Example

INPUT:

5

OUTPUT:

3

Explanation:

• binary representation of 5 is 101.
• removing ist bit it becomes 01 it's equivalent decimal value is 1
• removing ist bit it becomes 11 it's equivalent decimal value is 3
• removing ist bit it becomes 10 it's equivalent decimal value is 2

hence maximum among three integer is 3.Hence 3 is printed as output.

JAVA-code

Compiler

import java.util.*;

public class Hello {

    public static void main(String[] args) {

   Scanner sc=new Scanner(System.in);

  String str=Integer.toBinaryString(sc.nextInt());

  int max=0;

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

      StringBuilder s=new StringBuilder(str);

      s.deleteCharAt(i);

      int num=Integer.parseInt(s.toString(),2);

      if(num>max)

      max=num;

  }

  System.out.print(max);

 }

}

Print the desired pattern as shown below

Example

INPUT:

5233487

OUTPUT:

5433827

INPUT:

1002

OUTPUT:

1020

INPUT:

400352

OUTPUT:

2354

JAVA-code

Compiler

import java.util.*;

public class Hello {

    public static void main(String[] args) {

  Scanner sc=new Scanner(System.in);

 StringBuilder str=new StringBuilder(sc.next());

 int len=str.length(),k=0,flag=1,index=0;

 char temp='0';

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

 {

     if((str.charAt(i)-'0')%2==0&&flag==1)

     {

         temp=str.charAt(i);

         index=i;

         flag=0;

     }

     else if((str.charAt(i)-'0')%2==0&&flag==0)

     {

       char temp1=str.charAt(i);

         str.setCharAt(index,temp1);

         index=i;

     }

 }

 str.setCharAt(index,temp);

    System.out.print(Integer.parseInt(str.toString()));

}

}

Print the desired pattern as shown below

Example

INPUT:

5 5

OUTPUT:

* * * * *

* 0 * 0 *

* * * * *

* 0 * 0 *

* * * * *

C-code

Compiler

import java.util.*;

public class Hello {

    public static void main(String[] args) {

   Scanner sc=new Scanner(System.in);

  int x=sc.nextInt();

  int y=sc.nextInt();

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

  {

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

      {

          if(i%2==0||j%2==0)

          System.out.print("* ");

          else

          System.out.print("0 ");

      }

      System.out.println();

  }

 }

}

if x is odd print main diagonal else print opposite diagonal.

Example

INPUT:

4

77 48 51 82

13 53 76 68

64 71 45 53

87 87 64 24

45

OUTPUT:

77 53 45 24

C-code

Compiler

#include<stdio.h>

#include <stdlib.h>

int main()

{

int n;

scanf("%d",&n);

int mat[n][n];

int x;

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

{

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

    {

        scanf("%d ",&mat[i][j]);

    }

}

scanf("%d",&x);

int xrow,xcol;

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

{

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

    {

        if(mat[row][col]==x)

        {

        xrow=row;

        xcol=col;

        break;

        }

    }

}

if(x%2==0)

{

while(xrow>0&&xcol<n-1)

{

    xrow--;

    xcol++;

}

printf("%d %d\n",xrow,xcol);

while(xrow<n&&xcol>=0)

{

    printf("%d ",mat[xrow][xcol]);

    xrow++;

    xcol--;

}

}

else

{

    while(xrow>0&&xcol>0)

{

    xrow--;

    xcol--;

}

printf("%d %d\n",xrow,xcol);

while(xrow<n&&xcol<n)

{

    printf("%d ",mat[xrow][xcol]);

    xrow++;

    xcol++;

}

}

}

C-code

Compiler

#include<stdio.h>

#include <stdlib.h>

int main()

{

int n;

scanf("%d",&n);

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

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

{

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

    {

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

    }

}

int temp=n-1;

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

{

    for(int col=n/2;col<n;col++)

    {

        if(col<temp)

        printf("* ");

        else

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

    }

    printf("\n");

    if(row==n/2)

    {

        flag=0;

    }

    if(flag)

    temp--;

    else

    temp++;

}

}

Java-code

Compiler

import java.util.Scanner;

public class MyClass {

    static int count=0;

    static void fun(int num)

    {

        count+=Math.abs(num)/2;

        if(Math.abs(num)%2==1)

        count++;

    }

    public static void main(String args[]) {

        Scanner sc=new Scanner(System.in);

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

    if(Math.abs(x-y)>=5)

    {

        count=Math.abs(x-y)/5;

        if(Math.abs(x-y)%5>0)

        fun(Math.abs(x-y)%5);

    }

    else if(Math.abs(x-y)>=2)

    {

        fun(Math.abs(x-y));

    }

    else if(Math.abs(x-y)==1)

    count++;

    System.out.print(count);

    }

}

The program must accept an integer N as the input.The program must print the maximum possible sum S of integers which are having the same unit digit among the N integers.

Example Input/Output 1:

Input

5

60 69 39 98 20

Output

108

Explanation:

• The sum of integers which are having 9 as unit digit is 108(9+39).
• The sum of integers which are having 0 as unit digit is 80(60+20).
• The sum of integers which are having 8 as unit digit is 98.

Maximum sum is 108 hence printed as the output.

Example input/Output 2:

Input:

8

15 98 67 95 18 93 48 97

Output:

164

C-code

Compiler

#include<stdio.h>

#include <stdlib.h>

int main()

{

int n;

scanf("%d",&n);

int max=0,arr[n],fre[10]={0};

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

{

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

    fre[arr[i]%10]=1;

}

for(int i=0;i<10;i++)

{   int sum=0;

    if(fre[i]==1)

    {

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

        {

            if(i==arr[j]%10)

            sum+=arr[j];

        }

        if(sum>max)

        max=sum;

    }

}

printf("%d",max);

}

The program must accept an integer N as the input.The program must toggle the last occurring 1 in the binary representation of N.The the program must print the modified value of N as the output.

