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Map<String, List<String>> anagrammsMap = ....

public void addAnagramm(String str){
char[] strArr = str.toCharArray();
Arrays.sort( strArr );
String key = new String( strArr );
List<String> anagramms = anagrammsMap.get( key );
if( anagramms == null ){
anagrammsMap.put(key, anagramms);
}
anagramms.add( str );
}

ohh!!!!
sorry i did the palindrome..
thanks
:)

static int IsAnagrams(string str1, string str2)
{
str1 = str1.ToLower();
str2 = str2.ToLower();

if (str1.Length != str2.Length)
{
Console.WriteLine("Strings are not anagrams of each other the lengths are not equal");
return 0;
}

Hashtable ht1 = new Hashtable();
for (int i = 0; i < str1.Length; i++)
{
ht1.Add(i,str1[i]);
}

Hashtable ht2 = new Hashtable();
for (int i = 0; i < str2.Length; i++)
{
ht2.Add(i, str2[i]);
}
int countsareequal = 0;
foreach (char c in str1)
{
if (ht1[c] == ht2[c])
{
countsareequal += 1;
}
}

if (countsareequal == str1.Length)
{
Console.WriteLine("two strings are anagrams");
return 1;
}
else
{
Console.WriteLine("two strings are anagrams");
return 0;
}

}

Space: O(1)
Time: O(n)

#include <stdio.h>
int main(int argc, char *argv[])
{
        char map[255] = { 0 }, *answ = "YES";
        int i,len1 = strlen(argv[1]), len2 = strlen(argv[2]);
        if(len1 == len2) {
                for(i = 0; i < len1;i++)
                        map[argv[1][i]]++;
                for(i=0;i < len2;i++)
                        map[argv[2][i]]--;
                for(i = 0; i < sizeof(map)/sizeof(map[0]); i++)
                        if(map[i] != 0) {
                                answ = "NO";
                                break;
                        }
        }
        else
                answ = "NO";
        printf("%s\n",answ);
        return (0);
}

void anagram(char *d, char *c, int length)
{
int i=0;
while(length--)
{d[i]=c[length];
i++;}
d[i]='\0';
}

XOR All the characters of "Str1" and "Str2" , If result is zero then it is an anagram.