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23. Find Whether Two Strings are Permutation of each other
Objective: Given Two Strings, check whether one string is a permutation of other
Input: Two Strings Output: True or false based on whether strings are permutations of others 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.
Output:sumit and mtisu are permutation of each other? true xyzab and bayzxx are permutation of each other? false
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 exists in a hash map, if yes then increase its count by 1, and if it doesn't exist insert it into a hash map with the count as 1.
 Now navigate the second string taking each character at a time
 check if that character exists in the hash map, if yes then decrease its count by 1 and if it doesn't exist then return false.
 At the end navigate through the hash map and check if all the keys have 0 count against it if yes then return true else return false.
Output:
sumit and mtisu are permutation of each other? true xyzab and bayzxx are permutation of each other? false