Example Input/Output 1:

Input

10

Output

8

Explanation:

• The binary representation of 10 is 1010.
• The last occurring 1 in 1010 is toggled to 0.So the binary representation becomes 1000.
• Now the decimal equivalent of 1000 is 8.Hence the output is 8.

Example input/Output 2:

Input:

16

Output:

0

JAVA-code

Compiler

import java.util.*;

public class Hello {

    public static void main(String[] args) {

  Scanner sc=new Scanner(System.in);

 StringBuilder str=new StringBuilder(Integer.toBinaryString(sc.nextInt()));

 for(int i=str.length()-1;i>=0;i--)

 {

     if(str.charAt(i)=='1')

     {

     str.setCharAt(i,'0');

     break;

     }

 }

 System.out.print(Integer.parseInt(str.toString(),2));

 }

}

The program must accept an two integer as the input and print the number of common digits in it.

Example Input/Output 1:

Input

22

22222

Output

2

Example input/Output 2:

Input:

5465

47865

Output:

3

JAVA-code

Compiler

#include<stdio.h>

#include <stdlib.h>

int main()

{

int fre1[10]={0},fre2[10]={0};

int a,b;

scanf("%d%d",&a,&b);

while(a!=0)

{

    fre1[a%10]++;

    a/=10;

}

while(b!=0)

{

    fre2[b%10]++;

    b/=10;

}

int count=0;

for(int i=0;i<=9;i++)

{

    if(fre1[i]<=fre2[i]&&fre1[i]!=0)

   count=count+fre1[i];

   else if(fre1[i]>=fre2[i]&&fre1[i]!=0)

   count+=fre2[i];

}

printf("%d",count);

}

The program must accept an integer matrix of size RxC as the input.The program must print the number of rows containing the integers in ascending order in the matrix

Example Input/Output 1:

Input

4 5

4 6 7 8 19

1 3 5 7 8

3 5 6 12 6

5 7 8 9 13

Output

3

Example input/Output 2:

Input:

3 3

37 68 79

65 52 31

56 79 25

Output:

1

JAVA-code

Compiler

import java.util.*;

public class Hello {

    public static void main(String[] args) {

  Scanner sc=new Scanner(System.in);

 int r,c,count=0;

 r=sc.nextInt();

 c=sc.nextInt();

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

 {   int flag=0,temp=0;

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

     {

         int num=sc.nextInt();

         if(col!=0&&temp>num)

         {

             flag=1;

         }

         else{

             temp=num;

         }

     }

     if(flag==0)

     count++;

 }

 System.out.println(count);

 }

}

The program must accept N integers as the input and print the desired pattern as shown below

Example Input/Output 1:

Input

5

Output

%---%

---%-

--%--

-%---

%---%

Example input/Output 2:

Input:

7

Output:

%-----%

-----%-

----%--

---%---

--%----

-%-----

%-----%

C-code

Compiler

#include<stdio.h>

#include <stdlib.h>

int main()

{

int n;

scanf("%d",&n);

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

{

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

    {

        if(i==0||i==n-1)

            printf((j==0||j==n-1)?"%%":"-");

        else

           printf((j==n-1-i)?"%%":"-");

    }

    printf("\n");

}

}

The program must accept N integers as the input.The program must remove the first occurrence of 0 in each integer among the N integers.Then the program must print the sum of the N modified integers as the output.

Example Input/Output 1:

Input

3

100 320 10020

Output

1062

Explanation:

• 100 becomes 10
• 320 becomes 32
• 10020 becomes 1020
• The sum of 10 +32+1020=1062.

Example input/Output 2:

Input:

4

87 157 10 13

Output:

258

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++)

{   char str[100],tempstr[100];

    scanf("%s",str);

    int ctr=1,index=0;

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

    {

        if(str[j]=='0'&&ctr)

        {

            ctr--;

        }

        else

        {

          tempstr[index++]=str[j];  

        }

    }

    tempstr[index]='\0';

    sum=sum+atoi(tempstr);

}

printf("%d",sum);

}

The program must accept a N*N matrix as the input The program must modify the matrix based on the following conditions in the given order.

-The top-right quadrant of the matrix is replaced by the bottom-left quadrant.

-The bottom-right quadrant of the matrix is replaced by the top-left quadrant.

Finally the program must print the modified matrix as the output.

Example Input/Output 1:

Input

4

25 28 84 71

15 30 45 87

69 78 99 10

11 54 68 100

Output

25 28 69 78

15 30 11 54

69 78 25 28

11 54 15 30

Example input/Output 2:

Input:

6

25 28 84 71 88 77

15 30 45 87 22 33

69 78 99 10 27 18

11 54 68 100 19 08

69 78 99 10 27 18

10 20 30 40 50 60

Output:

25 28 84 11 54 68

15 30 45 69 78 99

69 78 99 10 20 30

11 54 68 25 28 84

69 78 99 15 30 45

10 20 30 69 78 99

C-code

Compiler

#include<stdio.h>

#include <stdlib.h>

int main()

{

int n;

scanf("%d",&n);

int mat[n][n];

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

{

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

    {

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

    }

}

int a=0;

for(int row=n/2;row<n;row++)

{   int b=0;

    for(int col=n/2;col<n;col++)

    {

        mat[row][col]=mat[a][b];

        b++;

    }

    a++;

}

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

{   int b=0;

    for(int col=n/2;col<n;col++)

    {

        mat[row][col]=mat[a][b];

        b++;

    }

    a++;

}

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

{

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

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

    printf("\n");

}

}

The program must accept a character matrix of size RxC containing only lower case alphabets as the input.The program must print the first occurring vowel from the top in each column of the matrix.

Note:

Example Input/Output 1:

Input

524 1200

Output

1724

Explanation:

• The first digit of 524 is 5.
• The last occurring nonzero digit of 1200 is 2.
• After swapping those two digits the integers become 224 and 1500.
• So their sum 1724 is printed as the output.

Example input/Output 2:

Input:

1000 2000

Output:

3000

C-code

Compiler

#include<stdio.h>

#include <stdlib.h>

#include<string.h>

int main()

{

char num1[100],num2[100];

scanf("%s %s",num1,num2);

for(int i=strlen(num2)-1;i>=0;i--)

{

    if(num2[i]!='0')

    {

        char temp=num1[0];

        num1[0]=num2[i];

        num2[i]=temp;

        break;

    }

}

printf("%d",atoi(num1)+atoi(num2));

}

The program must accept two integers X and Y as the input.The program must swap the first digit of X and the last occurring nonzero digit of Y.Then the program must print the sum of X and Y as the output.

Example Input/Output 1:

Input

524 1200

Output

1724

Explanation:

• The first digit of 524 is 5.
• The last occurring nonzero digit of 1200 is 2.
• After swapping those two digits the integers become 224 and 1500.
• So their sum 1724 is printed as the output.

Example input/Output 2:

Input:

1000 2000

Output:

3000

C-code

Compiler

#include<stdio.h>

#include <stdlib.h>

#include<string.h>

int main()

{

char num1[100],num2[100];

scanf("%s %s",num1,num2);

for(int i=strlen(num2)-1;i>=0;i--)

{

    if(num2[i]!='0')

    {

        char temp=num1[0];

        num1[0]=num2[i];

        num2[i]=temp;

        break;

    }

}

printf("%d",atoi(num1)+atoi(num2));

}

Print the desired pattern shown in he example

Example Input/Output 1:

Input

3

1 2 3

4 5 6

7 8 9

Output

1

2 4

3 5 7

6 8

9

Example input/Output 2:

Input:

5

1 2 3 4 5

6 7 8 9 10

11 12 13 14 15

16 17 18 19 20

21 22 23 24 25

Output:

1

2 6

3 7 11

4 8 12 16

5 9 13 17 21

10 14 18 22

15 19 23

20 24

25

C-code

Compiler

#include<stdio.h>

int main() {

    int n;

    scanf("%d",&n);

    int mat[n][n];

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

    {

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

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

    }

    int ctr=0;

    while(1)

    {

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

        {

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

        {

            if(row+col==ctr)

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

        }

        }

        printf("\n");

        ctr++;

        if(ctr==n+n-1)

        break;

    }

}

Mr.ABC has M boxes and N keys.Each box contains a certain number of gold coins and each key has a number written on it.The number of gold coins in ecah box and the number written on each key are passed as the input to the program.He wants to open as many boxes as possible based on the following conditions.

• If the number of gold coins in a box is even,the key with the odd number is the correct key to open the box.
• If the number of gold coins in a box is odd,the key with the even number is the correct key to open the box.
• A key cannot be used to open more than one box.

The program must print the maximum number of boxes he can open based on the given conditions.

Example Input/Output 1:

Input

4 3

8 2 1 7

2 9 5

Output

3

Explanation:

• The first box 8 gold coins.The possible keys to open the box are 9 and 5.here we can use the key 9.
• The second box 2 gold coins.The possible keys to open the box are 9 and 5,but 9 is already used so 5 is used to open.
• The third box 1 gold coins.The only possible keys is 2.
• There is no key to open the rest of the box .Hence 3 is printed as the output.

Example input/Output 2:

Input:

3 4

10 8 1

20 30 40 50

Output:

1

C-code

Compiler

#include<stdio.h>

#include <stdlib.h>

int main()

{

int b,k;

scanf("%d %d",&b,&k);

int box[b],key[k];

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

scanf("%d",&box[i]);

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

scanf("%d",&key[i]);

int count=0;

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

{

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

    {

        if((box[j]%2==0&&box[j]!=-1)&&key[i]%2==1)

        {

            count++;

            box[j]=-1;

            break;

        }

        else if((box[j]%2==1&&box[j]!=-1)&&key[i]%2==0)

        {

            count++;

            box[j]=-1;

            break;

        }

    }

}

printf("%d",count);

}

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

Example Input/Output 1:

Input

12345

Output

**1**

*2*2*

34543

Example input/Output 2:

Input:

skillrack

Output:

****s****

***k*k***

**i***i**

*l*****l*

lrackcarl

Example input/Output 3:

Input:

abcdefg

Output:

***a***

**b*b**

*c***c*

defgfed

C-code

Compiler

#include<stdio.h>

#include <stdlib.h>

int main()

{

char str[100];

int len;

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

int ctr=len/2,flag=1;

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

{   

    int index=i;

    if(ctr-i==-1)

    break;

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

    {

        if(ctr-i==0)

        {

            printf("%c",(flag)?str[index++]:str[index--]);

            if(index==len-1)

            flag=0;

        }

        else

             printf("%c",(j==ctr-i||j==ctr+i)?str[i]:'*');

    }

    printf("\n");

}

}

The program must accept a two lines of N integers each as the input.The program must print the odd integers from both the lines based on their order of occurrence as the output.if both lines having odd integers in the same position then print the odd integer from the first line followed by the odd integer from the second line

Example Input/Output 1:

Input

5

11 10 12 13 77

45 44 44 43 10

Output

11 45 13 43 77

Example input/Output 2:

Input:

4

3 1 4 8

6 2 7 9

Output:

3 1 7 9

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();

 int arr1[]=new int[n],arr2[]=new int[n];

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

 arr1[i]=sc.nextInt();

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

 arr2[i]=sc.nextInt();

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

 {

     if(arr1[i]%2==1)

     System.out.print(arr1[i]+" ");

     if(arr2[i]%2==1)

     System.out.print(arr2[i]+" ");

 }

 }

}

The program must accept a integer N as the input and print YES if all the digits in the integer are in ascending order else print NO.

Example Input/Output 1:

Input

25689

Output

YES

Example input/Output 2:

Input:

54789

Output:

NO

C-code

Compiler

#include<stdio.h>

int main() {

    int R,C,count=0;

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

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

    {

        int flag=0;

        char ch;

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

        {

            char str[2];

            scanf("%s",str);

            if(col==0)

            {

                ch=str[0];

                continue;

            }

            else if(str[0]<ch)

            {

                flag=1;

            }

            ch=str[0];

        }

        if(!flag)

        count++;

    }

    printf("%d",count);

}

The program must accept a integer N as the input and print YES if all the digits in the integer are in ascending order else print NO.

