 'True',type is equal to 'TECH' and attendeescount is greater than 50.

Note:The id of the events must be sorted in ascending order.

SQL-Query

select* from player where odimatches >10 and testmatches >10 and t20imatches >10 order by name;

flow chart

C-Code

Compiler

#include<stdio.h>

#include <stdlib.h>

int main()

{

int x,y;

scanf("%d %d",&x,&y);

int ctr=x;

while(ctr<=y)

{

    if(!(ctr%x))

    ctr++;

    else if(ctr%10==y%10)

    ctr++;

    else

    {

    printf("%d ",ctr);

    ctr++;

    }

}

}

Please convert the following C code to python so that the python program executes succesfully passing the test cases.

#include<stdio.h>

#include<stdlib.h>

int main()

{

int x,y;

scanf("%d%d",&x,&y);

char floatval[11];

double val;

sprintf(floatval,"%d.%d",x,y);

sscanf(flatval,"%lf",&val);

printf("%.2lf",val);

return 0;

python-Code

Compiler

x,y=input().split()

print('%.2f'%float(x+'.'+y))

Please convert the following Python code to C so that the C program executes succesfully passing the test cases.

N=int(input())

numList=list(map(int,input().split()))

numList=sorted(numList,key=lambda num:bin(num).count("1"))

print(*numList);

C-Code

Compiler

#include<stdio.h>

struct pair

{

    int num;

    int count1;

};

int bin(int num)

{

    int count1=0;

    while(num!=0)

    {

        if(num%2)

        count1++;

        num/=2;

    }

    return count1;

}

int* array(struct pair a[],int N,int max)

{

    int*arr=malloc(sizeof(int)*N),count=0,j=0;

    while(count<=max)

    {

        for(int i=0;i<N;i++)

        {

            if(a[i].count1==count)

            arr[j++]=a[i].num;

        }

        count++;

    }

    return arr;

}

int main()

{

    int N,max=0;

    scanf("%d",&N);

    struct pair a[N];

    for(int i=0;i<N;i++)

    {

        scanf("%d",&a[i].num);

        a[i].count1=bin(a[i].num);

        if(a[i].count1>max)

        max=a[i].count1;

    }

    int*newArray=array(a,N,max);

    for(int i=0;i<N;i++)

    printf("%d ",newArray[i]);

}

The program must accept N integers as the input.For each integer X among the N integers,the program must double the integer X.For each odd integers Y among the N integers,the program must print the sum of N integers as the output.

Example Input/Output 1:

Input:

4

5 2 8 7

Output:

68

Explanation:

• 5*4=20
• 2*2=4
• 8*2=16
• 7*4=28

Hence output is 68.

Example Input/Output 2:

Input:

5

4 9 1 1 8

Output:

68

C-Code

Compiler

#include<stdio.h>

#include <stdlib.h>

int main()

{

int N,sum=0;

scanf("%d",&N);

for(int i=0;i<N;i++)

{

    int num;

    scanf("%d",&num);

    if(num&1)

    sum+=(num*4);

    else

    sum+=(num*2);

}

printf("%d",sum);

}

The program must accept N integer and an integer X as the input.The program must print the sum of integers ending with X among the N integers as the output.If there is no such integer the program must print 0 as the output.

Example Input/Output 1:

Input:

5 4

87 54 15 984 104

Output:

1142

Example Input/Output 2:

Input:

4 87

187 787 154 978

Output:

95974

Example Input/Output :

Input:

6 400

6500 98 2 540 12 14

Output:

0

C-Code

Compiler

#include<stdio.h>

#include<math.h>

int main() {

    int N,X;

    scanf("%d %d",&N,&X);

    int sum=0,length=log10(X)+1,divide=pow(10,length);

    for(int i=0;i<N;i++)

    {

        int num;

        scanf("%d",&num);

        if(num%divide==X)

        sum+=num;

    }

    printf("%d",sum);

}

The length of one side S ( in cm) of a rectangle and its perimeter P(in cm) are passed as the input.The program must print the length of another side of the rectangle with the precision up to two decimal places as the output.

