Check if two strings are permutation of each other

String, Iterative approach

Write a function to check whether two given strings are Permutation of each other or not. A Permutation of a string is another string that contains same characters, only the order of characters can be different. For example, “abcd” and “dabc” are Permutation of each other.

Solutions

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".

Approach – 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.

Note: Please note that the program is case-sensitive.

Steps

1. First check if any string is null. If both null, then return true else any one is null then return false.

2. Check the length of string. If length is different, then return false.

3. Sort both strings alphabetically.

4. Run a loop till length on any string and compare each character of both strings index wise.

     public bool ArePermutation(string str1, string str2)
    {
        if (str1 is null && str2 is null)
            return true;

        if (str1?.Length != str2?.Length)
            return false;

        var char1 = str1!.ToArray();
        var char2 = str2!.ToArray();

        Array.Sort(char1);
        Array.Sort(char2);

        for (int i = 0; i < str1!.Length; i++)
            if (char1[i] != char2[i])
                return false;

        return true;
    }

Complexity

• Time complexity: O(N log N)

• Space complexity: O(1)

Approach – 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.

Note: Please note that the program is case-sensitive.

Steps

1. First check if any string is null. If both null, then return true else any one is null then return false.

2. Check the length of string. If length is different, then return false.

3. Create an integer array of ASCII characters length (256).

4. Run a loop till length on any string.

5. Get character from first string and increase that number count and get character from second string and decrease that number count.

6. Run a loop till length of an integer array and check if any number count is not equal to zero. If any number count is not zero, then return false else true.

   public bool ArePermutation(string str1, string str2)
    {
        int NO_OF_CHARS = 256;
        if (str1 is null && str2 is null)
            return true;

        if (str1?.Length != str2?.Length)
            return false;

        var temp = new int[NO_OF_CHARS];

        for (int i = 0; i < str1!.Length; i++)
        {
            temp[str1![i]]++;
            temp[str2![i]]--;
        }

        for (int i = 0; i < temp.Length; i++)
        {
            if (temp[i] != 0)
                return false;
        }
        return true;
    }

Complexity

• Time complexity: O(N)

• Space complexity: O(N)