**Objective:** Given Two Strings, check whether one string is permutation of other

**Input:** Two Strings

**Output: **True or false based on whether strings are permutation of other or not.

**Example: **

"sumit" and "tiums" are permutations of each other. "abcd" and bdea" are not permutations of each other.

**Approach:
**

**Method 1: Time Complexity – O(nlgn)**

Sort both the strings and compare it.

**Method 2 : Using Hash Map- Time Complexity – O(n)
**

- Check if both Strings are having the same length, if not , return false.
- Create a Hash Map, make character as key and its count as value
- Navigate the string one taking each character at a time
- check if that character already existing in hash map, if yes then increase its count by 1 and if it doesn’t exist insert into hash map with the count as 1.
- Now navigate the second string taking each character at a time
- check if that character existing in hash map, if yes then decrease its count by 1 and if it doesn’t exist then return false.
- At the end navigate through hash map and check if all the keys has 0 count against it if yes then return true else return false.

**Complete Code for Both Methods:**

**Output:**

sumit and mtisu are permutation of each other? true xyzab and bayzxx are permutation of each other? false