Example Input/Output 1:

Input

25689

Output

YES

Example input/Output 2:

Input:

54789

Output:

NO

C-code

Compiler

#include<stdio.h>

#include <stdlib.h>

int main()

{

int arr[8],index=0,flag=0;

while(scanf("%1d",&arr[index])>0)

{

    if(index!=0)

    {

        if(arr[index]<arr[index-1])

        flag=1;

    }

    index++;

}

printf(flag?"NO":"YES");

}

The program must accept a string S as the input.The program must print the desired pattern as shown in the Example Input/Output section

Example Input/Output 1:

Input

skillrack

Output

s********

ks*******

ski******

liks*****

skill****

rlliks***

skillra**

carlliks*

skillrack

Example input/Output 2:

Input:

malayalam

Output:

m********

am*******

mal******

alam*****

malay****

ayalam***

malayal**

alayalam*

malayalam

C-code

Compiler

#include<stdio.h>

#include <stdlib.h>

int main()

{

char str[101];

int len;

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

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

{

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

    {

        if(j<=i)

            printf("%c",(i%2==0)?str[j]:str[i-j]);

        else

           printf("*");

    }

    printf("\n");

}

}

A boy has M 1 rupee coins and N X rupee coins.He needs to pay a bill of K rupees .The values of X,M,N and K are passed as input to the program.The program must print YES if it is possible to pay exactly K rupees from the M 1 rupee coins and the N X rupee coins.Else the program must print NO as the output.

Example Input/Output 1:

Input

2 6 5 13

Output

YES

Explanation :

There are six 1 rupee coins and five 2 rupee coins.From that the sum of five 2 rupee coins (5*2=10) and three 1 rupee coins (3*1=3) is 13.So YES is printed as output.

Example input/Output 2:

Input:

3 1 8 14

Output:

NO

C-code

Compiler

#include<stdio.h>

#include <stdlib.h>

int main()

{

int X,M,N,K;

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

int check=K%N;

if(((X*N)+M)<K)

{

    printf("NO");

    return 0;

}

printf(M>=check?"YES":"NO");

}

The program must accept a string S containing only lower case alphabets and an integer Kas the input. The program must print YES if the first K English alphabets (lower case) are present at least once in the string S. Else the program must print NO as the output.

Example Input/Output 1:

Input

afehijklmnopzqrcstuvdwxgyb

5

Output

YES

Explanation :

The first 5 english alphabets (lower case) are a b c d and e which are present at least once in the string afehijklmnopzqrcstuvdwxgyb

So YES is printed.

Example input/Output 2:

Input:

bdhajrgst

3

Output:

NO

C-code

Compiler

#include<stdio.h>

#include <stdlib.h>

int main()

{

char str[101];

int len,x,fre[26]={0},flag=1;

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

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

fre[str[i]-'a']=1;

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

{

    if(fre[i]!=1)

    flag=0;

}

printf(flag==1?"YES":"NO");

}

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

Example Input/Output 1:

Input:

7 3 4 5

Output:

****

****

****

*-**

*--*

*---

*---

Example Input/Output 2:

Input:

5 6 1 4 2

Output:

*****

**-**

**-*-

**-*-

**---

-*---

Example Input/Output 3:

Input:

1 2 3 4 5 6

Output:

******

-*****

--****

---***

----**

-----*

C-code

Compiler

#include<stdio.h>

int main() {

   int index=0,max=0;

   int arr[100];

   while(scanf("%d",&arr[index])>0)

   {

       if(arr[index]>max)

       max=arr[index];

       index++;

   }

   char pattern[index];

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

   {

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

       {

           if(i==arr[j])

           pattern[j]='-';

           else if(pattern[j]!='-')

           pattern[j]='*';

       }

       pattern[index]='\0';

       printf("%s\n",pattern);

   }

}

The program must accept a string S with space as the input.The program convert the alphabets to uppercase in between the square brackets [] .And finally print the modified string as the output.

Example Input/Output 1:

Input:

[as[df]er]

Output:

[as[DF]er]

Example Input/Output 2:

Input:

[robert [was] a good [king]

Output:

[robert [WAS] a good [KING]

Example Input/Output 3:

Input:

uhuu [ntu e]tuhnt[uheo

Output:

uhuu [NTU E]tuhnt[uheo

C-code

Compiler

#include<stdio.h>

#include<string.h>

int main() {

    char str[1001];

    int len;

    scanf("%[^\n]%n",str,&len);

    int i=0;

    while(i<len)

    {

        int flag=0,j=i+1;

        if(str[i]=='[')

        {

            while(j<len)

            {

                if(str[j]==']')

                {

                    flag=1;

                    break;

                }

                else if(str[j]=='[')

                {

                    flag=0;

                    break;

                }

                j++;

            }

        }

        if(flag==1)

        {

            for(int index=i+1;index<j;index++)

            str[index]=toupper(str[index]);

            i=j;

        }

        else

        {

            i++;

        }

    }

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

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

}

The program must accept a string S as the input.The program must print the desired pattern as shown in the example Input/Output section.

Example Input/Output 1:

Input:

water

Output:

w * * * a

* t * e *

* * r * *

* t * e *

w * * * a

Example Input/Output 2:

Input:

OFFICE

Output:

O * * * * F

* F * * I *

* * C E * *

* * C E * *

* F * * I *

O * * * * F

C-code

Compiler

#include<stdio.h>

#include <stdlib.h>

int main()

{

char str[100];

int len,ctr=0;

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

char arr[len][len];

int end=(len%2)?(len/2)+1:len/2;

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

{

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

    {

        if(col==row||col==len-row-1)

        {

        arr[row][col]=str[ctr];

        ctr++;

        }

        else

        arr[row][col]='*';

    }

}

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

{

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

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

    printf("\n");

}

int end1=(len%2)?end-2:end-1;

for(int row=end1;row>=0;row--)

{

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

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

    printf("\n");

}

}

The program must accept a string S as the input.The program must print the desired pattern as shown in the example Input/Output section.