Formula:

• Perimeter of a rectangle =2*(length+breath)

Example Input/Output 1:

Input:

4 18

Output:

5.00

Example Input/Output 2:

Input:

3 15

Output:

4.50

C-Code

Compiler

#include<stdio.h>

int main() {

int side,peri;

scanf("%d%d",&side,&peri);

printf("%.2f",(peri/2.0)-side);

}

A table called event is created with the following DDL command:

CREATE TABLE event(id INT,title VARCHAR(100),type VARCHAR(10),attendeescount INT,active BOOLEAN);

Write the SQL query to select the id,title and attendeescount from the table event where active is equal to 'True',type is equal to 'TECH' and attendeescount is greater than 50.

Note:The id of the events must be sorted in ascending order.

SQL-Query

select id,title,attendeescount from event where active='true' and type='TECH' and attendeescount>50 order by id;

The program must accept two string values S1 and S2 are of equal length as the input. The string S1 contains only alphabets and the string S2 contains u or l in both lower case and upper case) For each alphabet CH in the string S2, the program must Toggle the case of the alphabet at the same position in the string S1 based on the following conditions

- If CH is u or U, the program must convert the corresponding alphabet in S1 to upper case

If CH is I or L the program must convert the corresponding alphabet in S1 to lower case

Finally, the program must print the modified string S1 as the output

Example Input/Output 1:

Input

book

ULIU SECE

Output

Book

Explanation:

The first alphabet in S2 is u. So the first alphabet in S1 must be in the upper case.

Now the S1 becomes Book

The second alphabet in S2 is L So the second alphabet in S1 must be in the lower case.

Now the S1 remains the same Book (as o is already in lower case)

The third alphabet in S2 ist. So the third alphabet in S1 must be in the lower case.

Now the S1 remains the same Book (as o is already in lower case)

The fourth alphabet in S2 is U So the fourth alphabet in S1 must be in the upper case

Now the S1 becomes Book

Hence the output is Book

Example Input/Output 2:

G00OOGLE

lululUUL

Output

gOoOoGLe

C-Code

Compiler

#include<stdio.h>

#include <stdlib.h>

int main()

{

char str1[1001],str2[1001];

scanf("%s %s",str1,str2);

for(int i=0;i<strlen(str2);i++)

{

    if(tolower(str2[i])=='u')

    str1[i]=toupper(str1[i]);

    else

    str1[i]=tolower(str1[i]);

}

printf("%s",str1);

}

You must implement the function getStringOfABC(int N) which accepts an integer N as the input.The function must form a string S of length N based on the following condition.

• The string S contains the alphabets a,b and c in circular fashion.
• Then the function must return the string S.

Example Input/Output 1:

Input:

5

Output:

abcab

Example Input/Output 2:

Input:

9

Output:

abcabcabc

C-Code

Compiler

#include<stdio.h>

#include<stdlib.h>

char* getStringOfABC(int N)

{

char *str,a='a';

str=malloc(sizeof(char)*N);

int i=0;

while(i<N)

{

    if(a=='c')

    {

    str[i]=a;

    a='a';

    }

    else if(a!='c'){

        str[i]=a;

        a++;

    }

    i++;

}

str[N]='\0';

return str;

}

int main() { 

    int N; 

    scanf("%d", &N); 

    char *str = getStringOfABC(N); 

    printf(str); 

    free(str); 

    return 0; 

}

The program for the given flow chart.

Java-Code

Compiler

import java.util.*;

public class Hello {

    public static void main(String[] args) {

   Scanner sc=new Scanner(System.in);

  int x=sc.nextInt(),y=sc.nextInt();

  int ctr1=1,ctr2=y;

  while(ctr1<=x||ctr2>=1)

  {

      if(ctr1<=x)

      {

          System.out.print(ctr1+" ");

          ctr1++;

      }

      if(ctr2>=1)

      {

          System.out.print(ctr2+" ");

          ctr2--;

      }

  }

 }

}

The program must accept an integer N as the input and print the desired pattern as shown in the Example

Example Input/Output 1:

Input:

5

Output:

1

2 5

3 4 1

4 3 2 5

5 2 3 4 1

Example Input/Output 2:

Input:

8

Output:

1

2 8

3 7 1

4 6 2 8

5 5 3 7 1

6 4 4 6 2 8

7 3 5 5 3 7 1

8 2 6 4 4 6 2 8

C-Code

Compiler

#include<stdio.h>

#include <stdlib.h>

int main()

{

int n;

scanf("%d",&n);

int arr[n][n],num=1;

for(int col=0;col<n;col++)

{

    for(int row=col;row<n;row++)

    {

        arr[row][col]=num;

        if(col%2)

        --num;

        else

          num++;

    }

    if(col%2)

    num=1;

    else

    num=n;

}

for(int row=0;row<n;row++)

{

    for(int col=0;col<=row;col++)

    {

        printf("%d ",arr[row][col]);

    }

    printf("\n");

}

}

The program must accept a integer N as the input and print the maximum number that can be formed only by means of rotating the number clockwise.

Example Input/Output 1:

Input:

4654

Output:

6544

Example Input/Output 2:

Input:

12458

Output:

81245

C-Code

Compiler

#include<stdio.h>

#include <stdlib.h>

int main()

{

int n,x,c,sum=0;

scanf("%d %d %d",&n,&x,&c);

for(int i=0;i<n;i++)

{

    int num;

    scanf("%d ",&num);

    sum+=(num>x?(num+c):num);

}

printf("%d",sum);

}

The program must accept an integer N as the input and make the integer as maximum possible number by rounding clockwise and print final maximum number output.

Example Input/Output 1:

Input:

4654

Output:

6544

Example Input/Output 2:

Input:

12458

Output:

81245

C-Code

Compiler

#include<stdio.h>

#include<stdlib.h>

#include<string.h>

int main() {

    char str[100];

    int len,max=0;

    scanf("%s%n",str,&len);

    while(len)

    {   char temp=str[0];

        for(int i=0;i<strlen(str)-1;i++)

        str[i]=str[i+1];

        str[strlen(str)-1]=temp;

        int num=atoi(str);

        if(num>max)

        max=num;

        len--;

    }

    printf("%d",max);

}

The program must accept N integers as the input.The program must print the integers which are having the same digit among the N integers as the output.If there is no such integer,the program must print -1 as the output.

Example Input/Output 1:

Input:

4

87 222 9999 1

Output:

222 9999 1

Example Input/Output 2:

Input:

5

887 56566 122 10000 56

Output:

-1

C-Code

Compiler

#include<stdio.h>

#include <stdlib.h>

int main()

{

int n,ctr=0;

scanf("%d\n",&n);

for(int i=0;i<n;i++)

{  int num;

    scanf("%d ",&num);

    int flag=0,check=num%10,temp=num;

    while(temp!=0)

    {

        if(temp%10!=check)

        flag=1;

        temp/=10;

    }

    if(!flag)

    {

    printf("%d ",num);

    ctr=1;

    }

}

if(!ctr)

printf("-1");

}

A table called bus is created with the following DDL command:

CREATE TABLE bus(id INT,busregistrationnum VARCHAR(20),source VARCHAR(30),destination VARCHAR(30),ac BOOLEAN,slpeer BOOLEAN,seater BOOLEAN);

Write the SQL query to select the id,,busregistrationnum from the table bus where ac is equal to 'False',sleeper is equal to 'True' and seater is equal to 'True'.

Note:The id of the bus must be sorted in alphabetical order.

SQL-Query

select id,busregistrationnum from bus where ac='False'and sleeper='True' and seater='True'order by id;

Please convert the following C code to Python so that the Python program executes successfully passing the test cases.

#include<stdio.h>

#include<stdlib.h>

int main()

{

int N;

scanf("%i",&N);

printf("%i",N);

return 0;

}

Python-Code

x=input()

if x[0]=='0':

    if x[1]=='x'or x[1]=='X':

        print('%d'%int(x,16))

    else:

        print('%d'%int(x,8))

else:

    print(x)

You must implement the function hidePassword(char str[]) which accepts a string str as the input.The function must calculate the number of alphabets A in the str and the number of digits D in str.Then the function must modify the string str by replacing each alphabet by *(asterisk) if A is greater than or equal to D.Else the function must modify the string str by replacing each digit by *(asterisk).

Example Input/Output 1:

Input:

s1r2a3ck

Output:

*1*2*3**

Example Input/Output 2:

Input:

123ABCD007

Output:

***ABCD***

C-Code

Compiler

void hidePassword(char str[])

{

    int countdigit=0,countalpha=0;

    for(int i=0;i<strlen(str);i++)

    {

        if(isalpha(str[i]))

        countalpha++;

        else if(isdigit(str[i]))

        countdigit++;

    }

    int j=0;

    if(countalpha>=countdigit)

    {

        j=1;

    }

    for(int i=0;i<strlen(str);i++)

    {

        if(j==1)

        {

            if(isalpha(str[i]))

            str[i]='*';

        }

        else

        {

            if(isdigit(str[i]))

            str[i]='*';

        }

    }

}

The program must accept a integer matrix of size RxC as the input.The program must print the sum of integers in the bottom-right quadrant of the matrix as the output.

Note: R and C are always even integers.

Example Input/Output 1:

Input:

4 4

1 2 3 4

5 6 7 8

9 1 2 3

4 5 6 7

Output:

18

Explanation:

The integer in the bottom -right quadrant of the matrix are 2 3 6 7.So the sum of those integers is 18(2+3+6+7)

Hence the output is 18.

Example Input/Output 2:

Input:

6 2

5 7

4 4

8 3

1 5

9 7

2 1

Output:

13

C-Code

Compiler

#include<stdio.h>

#include <stdlib.h>

int main()

{

int r,c;

scanf("%d %d\n",&r,&c);

int arr[r][c],sum=0;

for(int row=0;row<r;row++)

{

    for(int col=0;col<c;col++)

    {

        scanf("%d ",&arr[row][col]);

        if(row>=r/2&&col>=c/2)

        sum+=arr[row][col];

    }

}

printf("%d",sum);

}

The program must accept N integers as the input.The program must print the sum of digits in the octal representation of N as the output.

Example Input/Output 1:

Input:

12

Output:

5

Explanation:

The octal representation of 12 is 14.

The sum of digits in 14 is 5.

Hence the output is 5.

Example Input/Output 2:

Input:

9

Output:

2

C-Code

Compiler

#include<stdio.h>

#include <stdlib.h>

int sum=0;

void octal(int n)

{

    if(n!=0)

    {

        octal(n/8);

        sum+=n%8;

    }

}

int main()

{

int n;

scanf("%d",&n);

octal(n);

printf("%d",sum);

}

A table called donor is created with the following DDL command:

CREATE TABLE donor(id INT,name VARCHAR(20),bloodgroup VARCHAR(3));

Write the SQL query to select the id,name,bloodgroup from the table donor where bloodgroup is not equal to'AB+'.

Note:The name of the donors must be sorted in alphabetical order.

SQL-Query

select id,name,bloodgroup from donor where bloodgroup !='AB+' order by name;

The program must accept N integers as the input.The program must form a binary representation B of length N by concatenating 1s and 0s alternatively.Then the program must print the decimal equivalent of B as the output.

Example Input/Output 1:

Input:

5

Output:

21

Explanation:

The binary representation of B is 10101(alternative 1s and 0s of length N)

Example Input/Output 2:

Input:

8

Output:

170

Explanation:

The binary representation of B is 10101010(alternative 1s and 0s of length N)

C-Code

Compiler

#include<stdio.h>

#include <stdlib.h>

int main()

{

int num,count1,ctr=1;

long long int sum=0;

scanf("%d",&num);

count1=num/2;

if(num&1)

{

    count1=(num/2)+1;

    ctr=0;

}

while(count1)

{

    sum+=(long long int)(pow(2,ctr));

    ctr+=2;

    count1--;

}

printf("%lld",sum);

}

The program must accept N integers as the input.The program must sum the integers having odd digits.And the final sum is printed as output.

Note:If there is no integer having odd digit print 0 as output

Example Input/Output 1:

Input:

5

13 888 1005 102 8

Output:

998

Example Input/Output 2:

Input:

4

12 4568 32 758475

Output:

0

C-Code

Compiler

#include<stdio.h>

#include <stdlib.h>

int main()

{

int a;

scanf("%d\n",&a);

int num,len,sum=0;

for(int i=0;i<a;i++)

{

    scanf("%d%n\n",&num,&len);

    if(len%2)

        sum+=num;

}

printf("%d",sum);

}

The program must accept a string S as e input.The program mut print the desired pttern as shown in the eexample.

Example Input/Output 1:

Input:

skillrack

Output:

skillrack

*killrac*

**illra**

***llr***

****l****

***llr***

**illra**

*killrac*

skillrack

Example Input/Output 2:

Input:

logics

Output:

logics

*ogic*

**gi**

*ogic*

logics

C-Code

Compiler

#include<stdio.h>

#include <stdlib.h>

int main()

{

char str[1001];

int len;

scanf("%s%n",str,&len);

int start=-1,end=len,ctr=1;

while(1)

{

    for(int index=0;index<len;index++)

    {

        if(index<=start||index>=end)

            printf("*");

        else

            printf("%c",str[index]);

    }

    printf("\n");

    if(ctr)

    {

        start++;

        end--;

    }

    else

    {

        start--;

        end++;

    }

    if((start+3)>=end&&len%2==0)

        ctr=0;

    else if((start+2)>=end&&len%2==1)

        ctr=0;

    if(end-1==len)

        break;

}

}

The program must accept two string values U and P as the input U represents the username of a digital locker and P represents its password.The program must print VALID if the password does not contain any characters from the username U.Else the program must print INVALID as the output.

Example Input/Output 1:

Input:

uit047@demo

srack123

Output:

VALID

Example Input/Output 2:

Input:

AVNG007_bk47

c7sz0ct2o19

Output:

INVALID

C-Code

Compiler

#include<stdio.h>

#include <stdlib.h>

int main()

{

char user[51],pass[21];

scanf("%s\n%s",user,pass);

int flag=0;

for(int i=0;i<pass[i]!='\0';i++)

{

    for(int j=0;j<user[j]!='\0';j++)

    {

        if(pass[i]==user[j])

        {

            flag=1;

            break;

        }

    }

}

if(flag)

    printf("INVALID");

else

    printf("VALID");

}

The program must accept an integer N as the input.The program must split the integer N int three-digit integers(every three consecutive digits).Then the program must print the sum S of those three-digit integers as the output.

Example Input/Output 1:

Input:

210045

Output:

255

Explanation:

• Here N=210045
• The first three-digit integer of 210045 is 210.
• The last three-digit integer of 210045 is 045.
• So their sum 255(210+045) is printed as output.

Example Input/Output 2:

Input:

100100

Output:

200

C-Code

Compiler

#include<stdio.h>

int main() {

    int x;

    scanf("%d",&x);

    int sum=0;

    while(x!=0)

    {

        sum+=x%1000;

        x/=1000;

    }

    printf("%d",sum);

}

The program must accept an integer N as the input.The program must print N characters from the string"abcd" circularly in the clockwise direction as the output.

Example Input/Output 1:

Input:

5

Output:

abcda

Example Input/Output 2:

Input:

10

Output:

abcdabcdab

C-Code

Compiler

#include<stdio.h>

#include <stdlib.h>

int main()

{

int n;

char str[]="abcd";

scanf("%d",&n);

for(int i=0;i<n/4;i++)

  printf("%s",str);

for(int i=0;i<n%4;i++)

    printf("%c",str[i]);

}

The program must accept a time T(in 24hr format HH:MM) and an integer X as the input.The program must update the time T by adding X minutes to it .Then the program must print the updated time T in the same 24hr format as the output.

Example Input/Output 1:

Input:

08:15

Output:

08:35

Example Input/Output 2:

Input:

23:50

15

Output:

00:05

Example Input/Output 3:

Input:

13:45

102

Output:

15:27

C-Code

Compiler

#include<stdio.h>

#include <stdlib.h>

int main()

{

int h,m,x;

scanf("%d:%d",&h,&m);

scanf("\n%d",&x);

int minutes=(h*60)+m+x;

int hour=(minutes/60)%24;

int min=minutes%60;

if(hour<10)

    printf("0%d:",hour);

else

    printf("%d:",hour);

if(min<10)

    printf("0%d",min);

else

    printf("%d",min);

}

The program must accept two integers N and X as the input.The program must print the integers from 1 to N based on the following conditions.

• if X is a positive integer then repeat the first X integers twice among the integers from 1 to N.
• If X is a negative integer then repeat the last X integers twice among the integers from 1 to N.

Example Input/Output 1:

Input:

10 5

Output:

1 1 2 2 3 3 4 4 5 5 6 7 8 9 10

EXPLANATION:

Here X is positive integer.So the first 5 integers are repeated twice from 1 to 10.

Hence the output is 1 1 2 2 3 3 4 4 5 5 6 7 8 9 10

Example Input/Output 2:

Input:

10 -5

Output:

1 2 3 4 5 6 6 7 7 8 8 9 9 10 10

C-Code

Compiler

#include<stdio.h>

#include <stdlib.h>

int main()

{

int n,x;

scanf("%d %d",&n,&x);

for(int i=1;i<=n;i++)

{

    if(x>0&&i<=x)

        printf("%d %d ",i,i);

    else if(x<0&&i>(n+x))

        printf("%d %d ",i,i);

    else

        printf("%d ",i);

}

}

The program must accept three integers X,Y and Z as the input.The program must find and print the smallest positive integer N based on the following conditions:

• N is divisible by Z.
• N is not equal to any of the integers from X to Y(where X<Y).

Example Input/Output 1:

Input:

2 8 2

Output:

10

EXPLANATION:

10 is the smallest integer which is divisible by 2 and it is not equal to any integers from 2 to 8.So 10 is printed as the output.

Example Input/Output 2:

Input:

5 10 4

Output:

4

C-Code

Compiler

#include<stdio.h>

#include <stdlib.h>

int main()

{

int x,y,z;

scanf("%d %d %d",&x,&y,&z);

int n;

if(z<x)

    n=1;

else

    n=y+1;

while(1)

{

    if(n%z==0)

    {

        printf("%d",n);

        break;

    }

    n++;

}

}

There are five bags in a row where each bag contains gold coins.The number of gold coins in each bag is passed as the input.The program must print the maximum number of gold coins by choosing any two consecutive bags as the output.

Example Input/Output 1:

Input:

1 2 3 4 5

Output:

9

Example Input/Output 2:

Input:

9 0 8 4 1

Output:

12

C-Code

Compiler

#include<stdio.h>

#include <stdlib.h>

int main()

{

int max=0,prev=0,pre;

for(int i=0;i<5;i++)

{

    scanf("%d ",&pre);

    ((pre+prev)>max)&&(max=pre+prev);

    prev=pre;

}

printf("%d",max);

}

The program must accept the area of a square A(in cm2) as the input.The program must divide the square int four equal squares.Then the program must print the side(in cm) of any one of the squares with the precision up to two decimal as the output.

Example Input/Output 1:

Input:

16

Output:

2.00

Example Input/Output 2:

Input:

14

Output:

1.87

java-Code

Compiler

import java.util.*;

import java.lang.Math;

public class Hello {

    public static void main(String[] args) {

  Scanner sc=new Scanner(System.in);

 System.out.format("%.2f",Math.sqrt(sc.nextDouble()/4));

 }

}

The program must accept a string S containing only lower case alphabets as the input.For each vowel CH in the string S(from left to right),the program must remove the next character of CH from the string s.Then the program must print the modified string as the output.

Example Input/Output 1:

Input:

skillrack

Output:

skilrak

Example Input/Output 2:

Input:

aerospace

Output:

aropae

C-Code

Compiler

#include<stdio.h>

#include <stdlib.h>

int main()

{

char a[1001];

scanf("%s",a);

for(int i=0;i<strlen(a);i++)

{    char ch=tolower(a[i]);

    printf("%c",a[i]);

    if(ch=='a'||ch=='e'||ch=='i'||ch=='o'||ch=='u')

    {

        i++;

    }

}

}

The program must accept an integer matrix of size RxC as the input.The program must print the mirror image of the matrix as the output.

Example Input/Output 1:

Input:

2 3

1 2 3

4 5 6

Output:

3 2 1

6 5 4

Example Input/Output 2:

Input:

2 2

01 02

03 04

Output:

20 10

40 30

C-Code

Compiler

#include<stdio.h>

#include <stdlib.h>

#include<string.h>

int main()

{

int r,c;

scanf("%d %d\n",&r,&c);

char mat[r][c][100];

for(int row=0;row<r;row++)

{

    for(int col=0;col<c;col++)

    { 

        scanf("%s ",mat[row][col]);

    }

}

for(int row=0;row<r;row++)

{

    for(int col=c-1;col>=0;col--)

    {

        for(int i=strlen(mat[row][col])-1;i>=0;i--)

            printf("%c",mat[row][col][i]);

        printf(" ");

    }

    printf("\n");

}

}

A table called debtor is created with the following DDL command

CREATE TABLE debtor(id INT, accountnum INT,name VARCHAR(20),sanctionedamount INT,currentbalance INT);

Write the SQL query to delete the records of the debtor who have paid off the entire loan amount (i.e, Delete all the debtors who have their currentbalance as 0.)

SQL Query

delete debtor where currentbalance=0;

The program must accept four integers as the input.The program must print YES if the sum of any two integers is equal to the sum of the remaining integers.Else the program must print NO as the output.

Example Input/Output 1:

Input:

2 3 1 4

Output:

YES

Explanation:

The sum of 2 and 3 is 5.

