Anagram program in C
Anagram program in C to check whether two strings are anagrams or not. The strings are assumed to contain only lower case letters. The strings are anagrams of each other if the letters of one string can be rearranged to form the other string. So, in anagram strings, all characters occur the same number of times. For example, “ABC” and “CAB” are anagrams, as every character ‘A,’ ‘B,’ and ‘C’ occur the same number of times (one time here) in both the strings.
A user inputs two strings, we count how many times each letter (‘a’ to ‘z’) appear in them and then compare their corresponding counts. The frequency of an alphabet in a string is how many times it appears in the string. For example, the frequency of letter ‘m’ in the string “programming” is ‘2’ as it’s present two times in it.
C anagram program
#include <stdio.h>
int check_anagram(char [], char []);
int main()
{
char a[1000], b[1000];
printf(“Enter two strings\n“);
gets(a);
gets(b);
if (check_anagram(a, b))
printf(“The strings are anagrams.\n“);
else
printf(“The strings aren’t anagrams.\n“);
return 0;
}
int check_anagram(char a[], char b[])
{
int first[26] = {0}, second[26] = {0}, c=0;
// Calculating frequency of characters of the first string
while (a[c] != ‘\0‘) {
first[a[c]–‘a’]++;
c++;
}
c = 0;
while (b[c] != ‘\0‘) {
second[b[c]–‘a’]++;
c++;
}
// Comparing the frequency of characters
for (c = 0; c < 26; c++)
if (first[c] != second[c])
return 0;
return 1;
}
Output of C anagram program:
