**QUESTION**:

**Check Permutation**: Given two strings, write a method to decide if one is a permutation of the other.

**SOLUTION:**

- Like in many questions, we should confirm some details with our interviewer.
- We should understand if the permutation comparison is case sensitive. That is: is God a permutation of dog? Additionally, we should ask if whitespace is significant.
- We will assume for this problem that the comparison is case sensitive and whitespace is significant. So, “god ” is different from “dog”.
- Observe first that strings of different lengths cannot be permutations of each other. There are two easy ways to solve this problem, both of which use this optimization.

**Solution #1: Sort the strings.**

- If two strings are permutations, then we know they have the same characters, but in different orders.
- Therefore, sorting the strings will put the characters from two permutations in the same order.
- We just need to compare the sorted versions of the strings

String sort(String s) { char[] content= s.toCharArray(); java.util.Arrays.sort(content); return new String(content); } boolean permutation(String s, String t) { if (s.length() != t.length()) { return false; } return sort(s).equals(sort(t)); }

- Though this algorithm is not as optimal in some senses, it may be preferable in one sense: It’s clean, simple and easy to understand. In a practical sense, this may very well be a superior way to implement the problem.
- However, if efficiency is very important, we can implement it in a different way.

**Solution #2: Check if the two strings have identical character counts.**

- We can also use the definition of a permutation-two words with the same character counts-to implement this algorithm. We simply iterate through this code, counting how many times each character appears.
- Then, afterwards, we compare the two arrays.

boolean permutation(String s, String t) { if (s.length() != t.length()) { return false; } int[] letters = new int[128]; II Assumption char[] s_array = s.toCharArray(); for (char c : s_array) { // count number of each char in s. letters[c]++; } for (int i= 0; i < t.length(); i++) { int c= (int) t.charAt(i); letters[c]--; if (letters[c] < 0) { return false; } } return true; }

Note the assumption on line 6. In your interview, you should always check with your interviewer about the size of the character set. We assumed that the character set was ASCII.