not a big deal, but make sure to check the lengths for #3 (or for really any of the solutions). 1. sorting and strcmp is straightforward and no extra memory, but n log n. 2. not too bad for memory. O(n) 3. could get nasty with super long strings. O(n)

#2 and #3 are basically the same thing. the first thing in any algorithm here is to check lengths. the next thing i would do it make an array of ints the length of the range of input. then, i would read the first one in, incrementing the corresponding integer for its position. then compare this list of character frequency with the second string for a match. if you do not know the range of the input, or it is otherwise prohibitive, you can do a has from each character object to the integer keeping record of its frequency.

more clearly: boolean areAnagrams?( String s1, String s2) { int[] frequencies = new int[128]; //assuming ascii. make a hash table for unicode for( int i = 0; i < frequencies.length; i++) { frequencies[ i ] = 0; } for(int i = 0; i < s1.length(); i++) { frequencies[ (int)s1.charAt(i) ]++; }//now we have an int array corresponding to letter frequencies for(int i = 0; i < s2.length(); i++) { frequencies[ (int)s2.charAt(i) ]--; }//now, if they are anagrams, all will be zero for(int = 0; i < s1.length(); i++) { if( frequencies[ (int)s1.charAt(i) ] ) { return false; } } return true; }