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Write a C program to print all permutations of a given string

A permutation, also called an “arrangement number” or “order,” is a rearrangement of the elements of an ordered list S into a one-to-one correspondence with S itself. A string of length n has n! permutation.



Below are the permutations of string ABC.
ABC, ACB, BAC, BCA, CAB, CBA
Here is a solution using backtracking.

// C program to print all permutations with duplicates allowed #include <stdio.h> #include <string.h>   /* Function to swap values at two pointers */ void swap(char *x, char *y) {     char temp;     temp = *x;     *x = *y;     *y = temp; }   /* Function to print permutations of string    This function takes three parameters:    1. String    2. Starting index of the string    3. Ending index of the string. */ void permute(char *a, int l, int r) {    int i;    if (l == r)      printf("%s\n", a);    else    {        for (i = l; i <= r; i++)        {           swap((a+l), (a+i));           permute(a, l+1, r);           swap((a+l), (a+i)); //backtrack        }    } }   /* Driver program to test above functions */ int main() {     char str[] = "ABC";     int n = strlen(str);     permute(str, 0, n-1);     return 0; }

Output:

ABC
ACB
BAC
BCA
CBA
CAB

Algorithm Paradigm: Backtracking
Time Complexity: O(n*n!)

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programe to find date and time of system

 #include <time.h>
 #include <stdio.h> 

          int main(void)
 {
  time_t mytime;
  mytime = time(NULL);
  printf(ctime(&mytime));

  return 0;
 }

The string that is returned will have the following format:

 Www Mmm dd hh:mm:ss yyyy

 Www = which day of the week.
 Mmm = month in letters.
 dd = the day of the month.
 hh:mm:ss = the time in hour, minutes, seconds.
 yyyy = the year.

 Output example:
  Tue Feb 26 09:01:47 2009

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c programe to count no of set bit in an integer

#anki 



#include<stdio.h>

void main()
{
 int i,count=0;
 printf("enter the value");
 scanf("%d",&i);
 while(i)
 {
   if((i & 1)==1) count++;
  i=i>>1;
   }
   printf("count =%d",count);
   getch();   
}

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C Program to display current date

#include <dos.h>
#include <stdio.h>
int main(void)
{
struct date d;
getdate(&d);
printf(“The current year is: %d\n”, d.da_year);
printf(“The current day is: %d\n”, d.da_day);
printf(“The current month is: %d\n”, d.da_mon);
return 0;
}

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C Program to Set / Change current system date

#include <stdio.h>
#include <dos.h>
int main(void)
{
struct time t;
gettime(&t);
printf(“The current hour is: %d\n”, t.ti_hour);
printf(“The current min is: %d\n”, t.ti_min);
printf(“The current second is: %d\n”, t.ti_sec);
/* Add one to the hour,minute & sec struct element and then call settime */
t.ti_hour++;
t.ti_min++;
t.ti_sec++;
settime(&t);
printf(“The current hour is: %d\n”, t.ti_hour);
printf(“The current min is: %d\n”, t.ti_min);
printf(“The current second is: %d\n”, t.ti_sec);
return 0;
}

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Enter and print details of n employees using structures and dynamic memory allocation

 Enter and print details of n employees using structures and dynamic memory allocation

This program is used to show the most basic use of structures. The structure is made of a character array name[], integer age and float salary. We’ll make an array of the structure for the employees. We’ll also use dynamic memory allocation using malloc. You can also use a linked list to the same which is a better option. Let’s check the code.

?

#include #include typedef struct{ //structure of emp  char name[30];  int age;  float salary; }emp; int main(){  int n,i;  emp *employee;  printf("Enter no of employees: ");  scanf("%d",&n);  employee=(emp*)malloc(n*sizeof(emp)); //dynamic memory allocation using malloc()  for(i=0;i<n;i++){  printf("\n\nEnter details of employee %d\n",i+1);  printf("Enter name: ");  scanf("%s",employee[i].name);  printf("Enter age: ");  scanf("%d",&employee[i].age);  printf("Enter salary: ");  scanf("%f",&employee[i].salary);  }  printf("\nPrinting details of all the employees:\n");  for(i=0;i<n;i++){  printf("\n\nDetails of employee %d\n",i+1);  printf("\nName: %s",employee[i].name);  printf("\nAge: %d",employee[i].age);  printf("\nSalary: %.2f",employee[i].salary);  }  getch();  return 0; }

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Patterns (one of their favorites)

 Patterns (one of their favorites)

If the interviewer asks you a pattern and you don’t know how to do that you are screwed big time. Often you might make a very simple mistake which the interviewer was actually looking for. Here we’ll find out how to print this pattern

?

1 2 3 4 5

A B C D E D C B A   A B C D C B A     A B C B A       A B A         A

And here is the code.

?

01 02 03 04 05 06 07 08 09 10 11 12 13 14 15 16 17 18 19 20 21

//Printing pattern #include #include int main(){  int n,i,j;  printf("Enter no of lines: ");  scanf("%d",&n);  for(i=0;i<n;i++){  for(j=0;j<i;j++){ //for printing spaces  printf(" ");  }  for(j=0;j<n-i;j++){ //for printing the left side   printf("%c ",'A'+j); //the value of j is added to 'A'(ascii value=65)   }   for(j=n-i-2;j>=0;j--){ //for printing the right side     printf("%c ",'A'+j);   }  printf("\n");  }  getch(); }

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easiest way to sort

#include <stdio.h>
#include <stdlib.h>

int values[] = { 88, 56, 100, 2, 25 };

int cmpfunc (const void * a, const void * b)
{
   return ( *(int*)a - *(int*)b );
}

int main()
{
   int n;

   printf("Before sorting the list is: \n");
   for( n = 0 ; n < 5; n++ ) 
   {
      printf("%d ", values[n]);
   }

   qsort(values, 5, sizeof(int), cmpfunc);

   printf("\nAfter sorting the list is: \n");
   for( n = 0 ; n < 5; n++ ) 
   {   
      printf("%d ", values[n]);
   }
  
   return(0);
}

output

Before sorting the list is: 
88 56 100 2 25 
After sorting the list is: 
2 25 56 88 100 

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number of ways a particular no can formed..

#include<stdio.h>


// Returns the count of ways we can sum S[0...m-1] coins to get sum n

int count( int S[], int m, int n )
{
    // If n is 0 then there is 1 solution (do not include any coin)
    if (n == 0)
        return 1;
   
    // If n is less than 0 then no solution exists
    if (n < 0)
        return 0;

    // If there are no coins and n is greater than 0, then no solution exist
    if (m <=0 && n >= 1)
        return 0;

    // count is sum of solutions (i) including S[m-1] (ii) excluding S[m-1]
    return count( S, m - 1, n ) + count( S, m, n-S[m-1] );
}

// Driver program to test above function
int main()
{
    int i, j;
    int arr[] = {1, 3, 5};
    int m = sizeof(arr)/sizeof(arr[0]);
    printf("%d ", count(arr, m, 5));
    getchar();
    return 0;
}



It should be noted that the above function computes the same subproblems again and again. See the following recursion tree for S = {1, 2, 3} and n = 5.
The function C({1}, 3) is called two times. If we draw the complete tree, then we can see that there are many subproblems being called more than once.

C() --> count()
                              C({1,2,3}, 5)                     
                           /                \
                         /                   \              
             C({1,2,3}, 2)                 C({1,2}, 5)
            /     \                        /         \
           /        \                     /           \
C({1,2,3}, -1)  C({1,2}, 2)        C({1,2}, 3)    C({1}, 5)
               /     \            /    \            /     \
             /        \          /      \          /       \
    C({1,2},0)  C({1},2)   C({1,2},1) C({1},3)    C({1}, 4)  C({}, 5)
                   / \      / \       / \        /     \    
                  /   \    /   \     /   \      /       \ 
                .      .  .     .   .     .   C({1}, 3) C({}, 4)
                                               /  \
                                              /    \  

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C Program to delete a file.

#include<stdio.h>
#include<conio.h>
#include<stdlib.h>
void main()
{
 int status;
 char name[25];
 clrscr();
 printf("Enter the name of file you want to delete\n");
 gets(name);
 status=remove(name);
 if(status==0)
 {
   printf("%s file deleted successfully.\n",name);
 } 
 else
 {
   printf("Unable to delete the file\n");
 }
 getch();
}

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C Program to merge 2 files

#include<stdio.h>
#include<conio.h>
#include<stdlib.h>
void main()
{
 FILE *fs1,*fs2,*ft;
 char ch,file1[20],file2[20],file3[20];
 clrscr();
 printf("Enter name of first file\n");
 gets(file1);
 printf("Enter name of second file\n");
 gets(file2);
 printf("Enter name of file which'll store contents:\n");
 gets(file3);
 fs1=fopen(file1,"r");
 fs2 = fopen(file2,"r");
 if(fs1==NULL || fs2==NULL )
 {
  printf("Error opening file\n");
  getch();
  exit(1);
 }
 ft=fopen(file3,"w");
 if(ft==NULL )
 {
  printf("Error opening file\n");
  exit(1);
 }
 while((ch=fgetc(fs1))!=EOF)
  fputc(ch,ft);
 while((ch=fgetc(fs2))!=EOF)
  fputc(ch,ft);
 printf("Two files were merged into %s file \n",file3);
 fclose(fs1);
 fclose(fs2);
 fclose(ft);
 getch();
}