The sum of 1 and 4 is 5.

Here,both results are the same.Hence the output is YES.

Example Input/Output 2:

Input:

3 5 8 2

Output:

NO

C-Code

Compiler

#include<stdio.h>

#include <stdlib.h>

int main()

{

int num1,num2,num3,num4;

scanf("%d %d %d %d",&num1,&num2,&num3,&num4);

if(num1+num2==num3+num4)

    printf("YES");

else if(num1+num3==num2+num4)

    printf("YES");

else if(num1+num4==num2+num3)

    printf("YES");

else

    printf("NO");

}

The program must accept a string S containing only alphabets as the input.The program must print YES if all the vowels in the string S are same case(either upper case or lower case).Else the program must print NO as the output.

Example Input/Output 1:

Input:

SkillRack

Output:

YES

Explanation:

The string SkillRack contains two vowels i and a.Here i and a are in lower case.So YES is printed as the output.

Example Input/Output 2:

Input:

Information

Output:

NO

C-Code

Compiler

#include<stdio.h>

#include <stdlib.h>

int main()

{

char str[1001];

scanf("%s",str);

int scount=0,ccount=0;

for(int i=0;i<strlen(str);i++)

{   char a=str[i];

    if(a=='a'||a=='e'||a=='i'||a=='o'||a=='u')

        scount++;

    else if(a=='A'||a=='E'||a=='I'||a=='O'||a=='U')

        ccount++;

}

if(scount>0&&ccount==0)

    printf("YES");

else if(ccount>0&&scount==0)

    printf("YES");

else

    printf("NO");

}

A table called customer is created with the following DDL command

CREATE TABLE customer(id INT, accountnum INT,name VARCHAR(20),loanamount INT,branchname VARCHAR(30));

Write the SQL query to select the branchname and its sum of loanamount from the table customer.

Note: The total loanamount of the branches must be sorted in descending order.

SQL Query

select branchname,sum(loanamount) as a from customer group by branchname

order by a desc;

The program must accept two characters CH1 and CH2 as the input.The program must print all the ASCII values from the ASCII of CH1 to the ASCII of CH2 as the output.

Example Input/Output 1:

Input:

a e

Output:

97 98 99 100 101

Explanation:

Here CH1=a and CH2=e.

The ASCII value of a is 97 and the ASCII value of e is 101.

Hence the output is 97 98 99 100 101

Example Input/Output 2:

Input:

5 3

Output:

15

C-Code

Compiler

#include<stdio.h>

#include <stdlib.h>

int main()

{

char ch1,ch2;

scanf("%c %c",&ch1,&ch2);

for(;ch1<=ch2;ch1++)

{

    printf("%d ",ch1);

}

}

A table called student is created with the following DDL command

CREATE TABLE student (regnum INT,name VARCHAR(30),arrearscount INT,);

Write the SQL query to list all the arrearscount along with the number of students having that arrearscount from the table student.

Note: The arrearscount must be sorted in descending order.

SQL Query

select arrearscount,count(regnum) from student group by arrearscount order by arrearscount  desc;

A table called player is created with the following DDL command

CREATE TABLE player (id INT,name VARCHAR(30),medalscount INT,country VARCHAR(30));

Write the SQL query to list all the countries along with their total medals from the table player.

Note: The total medals of the countries must be sorted in descending order.

SQL Query

select country,sum(medalscount) as a from player group by country order by a desc;

The program must accept the factor of an integer N(excluding 1 and N)as the input.The program must find the integer N from its factors and print it as the output.

Example Input/Output 1:

Input:

8 2 6 12 4 3

16

Output:

24

Explanation:

The factors of 24 are 2,3,4,6,8 and 12(excluding 1 and 24).Here the given factors are the factors of 24.So 24 is printed as the output.

Example Input/Output 2:

Input:

5 3

Output:

15

C-Code

#include<stdio.h>

#include <stdlib.h>

int main()

{

int a,min,max=0;

while(scanf("%d",&a)>0)

{

    (max==0)&&(min=a);

    (a<min)&&(min=a);

    (a>max)&&(max=a);

}

printf("%d",min*max);

}

A table called employee is created with the following DDL command

CREATE TABLE employee(id INT,name VARCHAR(30),overtime INT,salary INT);

Write the SQL query to update the salary of all the employees by adding the overtime*50 to the salary.

SQL Query

UPDATE employee SET salary=salary+overtime*50;