# Find Whether Two Strings are Permutation of each other

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
```

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If you find anything incorrect or you feel that there is any better approach to solve the above problem, please write comment.
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