Example Input/Output 1:

Input:

skillrack

Output:

skillrack

kkillrack

iiillrack

lllllrack

lllllrack

rrrrrrack

aaaaaaack

cccccccck

kkkkkkkkk

Example Input/Output 2:

Input:

huge

Output:

huge

uuge

ggge

eeee

C-code

Compiler

#include<stdio.h>

#include <stdlib.h>

int main()

{

char str[101];

scanf("%s",str);

int n=strlen(str);

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

{

    int count=i;

    while(count--)

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

    printf("%s",&str[i]);

    printf("\n");

}

}

The program must accept an N integers as the input the program must swap the first occurred odd integers and second odd integer and so on and print the modified integers as the output.If the last integer is odd it should not be swapped.

Example Input/Output 1:

Input:

7

22 23 51 56 69 53 29

Output:

22 51 23 56 53 69 29

Example Input/Output 2:

Input:

7

22 24 51 56 68 52 29

Output:

22 24 51 56 68 52 29

C-code

Compiler

#include<stdio.h>

#include <stdlib.h>

int main()

{

int n;

scanf("%d",&n);

int arr[n],first=-1,ctr=1;

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

{

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

    if(ctr&&arr[i]%2)

    {

    first=i;

    ctr=0;

    }

    else if(i==n-1)

    {

        continue;

    }

    else if(arr[i]%2&&first!=-1)

    {

        int temp=arr[first];

        arr[first]=arr[i];

        arr[i]=temp;

        first=-1;

        ctr=1;

    }

}

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

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

}

The program must accept a string S as the input and pint the sub string from last to first.

Example Input/Output 1:

Input:

Brick

Output:

kckickrickBrick

Example Input/Output 2:

Input:

SkillRack

Output:

kckackRacklRackllRackillRackkillRackSkillRack

C-code

Compiler

#include<stdio.h>

#include <stdlib.h>

int main()

{

char str[101];

scanf("%s",str);

int len=strlen(str);

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

{

    for(int j=len-i-1;j<len;j++)

    {

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

    }

}

}

The program must accept N integers as the input.The program must form a binary representation B by concatenating the last two bits of each integer among the N integers(left to right).Then the program must print the decimal equivalent of B as the output.

Example Input/Output 1:

Input:

4

7 5 8 2

Output:

210

Explanation:

• The last two bits int the binary representation of 7 is 11.
• The last two bits int the binary representation of 5 is 01.
• The last two bits int the binary representation of 8 is 00.
• The last two bits int the binary representation of 2 is 10.
• So the obtained binary representation B is 11010010.
• The decimal equivalent of 11010010 is 210.
• Hence 210 is printed as output.

Example Input/Output 2:

Input:

3

12 10 32

Output:

8

C-code

Compiler

#include<stdio.h>

#include <stdlib.h>

int main()

{

int n;

scanf("%d",&n);

int bin[1000],index=0;

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

{   int num;

    scanf("%d ",&num);

    int ctr=0;

    while(ctr<2)

    {

       bin[index]=num%2;

       num/=2;

       index++;

       ctr++;

    }

    int temp=bin[index-1];

    bin[index-1]=bin[index-2];

    bin[index-2]=temp;

}

long long sum=0,count=0;

for(int i=index-1;i>=0;i--)

{

    if(bin[i]==1)

    {

    sum+=(long long)pow(2,count);

    }

    count++;

}

printf("%lld",sum);

}

The program must accept accept the string S containing only lower case alphabets as the input.The program must print Yes if all the consonants are present in the string S(in any order).Else the program must print No as the output.

Example Input/Output 1:

Input:

abcdefghijklmnopqrstuvwxyz

Output:

YES

Example Input/Output 2:

Input:

pqbchjrtvwmdgefdiugi

Output:

NO

C-code

Compiler

#include<stdio.h>

#include <stdlib.h>

int main()

{

char str[101];

scanf("%s",str);

int fre[26]={0};

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

{   if(str[i]!='a'&&str[i]!='e'&&str[i]!='i'&&str[i]!='o'&&str[i]!='u')

    fre[str[i]-'a']=1;

}

int count=0;

for(int i=0;i<26;i++)

{

    if(fre[i]==1)

    count++;

}

printf(count==21?"YES":"NO");

}

The program must accept N integers as the input.The program must print the values from 1 to N where the sum of odd digits is equal to the sum of even digits as the output

Example Input/Output 1:

Input:

160

Output:

112 121 134 143 156

Example Input/Output 2:

Input:

500

Output:

112 121 134 143 156 165 178 187 211 314 336 341 358 363 385 413 431

C-code

Compiler

#include<stdio.h>

#include <stdlib.h>

int main()

{

int N;

scanf("%d",&N);

for(int i=112;i<=N;i++)

{

    int oddsum=0,evensum=0,num=i;

    while(num!=0)

    {

        if((num%10)%2==1)

        oddsum+=(num%10);

        else

        evensum+=(num%10);

        num/=10;

    }

    if(oddsum==evensum)

    printf("%d ",i);

}

}

The program must accept N integers as the input.The program must print the smallest possible integer by removing exactly one digit from the last three digit.

Example Input/Output 1:

Input:

7654

Output:

754

Example Input/Output 2:

Input:

987987

Output:

98787

Example Input/Output 3:

Input:

100

Output:

0

C-code

Compiler

#include<stdio.h>

#include <stdlib.h>

#include<string.h>

#include<math.h>

int main()

{

char str[1001];

scanf("%s",str);

int maxindex,i;

char max='0';

i=strlen(str)-3;

for(;i<strlen(str);i++)

{

    if(str[i]>max)

    {

    maxindex=i;

    max=str[i];

    }

}

long long int num=0;

int len=pow(10,strlen(str)-2);

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

{

    if(j!=maxindex)

    {

    num=num+(str[j]-'0')*len;

    len/=10;

    }

}

printf("%lld",num);

}

The program must accept N integers as the input.The program must print the sum of the largest digit in each integer among the N integerrs as the output.

Example Input/Output 1:

Input:

5

87 1654 121 657 15

Output:

28

Explanation:

• The largest digit in 87 is 8.
• The largest digit in 1654 is 6.
• The largest digit in 121 is 2.
• The largest digit in 657 is 7.
• The largest digit in 15 is 5.
• Therefore,28(8+6+2+7+5)

Example Input/Output 2:

Input:

4

0 444 10 25

Output:

10

C-code

Compiler

#include<stdio.h>

#include <stdlib.h>

int main()

{

int n,num,sum=0;

scanf("%d",&n);

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

{   int max=0;

    scanf("%d",&num);

    while(num!=0)

    {

        if(num%10>max)

        max=num%10;

        num/=10;

    }

    sum+=max;

}

printf("%d",sum);

}

The program must accept a string S (encrypted string) containing only .(Dots) and-(Hyphens) as the input.The encryption algorithm is given below.

• Digit 0 is encrypted as . (Single Dot)
• Digit 1 is encrypted as -. (Hyphen and Dot)
• Digit 2 is encrypted as -- (Double Hyphen)

The program must decrypt the string s and print it as the output.

Example Input/Output 1:

Input:

.-.--

Output:

012

Example Input/Output 2:

Input:

--.

Output:

20

C-code

Compiler

#include<stdio.h>

#include <stdlib.h>

int main()

{

char str[1001];

scanf("%s",str);

int i=0;

while(i<=strlen(str)-1)

{

    if(str[i]=='.')

    {

        printf("0");

        i++;

    }

    else if(str[i]=='-'&&str[i+1]=='.')

    {

        printf("1");

        i+=2;

    }

    else if(str[i]=='-'&&str[i+1]=='-')

    {

        printf("2");

        i+=2;

    }

}

}

The program must accept an integer N as the input.The program generate a sequence of integers from 1 to N.Then the program must replace each integer X in the sequence based on the following conditions:

• if X is a multiple of 3,replace it by Skill.
• if X is a multiple of 5,replace it by Rack.
• if X is a multiple of both 3 and 5,replace it by SkillRack.
• else keep the integer as it is.

Example Input/Output 1:

Input:

15

Output:

1

2

Skill

4

Rack

Skill

7

8

Skill

Rack

11

Skill

13

14

SkillRack

Example Input/Output 2:

Input:

5

Output:

1

2

Skill

4

Rack

C-code

Compiler

#include<stdio.h>

int main() {

    int x;

    scanf("%d",&x);

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

    {

        if(i%3==0&&i%5==0)

        printf("SkillRack\n");

        else if(i%3==0)

        printf("Skill\n");

        else if(i%5==0)

        printf("Rack\n");

        else

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

    }

}

The program must accept a string S with spaces as the input.For each word W in the string S,the program must replace the last character of W by the first character of the next word.For the last word the program must replace the last character by the first word.Finally the program must print the modified string S as the output.

Example Input/Output 1:

Input:

Nice to meet you

Output:

Nict tm meey yoN

Example Input/Output 2:

Input:

k g s n i g j n

Output:

g s n i g j n k

4Example Input/Output 3:

Input:

tit for tat

Output:

tif fot tat

C-code

Compiler

#include<stdio.h>

#include <stdlib.h>

int main()

{

char str[1000];

scanf("%[^\n]",str);

char temp=str[0];

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

{

    if(str[i]==' ')

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

}

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

printf("%s",str);

}

The program must accept an integer N as the input.The program must print the desired pattern as shown in the example

Example Input/Output 1:

Input:

9165

Output:

9165

165

65

5

Example Input/Output 2:

Input:

108004

Output:

108004

8004

8004

4

4

4Example Input/Output 3:

Input:

8525849176

Output:

8525849176

525849176

25849176

5849176

849176

49176

9176

176

76

6

C-code

Compiler

#include<stdio.h>

#include <stdlib.h>

int main()

{

long int num;

int len;

scanf("%ld%n",&num,&len);

long int divi=pow(10,len);

while(divi!=1)

{

    printf("%ld\n",num%divi);

    divi/=10;

}

}

The program must accept two string as the input,the program must print sub string which is having the maximum length and occurring at the seams position in both the string values.if both sub string has same length print the first occurred one.(two string has same length)

Example Input/Output 1:

Input:

skillrack

SkillRack

Output:

kill

Explanation:

skillrack,SkillRack.since kill has maximum length kill is printed.

Example Input/Output 2:

Input:

abcxxyzmn

abdxyzkmn

Output:

ab

C-code

Compiler

#include<stdio.h>

#include <stdlib.h>

#include<string.h>

int main()

{

char str1[1000],str2[1000];

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

int i=0,j=0,start,end,max,ctr=0,maxstart,maxend,count=0;

while(i<strlen(str1))

{

    if(str1[i]==str2[j])

    {   

        if(ctr==0)

        {

            start=i;

        }

        end=i;

        ctr=1;

        count++;

    }

    if(count>max)

    {

        max=count;

        maxstart=start;

        maxend=end;

    }

    if(str1[i]!=str2[j])

    {

        ctr=0;

        count=0;

    }

    i++;

    j++;

}

for(int i=maxstart;i<=maxend;i++)

    printf("%c",str1[i]);

}

The program must accept N integers as the input.The program must print he sum of integers containing at least two odd digits among the N integers as the output.

