Why I replaced the solution?

The original solution works. but runs at complexity of O(N*log(N))
whereas N := max(s1.length(), s2,length) [maximum length of two Strings]
the complexity of N*log(N) is due to sorting the character array.
my solution runs at a linear complexity, O(N).
instead of sorting, I use an HashMap, and 3 successive loops.

My solution

1. I initialize the hash map.
2. first loop: I count the number of appearances in the first String and update the hashmap. I do that by iterating through the characters. if a character has not appeared yet, I put the value of 1. otherwise I add 1 to its value.
3. second loop: I iterate through each character in the second string. if a character does not appear in the hashmap. I return false. Otherwise, I reduce its value by 1.
4. third loop: I iterate over the values in the hashmap. If there is a different value rather than 0. I return false.
5. if we reached to the end of the function then return true.