Skip to main content

Wipro coding questions | Wipro Programming questions (Automata) -




Wipro Coding questions – set 1

1) Find the distinct elements in a given array. (Assume size of an array n<=20)
Sample Input:
  • 9 = size of an array 
  • 2 3 4 5 6 1 2 3 4  =  array elements
Sample Output:
  • 2 3 4 5 6 1
Program:
// C program to print all distinct elements in a given array
#include 
void distict_elements(int a[], int n);
int main()
{
    int size_array, i, arr[20];
    // Get the array size
    scanf(“%d”, &size_array) 
   // Get the array elements
    for(i=0; i<size_array; i++)
    {
        scanf(“%d”, &arr[i]);
    }
    // Function call to print the distinct elements in an array
    distict_elements(arr, size_array);   
    return 0;
}
void distict_elements(int a[], int n)
{
    int i, j;
    // Pick all elements one by one
    for (i=0; i<n; i++)
    {
        // Check if the picked element is already printed
          for (j=0; j<i; j++)
          {
              if (a[i] == a[j])
                 break;
          }
        // If not printed earlier, then print it
        if (i == j)
        printf(“%d “, a[i]);
    }
}
2) Program to sort array in ascending & descending order.
Input:
5
8 6 9 2 7
Output:
2 6 7 8 9
9 8 7 6 2
Program:
// C program to sort the given array elements in ascending and descending order
#include 
int main(void)
{
              int arr[10], i=0, j=0, size, temp;
              // Get the size of an array
              scanf (“%d”, &size);
    // Get the array elements as an input
              for (i = 0; i <size; i++)
              {
                             scanf (“%d”, &arr[i]);
              }
    // Sorting elements in ascending order
              for (j=0 ; j<(size-1) ; j++)
              {
                             for (i=0 ; i<(size-1) ; i++)
                             {
                                           if (arr[i+1] < arr[i])
                                           {
                                                  temp = arr[i];
                                                  arr[i] = arr[i + 1];
                                                  arr[i + 1] = temp;
                                           }
                             }
              }
              // Print the elements from index value 0 to (size-1) –> ascending order
              for (i=0 ; i
{             printf (“%d “, arr[i]);
              }
              printf(“\n”);
               // Print the elements from the index value (size-1) to 0  –> descending order
              for (i=size-1; i>=0 ; i–)
              {
                             printf (“%d “, arr[i]);
              }
      return 0;
}
3)  Sort first half of an array in ascending and second half in descending order.
Example 1:
Input:
 8
2 4 7 9 3 1 6 8
Output:
 1 2 3 4 9 8 7 6
Example 2:
Input:
6
1 2 3 4 5 6
Output:
1 2 3 6 5 4
Algorithm:
i) Sort the given array.
ii) Run a loop up to half the length of the array and print the elements of the sorted array.
iii) Run a loop from the last index of the array to the middle of the array and print the elements in reverse order.
Program:
#include 
void sorting_elements(int arr[], int n);
void display(int arr[], int n);
int main()
{
    int size, arr[20], i;
    scanf(“%d”, &size);
    for(i=0; i<size; i++)
    {
        scanf(“%d”, &arr[i]);
    }
    display(arr, size);
    return 0;
}
// Sort the elements in the ascending order
void sorting_elements(int arr[], int n)
{
    int i,j,temp;
    for (j=0 ; j<(n-1) ; j++)
              {
                             for (i=0 ; i<(n-1) ; i++)
                             {
                                           if (arr[i+1] < arr[i])
                                           {
                                                  temp = arr[i];
                                                  arr[i] = arr[i + 1];
                                                  arr[i + 1] = temp;
                                           }
                             }
              }
}
// Display the sorted elements
void display(int arr[], int n)
{
    sorting_elements(arr, n);
    int i, j
    // Print the first half as such (i.e. from index 0 to midlle)
    for (i=0; i<n/2; i++)
    {
        printf(“%d “, arr[i]);   
    }
    // Print the second half in the reverse order (i.e. from n-1 to midlle)
    for (j=n-1; j>=n/2; j–)
    {
        printf(“%d “, arr[j]);
    }
}

