C Program to Count the Occurrences of a Word in a String

This article will teach you and provide you with code to count the occurrences of a word in a string using the C programming language. The user must enter the word and string here (at run-time). For example, if the given string is:

It was fresherearth. It is fresherearth. It will be fresherearth

then the occurrence of the word

fresherearth is 3
It is 3
was is 1
is is 1
will is 1
be is 1

Now let's move on and implement it in a C program.

Counting Words in a String

To count the occurrence of a word in a string or sentence in C programming, you have to ask the user to enter the string and a word, and then count how many times the given word is available in the given string, as shown in the program given below:

#include<stdio.h>
#include<conio.h>
#include<string.h>
int main()
{
   char str[100], word[20];
   int i, j, ls, lw, temp, countW=0, chk;
   printf("Enter the String: ");
   gets(str);
   printf("Enter a Word: ");
   gets(word);
   ls = strlen(str);
   lw = strlen(word);
   for(i=0; i<ls; i++)
   {
      temp = i;
      for(j=0; j<lw; j++)
      {
         if(str[i]==word[j])
            i++;
      }
      chk = i-temp;
      if(chk==lw)
         countW++;
      i = temp;
   }
   printf("\nOccurrence = %d", countW);
   getch();
   return 0;
}

The program was written in the Code::Blocks IDE. Here is its sample run after a successful build and run:

c program count number of words in sentence

Now supply any string, say

It was fresherearth. It is fresherearth. It will be fresherearth

And then enter a word, say "fresherearth," to count how many times this word occurred in the given string:

Program Explained

The two variables, ls and lw, are used to hold the length of string and word, respectively.
The first for loop runs from the first character to the last character of the string.
And the second for loop runs from the first character to the last character of the given word.
In the second for loop, we have checked whether the given word is present inside the string or not, using character-by-character matching.
That is, if the first character of a word gets found in the string at any index, then it checks for a second match. That is, match the second character of a word with the character just after the previous index, and so on.
For example, let's suppose the given word is "fresherearth." So its length is 12. Therefore, if all characters continuously get matched in the given string, then the value of i will be (its previous value + 12).
And after exiting the second for loop, using i-temp or (i's previous value + 12)-temp gets initialized to chk.
If the previous value of i was 7, the value of i will be 7 + 12 or 19.
And 19-temp or 19-7 equals 12. As a result, 12 is set up to chk.
Now check whether the value of chk is equal to the length of the word or not.
If it is equal, then the word gets found, so we have to increment the value of the countW variable, which holds the occurrence of the word.
Finally, initialize the value of temp (the value of the previous i before the second for the loop).
And so on until you reach the last character of the string.

This program has a limitation: if you enter a string, such as "this is fresherearth," and a word, such as "is," the occurrence will be displayed as 2. Because "is" from "this" will be counted as well. To overcome this problem, let me create another program for you.

Complete Version of the Counting a Word in a String Program

Here is the complete version of the previous program:

#include<stdio.h>
#include<conio.h>
#include<string.h>
int main()
{
   char str[100], word[20];
   int i, j, ls, lw, temp, countW=0, chk, doIncrement;
   printf("Enter the String: ");
   gets(str);
   printf("Enter a Word: ");
   gets(word);
   ls = strlen(str);
   lw = strlen(word);
   for(i=0; i<ls; i++)
   {
      temp = i;
      doIncrement = 0;
      for(j=0; j<lw; j++)
      {
         if(str[i]==word[j])
         {
            if(temp>0 && (temp+lw)<ls)
            {
               if(str[temp-1]== ' ' && str[temp+lw]==' ')
                  doIncrement=1;
            }
            else if(temp==0 && (temp+lw)<ls)
            {
               if(str[temp+lw]==' ')
                  doIncrement=1;
            }
            else if(temp>0 && (temp+lw)==ls)
            {
               if(str[temp-1]== ' ')
                  doIncrement=1;
            }
            if(doIncrement==1)
               i++;
            else
               break;
         }
      }
      chk = i-temp;
      if(chk==lw)
         countW++;
      i = temp;
   }
   printf("\nOccurrence = %d", countW);
   getch();
   return 0;
}

Now let's check it out with string input as "this is is is isth fresherearth," and word input as "is."

The same program in different languages

C Quiz

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