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C Program to count characters, lines, spaces & tabs in a file

#include<stdio.h>
#include<conio.h>
#include<stdlib.h>
void main()
{
 FILE *fp;
 char a[20];
 int nol=0,not=0,nob=0,noc=0;
 char c;
 clrscr();
 printf("Enter the name of File:\n");
 gets(a);
 if((fp=fopen(a,"r"))==NULL)
 {
  printf("File dosen't exist.");
 }
 else
 {
  while(1)
  {
   c=fgetc(fp);
   if(c==EOF)
      break;
   noc++;
   if(c==' ')
   nob++;
   if(c=='\n')
   nol++;
   if(c=='\t')
   not++;
  }
 }
 fclose(fp);
 printf("Number of characters = %d\n",noc);
 printf("Number of blanks = %d\n",nob);
 printf("Number of tabs = %d\n",not);
 printf("Number of lines = %d\n",nol);
 getch();
}

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How to find size of integer data type without using sizeof operator in c programming language

How to find size of integer data type without using sizeof operator in c programming language  

#include<stdio.h>

int main(){

  int *ptr = 0;

  ptr++;

  printf("Size of int data type:  %d",ptr);

  return 0;

}

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How to read STRING FROM text file by c program

#include<stdio.h>

int main(){

char str[70];

FILE *p;

if((p=fopen("string.txt","r"))==NULL){

printf("\nUnable t open file string.txt");

exit(1);

}

while(fgets(str,70,p)!=NULL)

puts(str);

fclose(p);

return 0;

}

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COPY DATA FROM ONE FILE TO ANOTHER FILE USING C PROGRAM

#include<stdio.h>

int main(){

  FILE *p,*q;

  char file1[20],file2[20];

  char ch;

  printf("\nEnter the source file name to be copied:");

  gets(file1);

  p=fopen(file1,"r");

  if(p==NULL){

      printf("cannot open %s",file1);

      exit(0);

  }

  printf("\nEnter the destination file name:");

  gets(file2);

  q=fopen(file2,"w");

  if(q==NULL){

      printf("cannot open %s",file2);

      exit(0);

  }

  while((ch=getc(p))!=EOF)

      putc(ch,q);

  printf("\nCOMPLETED");

  fclose(p);

  fclose(q);

 return 0;

}

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C code which prints initial of any name

C code which prints initial of any name

#include<stdio.h>

int main(){

   char str[20];

   int i=0;

   printf("Enter a string: ");

   gets(str);

   printf("%c",*str);

   while(str[i]!='\0'){

       if(str[i]==' '){

            i++;

            printf("%c",*(str+i));

       }

       i++;

   }

   return 0;

}

Sample output:

Enter a string: ANKITHA GOWDA

 AG

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Program for sorting of string in c language

Program for sorting of string in c language

#include<stdio.h>

int main(){

  int i,j,n;

  char str[20],temp[20];

  puts("Enter the no. of string to be sorted");

  scanf("%d",&n);

  for(i=0;i<=n;i++)

      gets(str[i]);

  for(i=0;i<=n;i++)

      for(j=i+1;j<=n;j++){

           if(strcmp(str[i],str[j])>0){

               strcpy(temp,str[i]);

              strcpy(str[i],str[j]);

              strcpy(str[j],temp);

           }

      }

  printf("The sorted string\n");

  for(i=0;i<=n;i++)

      puts(str[i]);

  return 0;

}

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Write a c program to find largest among three numbers using conditional operator

Write a c program to find largest among three numbers using conditional operator

#include<stdio.h>

int main(){

  int a,b,c,big;

  printf("\nEnter 3 numbers:");

  scanf("%d %d %d",&a,&b,&c);

  big=(a>b&&a>c?a:b>c?b:c);

  printf("\nThe biggest number is: %d",big);

  return 0;

}

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Write a c program to find largest among three numbers using binary minus operator

 

 Write a c program to find largest among three numbers using binary minus operator.

 

#include<stdio.h>

int main(){

    int a,b,c;

    printf("\nEnter 3 numbers: ");

    scanf("%d %d %d",&a,&b,&c);

    if(a-b>0 && a-c>0)

         printf("\nGreatest is a :%d",a);

    else

         if(b-c>0)

             printf("\nGreatest is b :%d",b);

         else

             printf("\nGreatest is c :%d",c);

    return 0;

}

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