Example Input/Output 1:

Input:

5

78 549 123 877 214

Output:

1549

Explanation:

• 549,123 and 877 are having at least two odd digits.So the sum is 1549(549+123+877)

Example Input/Output 2:

Input:

4

124 4266 184 42

Output:

0

C-code

Compiler

#include<stdio.h>

#include <stdlib.h>

int main()

{

int n,sum=0;

scanf("%d",&n);

int num;

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

{

    scanf("%d ",&num);

    int a=num,count=0;

    while(a!=0)

    {

        if(a%2==1)

            count++;

        a=a/10;

    }

    if(count>=2)

        sum=sum+num;

}

printf("%d",sum);

}

The program must accept a string S containing only lower case alphabets as the input.For each sub string containing only vowels in the string S,the program must encloses the sub string by a pair of curly braces.Then the program must print the modified string S as the output.

Example Input/Output 1:

Input:

bookreadin

Output:

b{oo}kr{ea}d{i}ng

Example Input/Output 2:

Input:

aeiou

Output:

{aeiou}

Example Input/Output 3:

Input:

apple

Output:

{a}ppl{e}

C-code

Compiler

#include<stdio.h>

#include <stdlib.h>

#include<string.h>

int isvowel(char ch)

{

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

        return 1;

    else

        return 0;

}

int main()

{

char str[101];

int ctr=0;

scanf("%s",str);

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

{

    if(isvowel(str[i])&&ctr==0)

    {

      printf("{");

      ctr=1;

    }

    else if(ctr==1&&!isvowel(str[i]))

    {

        printf("}");

        ctr=0;

    }

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

    if((i==strlen(str)-1)&&isvowel(str[i]))

        printf("}");

}

}

The program must accept a fraction X/Y as the input.X represents the size of the pattern and Y represents the number of digits required to represent each integer in the pattern.The program must print the desired pattern as shown in the Example Input/Output section.

Example Input/Output 1:

Input:

5/3

Output:

001 002 003 004 005

006 007 008 009 010

011 012 013 014 015

016 017 018 019 020

021 022 023 024 025

Example Input/Output 2:

Input:

3/5

Output:

00001 00002 00003

00004 00005 00006

00007 00008 00009

Example Input/Output 3:

Input:

10/

Output:

1 2 3 4 5 6 7 8 9 0

1 2 3 4 5 6 7 8 9 0

1 2 3 4 5 6 7 8 9 0

1 2 3 4 5 6 7 8 9 0

1 2 3 4 5 6 7 8 9 0

1 2 3 4 5 6 7 8 9 0

1 2 3 4 5 6 7 8 9 0

1 2 3 4 5 6 7 8 9 0

1 2 3 4 5 6 7 8 9 0

1 2 3 4 5 6 7 8 9 0

C-code

Compiler

#include<stdio.h>

#include <stdlib.h>

int main()

{

int n,len,k=1;

scanf("%d/%d",&n,&len);

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

{

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

    {   if(len==1&&k==10)

        {

            k=0;

        }

        printf("%0*d ",len,k);

        k++;

    }

    printf("\n");

}

}

The program must accept a string S as the input.The program must find and print the number of occurrences of last two characters of S in the same string S as the output.

Example Input/Output:

Input:

hiabchi

Output:

2

Explanation:

The last two characters of the string hiabchi are h and i.Here the hi has occurred 2 times in the string hiabchi.

Hence the output is 2.

Example Input/Output:

Input:

MOOONOO

Output:

3

C-code

Compiler

#include<stdio.h>

#include <stdlib.h>

int main()

{

char str[10000];

int len,count=0;

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

char first=str[len-2],sec=str[len-1];

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

{

    if(str[i]==first&&str[i+1]==sec)

        count++;

}

printf("%d",count);

}

The program must accept an integer N as the input.The program must remove the tenth digit of all the N integers.Then the program must print the sum of N modified integers as the output.

Example Input/Output:

Input:

4

180 212 5295 5001

Output:

1058

Explanation:

After removing tenth digit the integer becomes 10 22 525 501 and their sum is 1058.Hence 1058 is printed as output.

Example Input/Output:

Input:

5

212 481 548 6545 1

Output:

777

C-code

Compiler

#include<stdio.h>

#include <stdlib.h>

int main()

{

int n,num,sum=0;

scanf("%d",&n);

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

{

    scanf("%d ",&num);

    sum=sum+((num%10)+(num/100)*10);

}

printf("%d",sum);

}

The program must accept an integer N as the input.The program must convert the decimal value to its equivalent binary values and print the binary value in desired pattern as shown below.

Example Input/Output:

Input:

463

Output:

111

00

1111

Explanation:

The binary value of 463 is 111001111.

Example Input/Output:

Input:

81

Output:

1

0

1

000

1

C-code

Method-1(Recursion)

Compiler

#include<stdio.h>

#include <stdlib.h>

long int b,k=0;

void binary(int n)

{

    if(n!=0)

    {

        binary(n/2);

        int a=n%2;

        if(k==0)

        {

            b=a;

            k=1;

        }

        if(b!=a)

            printf("\n");

        printf("%d",a);

        b=a;

    }

}

int main()

{

long long int N;

scanf("%lld",&N);

binary(N);

}

Method-2

Compiler

#include<stdio.h>

#include <stdlib.h>

int main()

{

long long int N;

scanf("%lld",&N);

long int arr[1001],i=0;

while(N>0)

{

    arr[i++]=N%2;

    N=N/2;

}

for(int j=i-1;j>=0;j--)

{

    printf("%ld",arr[j]);

    if(arr[j]!=arr[j-1])

        printf("\n");

}

}

The program must accept an integer N as the input.The program must remove the last occurring consecutive odd digits of N if the last digit of N is odd.Else the program must remove the last occurring consecutive even digits of N.Finally the program must print the modified integer as the output.

Example Input/Output:

Input:

2315642

Output:

2315

Example Input/Output:

Input:

92423

Output:

9242

C-code

Compiler

#include<stdio.h>

#include <stdlib.h>

int main()

{

char str[10001];

int len,ctr;

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

if((str[len-1]-'0')%2==0)

    ctr=0;

else

    ctr=1;

for(int i=len-1;i>=0;i--)

{

    int digit=str[i]-'0';

    if(digit%2!=ctr)

    {

        str[i+1]=NULL;

        break;

    }

}

printf("%s",str);

}

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

Example Input/Output:

Input:

5

Output:

1 * 1

2 2 *

3 * *

4 4 *

5 * 5

Example Input/Output:

Input:

7

Output:

1 * * 1

2 * 2 *

3 3 * *

4 * * *

5 5 * *

6 * 6 *

7 * * 7

C-code

Compiler

#include<stdio.h>

#include <stdlib.h>

int main()

{

int N;

scanf("%d",&N);

int ctr=(N/2)+1;

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

{   

    for(int j=1;j<=(N/2)+1;j++)

    {

        if(j==1||j==ctr)

            printf("%d ",i);

        else

            printf("* ");

    }

    printf("\n");

    if(i>=(N/2)+1)

      ctr++;

     else

        ctr--;

}

}

The program must accept a integer N which contains 1's and 0's .The program is to find the decimal value by rearranging the 1's first and 0's followed by it.

Example Input/Output:

Input:

4

1 0 0 1

Output:

12

Explanation:

It is rearranged as 1100 and the decimal value of 1100 is 12 which is printed as output.

Example Input/Output:

Input:

42

1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 1

Output:

4398046248960

C-code

Compiler

#include<stdio.h>

#include <stdlib.h>

int main()

{

int N;

scanf("%d",&N);

int a,count1=0,count0=0;

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

{

    scanf("%d",&a);

    if(a==1)

        count1++;

    else

        count0++;

}

long long int sum=0;

for(int i=0;i<count1;i++)

{

    sum=sum+((long int)1<<count0);

    count0++;

}

printf("%lld",sum);

}

The program must accept a string s where the length of s is always a multiple of 4 as the input.The program must split the string S into four equal parts.Then the program must reverse the characters in each part.Then the program must print the modified four parts of string

Example Input/Output:

Input:

paraacetoaminophenol

Output:

aarap aotec ponim loneh

Example Input/Output:

Input:

FAST

Output:

F A S T

C-code

Compiler

#include<stdio.h>

#include <stdlib.h>

char str[1001];

void print(int from,int to)

{

    while(from>=to)

    {

        printf("%c",str[from--]);

    }

    printf(" ");

}

int main()

{

int len;

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

int start=0,end=(len/4)-1;

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

{

    print(end,start);

    start+=len/4;

    end+=len/4;

}

}

Example Input/Output:

Input:

National

Output:

Natainol

Explanation:

National contains a,i,o,a as vowels after rotating the vowels in clockwise direction for 1 time the string become Natainol. Hence it is printed as output

Example Input/Output:

Input:

AEiou

Output:

uAEio

C-code

Compiler

#include<stdio.h>

#include <stdlib.h>

int isvowel(char ch)

{

    ch=tolower(ch);

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

       return 0;

    return 1;

}

void main()

{

char str[1001],vowel[1001];

int len,index=0;

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

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

{   

    if(!isvowel(str[i]))

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

}

char temp=vowel[index-1];

for(int i=index-1;i>0;i--)

{

    vowel[i]=vowel[i-1];

}

vowel[0]=temp;

int ctr=0;

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

{

    if(!isvowel(str[i]))

        str[i]=vowel[ctr++];

}

printf("%s",str);

}

Example Input/Output:

Input:

5

1 1 2 3 1

Output:

YES

Explanation:

The sequence 1 2 3 is occurred in the 5 integers. Hence,the output will be YES

Example Input/Output:

Input:

6

1 2 1 2 13

Output:

NO

C-code

Compiler

#include<stdio.h>

#include <stdlib.h>

int main()

{

int n;

scanf("%d",&n);

int arr[n],count=0;

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

{

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

    if(arr[i]==3&&arr[i-1]==2&&arr[i-2]==1)

        count=1;

}

if(count)

    printf("YES");

else

    printf("NO");

}

Example Input/Output:

Input:

5

Output:

1 5 2 4 3 3 4 2 5 1

* 1 4 2 3 3 2 4 1 *

* * 1 3 2 2 3 1 * *

* * * 1 2 2 1 * * *

* * * * 1 1 * * * *

Example Input/Output:

Input:

3

Output:

1 3 2 2 3 1

* 1 2 2 1 *

* * 1 1 * *

C-code

Compiler

#include<stdio.h>

int main() {

   int num,count=0;

   scanf("%d",&num);

   int a=1,b=num;

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

  {

      for(int j=0;j<num*2;j++)

      {

          if(j<i||j>=(num*2)-i)

            printf("* ");

        else

        {

            if(j>=i)

            {   

                if(count==0)

                  {

                      printf("%d ",a);

                      a++;

                      count=1;

                  }

                  else

                  {

                      printf("%d ",b);

                      b--;

                      count=0;

                  }

            }

        }

      }

      printf("\n");

      a=1;

      b=num-i-1;

  }

}

Example Input/Output:

Input:

abcde

Output:

abcde

abcd*

*bcde

abc**

**cde

ab***

***de

a****

****e

*****

*****

C-code

Compiler

#include<stdio.h>

#include <stdlib.h>

int main()

{

char str[1001];

int len;

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

int a=len,b=-1;

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

{

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

    {

        if(i%2==1)

        {

            if(j<a)

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

            else

            printf("*");

        }

        else

        {   if(j>b)

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

             else

            printf("*");

        }

    }

    printf("\n");

    if(i%2==0)

         a--;

    else

        b++;

}

}

Example Input/Output:

Input:

3

all is well

Output:

aiw

lse

l*l

**l

C-code

Compiler

#include<stdio.h>

#include<string.h>

int main() {

   int N;

   scanf("%d",&N);

   char str[N][101];

   int len[N],max=0;

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

    {

        scanf("%s",str[i]);

        len[i]=strlen(str[i]);

        if(len[i]>max)

            max=len[i];

    }

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

    {

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

        {   if(i>(len[j]-1))

                printf("*");

            else

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

        }

        printf("\n");

    }

}