4) Print the below pattern
Input:
              3  4
Output:
               3
              44
              555
              6666
              555
              44
              3
Input :
              4  4
Output:
              4
              55
              666
              7777
              666
              55
              4
Program:
#include 
int main()
{
    int i,j,s,N,count=0;
    scanf(“%d%d”,&s,&N);
    for(i=s;count<4;count++)
    {
        for(j=0;j<count+1;j++)
            printf(“%d”,i);
        printf(“\n”);
        i=i+1;
    }
    for(i=s+N-2;count>0;count–)
    {
        for(j=0;j<count-1;j++)
            printf(“%d”,i);
        printf(“\n”);
        i=i-1;
    }
    return 0;
}
5) Print the following pattern
Input :
3
Output:
1
2*2
3*3*3
3*3*3
2*2
1
Input :
4
Output:
1
2*2
3*3*3
4*4*4*4
4*4*4*4
3*3*3
2*2
1
Program:
#include 
int main()
{
    int i,j,k,N,count=0;
    scanf(“%d”,&N);
    for(i=1;i<=N;i++)
    {
        k=1;
        for(j=0;j<i;j++)
        {
            printf(“%d”,i);
            if(k<i)
            {
                printf(“*”);
                k=k+1;
            }
        }
        printf(“\n”);
    }
    for(i=N;i>0;i–)
    {
        k=1;
        for(j=0;j<i;j++)
        {
            printf(“%d”,i);
            if(k<i)
            {
                printf(“*”);
                k=k+1;
            }
        }
        printf(“\n”);
    }
    return 0;
}
6) Print the below pattern
Input:
4
Output:
1
2*3
4*5*6
7*8*9*10
7*8*9*10
4*5*6
2*3
1
Program:
#include
int main()  {
  int i,j,count=1,n;
  printf(“Enter a number\n”);
  scanf(“%d”,&n);
 for(i=1;i<=n;i++)
  {     
for(j=1;j<=i;j++)
       {
           if(j<i)
            printf(“%d*”,count++);
           else
            printf(“%d”,count++);
       }         printf(“\n”);
      }
count=count-n;
 for(i=n;i>=1;i–)
   {        for(j=1;j<=i;j++)
       {
           if(j<i)
            printf(“%d*”,count++);
           else
            printf(“%d”,count++);
       }
       count=(count+1)-2*i;
       printf(“\n”);
     }
   return 0;
 }

Wipro Coding questions – set 2

7) Print the following pattern
Input: 
3  4
Output:
3
44
555
6666
6666
555
44
3
Program:
#include 
int main()
{
    int i,j,s,N,count=0;
    scanf(“%d%d”,&s,&N);
    for(i=s;count<4;count++)
    {
        for(j=0;j<count+1;j++)
            printf(“%d”,i);
        printf(“\n”);
        i=i+1;
    }
 for(i=s+N-2;count>0;count–)
    {
        for(j=0;j<count-1;j++)
            printf(“%d”,i);
        printf(“\n”);
        i=i-1;
    }
    return 0;
}
8) Print the below pattern
Input:
5
Output:
1
3*2
4*5*6
10*9*8*7
11*12*13*14*15
Program:
#include 
int main()
{
    int i,j,k,l=1,N,d,r,count=0;
    scanf(“%d”,&N);
    for(i=1;i<=N;i++)
    {
        k=1;
        d=i%2;
        r=l+i-1;
        for(j=0;j<i;j++)
        {
 if(d==0)
            {
                printf(“%d”,r);
                r–;
                if(k<i)
                {
                    printf(“*”);
                    k=k+1;
                }
                l++;
                continue;
            }
            printf(“%d”,l);
            l++;
            if(k<i)
            {
                printf(“*”);
                k=k+1;
            }
        }
        printf(“\n”);
    }
    return 0;
}
9) Print the below pattern
Input:
4
Output:
1*2*3*4*17*18*19*20
– -5*6*7*14*15*16
– – – -8*9*12*13
– – – – – -10*11
Program:
#include 
void pattern(int);
int main()
{
               int n;
              scanf(“%d”, &n);
              pattern(n);
              return 0;
}
void pattern(int n)
{            
              int i, j, k, s, a = 1,b = n*n + 1;
              for (i = n; i >= 1; i–) {                    
                             for (s = 0; s < n – i; s++)
                                           printf(“–“);
                             for (j = 0; j < i; j++)
                                           printf(“%d*”, a++);
                             for (k = 0; k < i – 1; k++)
                                           printf(“%d*”, b++);
                             printf(“%d\n”, b);                                         // last b should without *
                             b -= 2*(i – 1);
              }            
}

10) Print the below pattern
Input:
3
Output:
3 3 3
3 1 3
3 2 3
3 3 3
Program:
#include
int main()
{
    int i, j, n, c=1;
    scanf(“%d”, &n);
    for(i=1; i<=n+1; i++)
    {
        for(j=1; j<=n; j++)
        {
            if(i!=1 && j==n-1)
            {
            printf(“%d “, c);
                             c++;
            }
            else
            printf(“%d “, n);
               }
                   printf(“\n”);
                 }
                return 0;
}

11) Paranthesis checker: Check whether the given expression is valid or not(only parenthesis symbol). 
Test Case: 1
Input:  “(( ))”
Output: Valid
Test Case: 2
Input: “()(“
Output: Invalid
Program:
#include 
#include 
#include 
int top = -1;   char stack[100];
void push(char);                           
void pop();
void find_top();
void main()
{
int i;
char a[100];
                     scanf(“%s”, &a);
                     for (i = 0;  a[i] != ‘\0’; i++)
                     {
                             if (a[i] == ‘(‘)
                                           push(a[i]);
                             else if (a[i] == ‘)’)
                                           pop();
                      }
                 find_top();
                  }
 // to push elements in stack
void push(char a)
{
              top++;
              stack[top] = a;
}
 // to pop elements from stack
void pop()
{
if (top == -1)
    {
printf(“Invalid”);
exit(0);
     }       
else
      top–;
}
// to find top element of stack
void find_top()
{
if (top == -1)
printf(“Valid”);
else
printf(“Invalid”);
}
12) Print the transpose of a Matrix:
#include 
int main()
{
    int a[10][10], transpose[10][10], r, c, i, j;
    printf(“Enter rows and columns of matrix: “);
    scanf(“%d %d”, &r, &c);
    // Storing elements of the matrix
    printf(“\nEnter elements of matrix:\n”);
    for(i=0; i<r; ++i)
        for(j=0; j<c; ++j)
        {
            printf(“Enter element a%d%d: “,i+1, j+1);
            scanf(“%d”, &a[i][j]);
        }
// Displaying the matrix a[][] */
    printf(“\nEntered Matrix: \n”);
    for(i=0; i<r; ++i)
        for(j=0; j<c; ++j)
        {
            printf(“%d  “, a[i][j]);
            if (j == c-1)
                printf(“\n\n”);
        }
    // Finding the transpose of matrix a
    for(i=0; i<r; ++i)
        for(j=0; j<c; ++j)
        {
            transpose[j][i] = a[i][j];
        }
 // Displaying the transpose of matrix a
    printf(“\nTranspose of Matrix:\n”);
    for(i=0; i<c; ++i)
        for(j=0; j<r; ++j)
        {
            printf(“%d  “,transpose[i][j]);
            if(j==r-1)
                printf(“\n\n”);
        }
    return 0;
}
13) Matrix Addition:
Program:
#include 
int main()
{
    int r, c, a[100][100], b[100][100], sum[100][100], i, j;
    printf(“Enter number of rows (between 1 and 100): “);
    scanf(“%d”, &r);
    printf(“Enter number of columns (between 1 and 100): “);
    scanf(“%d”, &c);
    printf(“\nEnter elements of 1st matrix:\n”);
for(i=0; i<r; ++i)
     for(j=0; j<c; ++j)
        {
            printf(“Enter element a%d%d: “,i+1,j+1);
            scanf(“%d”,&a[i][j]);
        }
    printf(“Enter elements of 2nd matrix:\n”);
    for(i=0; i<r; ++i)
        for(j=0; j<c; ++j)
        {
            printf(“Enter element a%d%d: “,i+1, j+1);
            scanf(“%d”, &b[i][j]);
        }
 // Adding Two matrices
    for(i=0;i<r;++i)
        for(j=0;j<c;++j)
        {
            sum[i][j]=a[i][j]+b[i][j];
        }
 // Displaying the result
    printf(“\nSum of two matrix is: \n\n”);
    for(i=0;i<r;++i)
   {     for(j=0;j<c;++j)
        {
            printf(“%d   “,sum[i][j]);
            if(j==c-1)
            {
                printf(“\n\n”);
            }
}
return 0;
}

Comments

  1. I have found that this site is very informative, interesting and very well written. keep up the nice high quality writing Planning and Administering Microsoft Azure for SAP Workloads AZ-120

    ReplyDelete

Post a Comment

Popular posts from this blog

Microsoft Virtual Academy (Free Certification course )

Microsoft Virtual Academy (Free Certification course ) he  Microsoft Virtual Academy  (MVA) is a free online school with courses that cover  Microsoft -related topics and specific  Microsoft  products. The MVA offers a mix of on-demand courses and live events; each course contains a video and PDF download of the video transcript. An online  school  ( virtual school  or e- school  or cyber- school ) teaches students entirely or primarily online or through the internet. ... Online education exists all around the world and is used for all levels of education (K-12, college, or graduate school ). How do I get Microsoft certified? Let's take a look at the steps to becoming a MOS. Step 1: Obtain Basic Computer Skills. Before becoming a certified Microsoft Office Specialist (MOS), individuals must obtain basic computer skills. ... Step 2: Enroll in Microsoft Office Courses. ... Step 3: Choose a Certification Program. ... S...

Top Web Site Which Provide Free Courses With II Certification

Top Web Site Which Provide Free Courses With Certification  In this I'm going to write about the web sites which provides free course with certification with their links. Solo Learn : Learn to code for FREE! Anytime and Anywhere, on Any Device. Solo Learn provides many IT field courses which is written as below : C++ C# JAVA jQUERY JAVA SCRIPT HTML PHP SWIFT RUBY SQL PYTHON         Web site link is https://www.sololearn.com    Coursera :  Coursera.org is a website that partners with universities and organizations around the world. This brings a wide variety of topics and perspectives to one searchable database. Coursera is a powerful tool for free online education, and includes courses from many top universities, museums and trusts. This gives the site an extremely wide range of in-depth courses. Coursera is extremely useful if you’re looking to study many different topics, or want courses ...

Very Important Question of Artificial Intelligence all Unit covered Must read scoring Good Marks in Exam as per AKTU Pattern

INTRODUCTION Define AI Its advantage foundation and branches of AI Define agent and Intelligent agent Also describe kinds of agent programs  NLP Natural Language Processing Turing Test  Artificial intelligence In computer science, artificial intelligence, sometimes called machine intelligence, is intelligence demonstrated by machines, in contrast to the natural intelligence displayed by humans and other animals Artificial Intelligence  is a way of making a computer, a computer-controlled robot, or a software think intelligently, in the similar manner the  intelligent  humans think. Consumer goods. Using natural language processing, machine learning and advanced analytics, Hello Barbie listens and responds to a child. ... Creative Arts. ... Energy. ... Financial Services. ... Healthcare. ... Manufacturing. ... Media. ... Retail.   intelligent agent In  artificial intelligence , an  intelligent agent  (IA) ...