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Anagrams are well defined and by that definition AABBC and CBBAA are anagrams.

Please define the question properly.

Why are last two strings not anagrams ?

i think the point of the question is to find only anagrams that are not palindromes

Please check the compilation errors. The logic is to take a Map Data structure, Keys will be the characters in Strings and value will be the frequency of characters. In the first run i will populate the map and in the second run i will depopulate it. If the map is empty then the strings are empty.

public boolean checkAnagram(String a, String b)
{
if(a.length()!= b.length())
return false;
HashMap<String, Integer> hm = new HashMap<String, Integer>();
for(int i = 0; i<a.length();i++)//Sorry for not using enahanced for loop.
{
if(!(hm.containsKey(a.charAt[i])))
hm.put(a.charAt[i],1); Putting the frequency as 1 for the first time
else
{
int val = hm.get(a.charAt[i]);// Not casting to Integer as it is automatically done by Java
hm.put(a.charAt[i],++val)// Increased the frequency by 1
}
}// close the first for loop
for (int i= 0; i <b.length();i++)
{
if(hm.containsKey(b.charAt[i])
{
int val = hm.get(b.charAt[i])
if (val == 1)
hm.remove(b.charAt[i])
else
hm.put(b.charAt[i],--val)
}
}// close the 2nd for loop
return hm.isEmpty()
}// Method close

Sort the two string and check if they are equal. We can use a variant of count sort if the characters are ASCII. Btw, last two are anagrams (by general definition of anagram)

1. Store the count of each charcter in S1 into a seperate Hashtable
2. Iterate through each character of s2 and decrement the count of the corresponding key, once the count=1 remove that element. Also check if that key exists in that hastable, if not its not an anagram.
Orde is O(n)

-- Psuedo code explained on Nishank Rajvanshi Program

It's simple, I guess wats intended is that the relative distance or displacement of characters should remain the same. So the soln for this would be to append the same string in itself and now start checking if the original string is present in this String or not.

what is the meaning of Anagram?

bool IsAnagram(char* s1, char* s2) {
char hash[256] = { 0 };
if (strlen(s1) != strlen(s2)) return false;
for(int i=0;i<strlen(s1);i++)
hash[s1[i]]++;
for(int i=0;i<strlen(s2);i++) {
if(s1[i] == s2[i] ||
--hash[s2[i]] < 0) return false;
}
return true;
}

from collections import Counter,OrderedDict
def anagramcheck(str1,str2):

if OrderedDict(Counter(str1))==OrderedDict(Counter(str2)):
return True
else:
return False

answer=anagramcheck('AABaaBC','CBBAAaa')
print answer

from collections import Counter,OrderedDict
def anagramcheck(str1,str2):

    if OrderedDict(Counter(str1))==OrderedDict(Counter(str2)):
        return True
    else:
        return False

answer=anagramcheck('AABaaBC','CBBAAaa')
print answer

from collections import Counter,OrderedDict
def anagramcheck(str1,str2):

    if OrderedDict(Counter(str1))==OrderedDict(Counter(str2)):
        return True
    else:
        return False

answer=anagramcheck('AABaaBC','CBBAAaa')
print answer

1. Store the count of each charcter in S1 into a seperate
2. Iterate through each character of s2 and decrement the count of the corresponding key, once the count=1 remove that element. Also check if that key exists in that hastable, if not its not an anagram.
Orde is O(n)

-- Psuedo code explained on Nishank Rajvanshi Program