Posts

Implementing linked list in C language

#include <stdio.h> #include <stdlib.h> // Define the structure for a node in the linked list struct node {     int data; // The data stored in the node     struct node *next; // Pointer to the next node in the list }; struct node* head = NULL; // Initialize the head of the list as NULL // Function to insert nodes at the end of the list void insert(struct node **head) {     int num = 0, i, data;     printf("\nEnter the number of elements to be inserted: ");     scanf("%d", &num); // Get the number of nodes to be inserted     // Loop to insert the nodes     for(i = 0; i<num; i++) {         // Allocate memory for a new node and get the data for it         struct node* newNode = malloc(sizeof(struct node));         struct node *last = *head;         printf("\nEnter the element %d: ", i+1);         scanf("%d", &data); // Get the data for the node         // Initialize the new node with the given data and set its next pointer to
Post a Comment
Read more

Implementation of stack using array

#include <stdio.h> int val, top=-1, n, i, j, t, choice=0, stack[100]; void push(); void pop(); void show(); int main() {     printf("\nEnter the capacity of the stack\n");     scanf("%d", &n);     while(choice != 4)     {         printf("\nEnter your choice\n1.Push\n2.Pop\n3.Show\n4.Exit\nYou choose: ");         scanf("%d",&choice);         switch(choice)         {             case 1:             {                 push();                 t += 1;                 break;             }             case 2:             {                 pop();                 break;             }             case 3:             {                 show();                 break;             }             case 4:             {                 printf("\nExited\n");                 break;             }             default:             {                 printf("\nInvalid choice.\nEnter a valid choice.\n");             }         }     }     re
Post a Comment
Read more

The number of required [n x 1] multiplexer to implement one [m x 1] multiplexer.

  /****************************************************************************** This is a program to find the number of required (m x 1) multiplexer to implement one (n x 1) multiplexer. And m is always even, otherwise we are wasting the bit(s). If your m is odd then increase it by one. *******************************************************************************/ #include <stdio.h> int main() {     int n, m, x, y = 0;      printf("This is a program to find the number of required (m x 1) multiplexer to implement one (n x 1) multiplexer. And m is always even, otherwise, we are wasting the bit(s). If your m is odd then increase it by one.\n");     printf("Enter the value of n:\t");     scanf("%d", &n);     printf("Enter the value of m:\t");     scanf("%d", &m);     if(n>m){         x = n;         while(x > m){             //If x is odd then we need an another multiplexer.             if( x%2 == 0 ){             x =
Post a Comment
Read more

The number of required [n x 1] multiplexer to implement one [m x 1] multiplexer. Only for even [n/m].

/****************************************************************************** This is a program to find the number of required (n x 1) multiplexer to  implement one (m x 1) multiplexer, only when n/m never gives odd or decimal number . *******************************************************************************/ #include <stdio.h> int main() {     int n, m, x, y = 0;     printf("Enter the value of n:");     scanf("%d", &n);     printf("Enter the value of m:");     scanf("%d", &m);     if(n>m){         x = n;         while(x > m){             x = x/m;             y = y + x;         }         printf("%d", y+1);     }     else         printf("1");     return 0; }  
Post a Comment
Read more

Second largest element in an array

Pre defined array: #include <stdio.h> int main() {     int a[10] = {1, 11, 3, 4, 150, 69, 7, 8, 9, 10};     int i, l, s=0;     l=a[0];     for(i=0; i<10; i++) {         if(l<a[i])             l = a[i];     }     for(i=0; i<10; i++) {         if(a[i]>a[i+1] && a[i]!=l)             s = a[i];     }     printf("Largest - %d \nSecond - %d \n", l, s);     return 0; } Taking array as an input: #include <stdio.h> int main() {     int n, i, l, s=0;     printf("Enter the number of element(s):\n");     scanf("%d", &n);     int a[n];     for(i=0;i<n;i++)     {         printf("\nEnter the element %d:\t", i);         scanf("%d", &a[i]);     }     l=a[0];     for(i=0; i<n; i++)     {         if(l < a[i])         {             s = l;             l = a[i];         }         else if(s < a[i])             s = a[i];     }     printf("\nLargest - %d \nSecond largest- %d \n", l, s);     return 0;
Post a Comment
Read more

Degree of Polynomial | Problem - 52 | Beginner level | C language | CodeChef

Link to the problem statement:   https://www.codechef.com/submit/DPOLY Solution:   #include <stdio.h> int main(void) { int t; scanf("%d", &t); while(t--) {     int i, n, d;     scanf("%d",  &n);     int a[n];     for(i=0;i<n;i++)     {         scanf("%d", &a[i]);         if(a[i] != 0)         {             d = i;         }              }     printf("%d\n", d); } return 0; } Hope this will help you, if you want anything else leave a comment. Thank you  
Post a Comment
Read more

Battery Low | Problem - 51 | Beginner level | C language | CodeChef

Link to the problem statement:   codechef.com/submit/BATTERYLOW Solution:   #include <stdio.h> int main(void) { int t, x; scanf("%d", &t); while(t--) {     scanf("%d", &x);     if(x > 15)         printf("NO\n");     else         printf("YES\n"); } return 0; } Hope this will help you, if you want anything else leave a comment. Thank you  
Post a Comment
Read more

Small factorials | Problem - 50 | Beginner level | C language | CodeChef

    Link to the problem statement:   https://www.codechef.com/submit/FCTRL2 Solution:   The solution is already given, you just have to click on  SUBMIT CODE  button. //We have populated the solutions for the 10 easiest problems for your support. //Click on the SUBMIT button to make a submission to this problem. #include<stdio.h> int main() {     int t,n,a[200],i,j,k,l,m;     scanf("%d",&t);     while(t--)     {     scanf("%d",&n);         m=1;         a[0]=1;         for(j=2;j<=n;j++)         {             l=0;             for(k=0;k<m;k++)             {                 a[k]=a[k]*j+l;                 l=a[k]/10;                 a[k]=a[k]%10;             }             while(l)             {             a[k]=l%10;                 k++;                 m++;                 l=l/10;             }         }         for(i=m-1;i>=0;i--)             printf("%d",a[i]);         printf("\n");     }     return 0; } Hope this will help
Post a Comment
Read more

Increase IQ | Problem - 49 | Beginner level | C language | CodeChef

  Link to the problem statement:  https://www.codechef.com/submit/INCRIQ Solution:   #include <stdio.h> int main(void) { int x;     scanf("%d",&x);     if(x>=100 && x <= 169)     {         if(x+7 > 170)             printf("YES\n");         else             printf("NO\n");     } return 0; } Hope this will help you, if you want anything else leave a comment. Thank you  
Post a Comment
Read more

Problems in your to-do list | Problem - 48 | Beginner level | C language | CodeChef

  Link to the problem statement:  https://www.codechef.com/submit/TODOLIST Solution:   #include <stdio.h> int main(void) {     int t, n, i, j; scanf("%d", &t); for(i=0;i<t;i++) {     scanf("%d", &n);     int b=0, a[n];     for(j=0;j<n;j++)     {         scanf("%d", &a[j]);         if(a[j]>=1000)             b++;         else            continue;     }     printf("%d\n", b); } return 0; } Hope this will help you, if you want anything else leave a comment. Thank you  
Post a Comment
Read more