CareerCup

i think xoring of all integers values of characters should be 0 if they are anagram.

// Write a method to decide if two strings are anagrams or not.
	// Complexity O ( 2n + k)  avg, since s1 and s2 should contains the same length otherwise It will return false immediate
	public static boolean isAnagram(char s1[], char s2[]){
		if(s1.length != s2.length) return false;
		int letters [] = new int[256];
		for( int i = 0; i < s1.length; i++){
			letters[toUpperCase(s1[i])]++;
		}
		for( int j = 0; j < s2.length; j++){
			letters[toUpperCase(s2[j])]--;
			if(letters[toUpperCase(s2[j])] < 0) return false;
		}
		for(int x = 0; x < letters.length; x++){
			if(letters[x] != 0 ) return false;
		}
		return true;
	}

Just sort the 2 strings and compare, O(nlogn)

public static boolean isAnagram(String word, String anagram) { 
		if(word.length() != anagram.length())
			return false; 
		char[] chars = word.toCharArray(); 
		for(char c : chars)
		{ 
			int index = anagram.indexOf(c); 
			if(index != -1){ 
				anagram = anagram.substring(0,index) + 
						anagram.substring(index +1, anagram.length()); 
				}
			else { 
				return false; 
				} 
		} 
		return anagram.isEmpty(); 
	}

Time Complexity: O(n)

public static boolean IsAnagram(String str1, String str2){
if(str1.length() != str2.length()){
return false;
}
char[] charArr1 = str1.toLowerCase().toCharArray();
char[] charArr2 = str2.toLowerCase().toCharArray();

Character c;
int hashCodeValue1 =0;
int hashCodeValue2 =0;
for(int i=0;i<charArr1.length;i++){
c = charArr1[i];
hashCodeValue1+=c.hashCode();
}
for(int i=0;i<charArr2.length;i++){
c = charArr2[i];
hashCodeValue2+=c.hashCode();
}

if(hashCodeValue1 == hashCodeValue2)
return true;

return false;
}

i don't know how to code yet, but here are the steps i think are needed:
1) take both strings in two seperate variables
2) compare length of variables if yes then goto next step
3) reverse any string, then compare with the other string. if yes then anagram