### 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: