DZone Snippets is a public source code repository. Easily build up your personal collection of code snippets, categorize them with tags / keywords, and share them with the world

Snippets has posted 5883 posts at DZone. View Full User Profile

Fast Anagram Determination

05.14.2008
| 17148 views |
  • submit to reddit
        // This function will determine whether the two C strings provided are anagrams or not. Only alphabetical strings are considered.

/******************************************************************************
 * This function is a code snippet. It can be freely used and distributed.
 * Author: irfan[dot]hamid[at]gmail[at]com
 *
 * This function takes two ASCII strings in C syntax (null-terminated
 * character arrays) and determines whether they are anagrams or not.
 *
 * Implementation notes:
 * The function contains a histogram that stores the frequency of each 
 * character it encounters in the first string. For each character of the 
 * second string, it decrements the corresponding entry in the histogram. If 
 * before every decrement, the value of the histogram bucket reaches zero, 
 * then the two strings are not anagrams, as there is some character that has
 * more occurrences in the second string than in the first string.
 *
 * If either of the two strings contains a non-alphabetic character, the
 * anagram property is considered to be violated.
 *
 * Remarks:
 * It is an efficient implementation because:
 *   o It is in-place, the original strings are not copied, no alloc or copy
 *   o Does not clobber the original strings
 *   o It is a linear time implementation
 *
 * Returns:
 * True if the strings are anagrams, false otherwise.
******************************************************************************/

bool is_anagram (char w1[], char w2[])
{
	unsigned int i, sz;

	/* The histogram */
	int freqtbl[26];

	/* Sanity check */
	if ((sz = strlen(w1)) != strlen(w2))
		return false;

	/* Initialize the histogram */
	bzero(freqtbl, 26*sizeof(int));

	/* Read the first string, incrementing the corresponding histogram entry */
	for (i = 0; i < sz; i++) {
		if (w1[i] >= 'A' && w1[i] <= 'Z')
			freqtbl[w1[i]-'A']++;
		else if (w1[i] >= 'a' && w1[i] <= 'z')
			freqtbl[w1[i]-'a']++;
		else
			return false;
	}

	/* Read the second string, decrementing the corresponding histogram entry */
	for (i = 0; i < sz; i++) {
		if (w2[i] >= 'A' && w2[i] <= 'Z') {
			if (freqtbl[w2[i]-'A'] == 0)
				return false;
			freqtbl[w2[i]-'A']--;
		} else if (w2[i] >= 'a' && w2[i] <= 'z') {
			if (freqtbl[w2[i]-'a'] == 0)
				return false;
			freqtbl[w2[i]-'a']--;
		} else {
			return false;
		}
	}

	return true;
}