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Check Permutation

Easy

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Average time to solve is 15m

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Problem statement

You have been given two strings 'STR1' and 'STR2'. You have to check whether the two strings are anagram to each other or not.

Note:

Two strings are said to be anagram if they contain the same characters, irrespective of the order of the characters.

Example :

If 'STR1' = “listen” and 'STR2' = “silent” then the output will be 1.

Both the strings contain the same set of characters.

Detailed explanation ( Input/output format, Notes, Images )

Input Format:

The first line contains an integer ‘T’ which denotes the number of test cases. 

The first and only line of each test case contains two space-separated strings 'STR1' and 'STR2', respectively.

Output Format:

For each test case, return true if the two strings are anagrams of each other else return false.

Print the output of each test case in a separate line.

Constraints:

1 <= T <= 100
1 <= |STR1|, |STR2| <= 10^3

Where |STR1| and |STR2| are the lengths of the string 'STR1' and 'STR2' respectively.

Time limit: 1 sec

Sample Input 1:

2
listen silent
east eats

Sample Output 1:

1
1

Explanation for Sample Output 1:

In test case 1, "listen" and "silent" has same set of characters.

In test case 2, "east" and "eats" has same set of characters.

Sample Input 2:

2
triangle integral
hearts earth

Sample Output 2:

1
0

Explanation for Sample Output 1:

In test case 1, "triangle" and "integral" has same set of characters.

In test case 2, "hearts" and "earth" does not have same set of characters.

Hint 1

Hint 2

Can sorting the strings help?

Approaches (2)

Approach 1

Approach 2

Brute-Force

The basic idea of this approach is to sort both the strings and compare the characters of the sorted strings.

Sort both strings.
Compare the characters of the sorted strings.

Time Complexity

O(N * log(N)), Where ‘N’ is the length of the strings.

Since we are sorting the strings which have time complexity O(N * log(N)).

Space Complexity

O(1)

Since we not using any extra space, the overall space complexity will be O(1).

(75% EXP penalty)

(100% EXP penalty)

Check Permutation

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