Program 83:To know the frequency of each character in sentence

Program 83:

#include<stdio.h>
#include<string.h>
main()
{
    int i,k=0,count[26]={0},x;
    char str1[100];
    printf("Enter a sentence\n");
    gets(str1);
    
    while(str1[k]!='\0')
    {
        if(str1[k]>='a'&&str1[k]<='z')
        {
            x=str1[k]-'a';
            count[x]++;
        }
        if(str1[k]>='A'&&str1[k]<='Z')
        {
           x=str1[k]-'A';
            count[x]++;
        }
        k++;
    }
    for(i=0;i<26;i++)
    {
        if(count[i]!=0)
        {
            printf("%c occured %d times\n",i+'a',count[i]);
        }
       
    }

}

Program to print Frequency of digits in an Array

Explanation:

1. This program starts with initializing :
   • str1[100] → To store string with length of 100 which means it can store 100 letters
   • i ,k,x→Used as helping variable
   • count[26]→ To count number of times each letter is repeated from a-z

2.  printf("Enter a sentence\n");
       gets(str1);

   Taking string from the user. 
3.  

   while(str1[k]!='\0')
       {
           if(str1[k]>='a'&&str1[k]<='z')//1st if part
           {
               x=str1[k]-'a';
               count[x]++;
           }
           if(str1[k]>='A'&&str1[k]<='Z')//second if part
           {
              x=str1[k]-'A';
               count[x]++;
           }
           k++;
       }

   Lets take a small example str1="Hello".count[26] means each one from count[0] to count[26] are initialized to zero.
   • Iteration 1:k=0,str1[0]=H which is not '\0'(i.e. not end of string) so the loop is executed
     • 'H' lies between 'A' and 'Z' so the second if part is executed.(refer ascii sheet)
     • x=str1[k]-'A' →str1[0]-'A'→'H'-'A'→x=72-65=7.
     • 'H'-'A' will give integer value as output as the output of the result is stored in integer variable.Where 'H' ascii value is 72 and that of 'A' is 65 so 72-65 is 7
     • Therefore,x=7
     • count[7]++→count[7]=1 as it is previously initilized to 0.
     • now k++ so k is incemented by 1,then k=1.To move to the next character/letter.
   • Iteration 2:k=1,str1[1]=e which is not '\0'(i.e. not end of string) so the loop is executed
     • 'e' lies between 'a' and 'z' so the first if part is executed.
     • x=str1[k]-'a' →str1[1]-'a'→'e'-'a'→x=101-97=4.
     • 'e'-'a' ('e' ascii value is 101 and that of 'a' is 97 so 101-97 is 4)
     • Therefore,x=4
     • count[4]++→count[4]=1 as it is previously initilized to 0.
     • now k++ so k is incemented by 1,then k=2.To move to the next character/letter.
   • Iteration 3:k=2,str1[2]=l which is not '\0'(i.e. not end of string) so the loop is executed
     • 'l' lies between 'a' and 'z' so the first if part is executed.
     • x=str1[k]-'a' →str1[2]-'a'→'l'-'a'→x=108-97=11.
     • 'l'-'a' ('l' ascii value is 108 and that of 'a' is 97 so 108-97 is 11)
     • Therefore,x=11
     • count[11]++→count[11]=1.
     • now k++ so k is incemented by 1,then k=3.
   • Iteration 4:k=3,str1[3]=l (again) which is not '\0'(i.e. not end of string) so the loop is executed
     • 'l' lies between 'a' and 'z' so the first if part is executed.
     • x=str1[k]-'a' →str1[3]-'a'→'l'-'a'→x=108-97=11.
     • 'l'-'a' ('l' ascii value is 108 and that of 'a' is 97 so 108-97 is 11)
     • Therefore,x=11
     • count[11]++→count[11]=1+1=2.
     • now k++ so k is incemented by 1,then k=4.
   • Iteration 5:k=4,str1[4]=o which is not '\0'(i.e. not end of string) so the loop is executed
     • 'o' lies between 'a' and 'z' so the first if part is executed.
     • x=str1[k]-'a' →str1[4]-'a'→'o'-'a'→x=111-97=14.
     • 'o'-'a' ('o' ascii value is 111 and that of 'a' is 97 so 111-97 is 14)
     • Therefore,x=14
     • count[14]++→count[14]=1.
     • now k++ so k is incemented by 1,then k=5.
   • Iteration 6:k=5,str1[5]is the end of string '\0'(i.e. not end of string) so the loop is terminated.
   • Finally values are 
     
     • count[0], count[1].....count[3]=0
     •  count[4]=1
     •  count[5]...count[6]=0
     •  count[7]=1
     •  count[8]....count[10]=0
     • count[11]=2 
     • count[12]..count[13]=0
     •  count[14]=1
     •  count[15]..count[25]=0
     • *-Remember either the letter is 'H' or 'h' both their count is recorded in same array i.e if 'h' or 'H' both their total count occurs 4 times then count[7]=4 as in (hHhH) where h-2 times and H-2 times so a total of 4 times

4.  for(i=0;i<26;i++)
       {
           if(count[i]!=0)
           {
               printf("%c occured %d times\n",i+'a',count[i]);
           }
          
       }

   count array whose values are not zero is printed by comparing one by one from 0 to 25
5. For example count[7]=1 then
   

   "%c occured %d times\n",i+'a',count[i]

   i+'a' means 7+'a'=7+97=104 whose value in ascii is 'h'.So h occured count[7] times.Which means h occured 1 times. In this way it will check from count[0] to count[25] and if they are not zero then that value is printed

 Output: