Final string after performing given operations

https://www.geeksforgeeks.org/final-string-after-performing-given-operations/

Given a string str containing only characters x and y, the task is to perform the following operations while possible:
Find an index such that s[i] = ‘x’ and s[i+1] = ‘y’ and delete both the characters s[i] and s[i+1], if no such index is found then find an index such that s[i] = ‘y’ and s[i+1] = ‘x’ and swap(s[i], s[i+1]).
Print the final string after performing the given operation.

Examples:

Input: str = “xyyxx”
Output: x
Step 1: yxx (xy got deleted)
Step 2: xyx (yx got swapped)
Step 3: x (xy got deleted)

Input: str = “xxyyxyy”
Output: y

Approach: In the final string there will be either only x or only y because if we have both xand y in the string then there would be a point where we have either xy or yx as sub-string.

• If it’s xy, we delete it straight away.
• If it is yx, we reverse it to get xy and then delete it.

Since in each deletion step, one x and one y gets deleted, the final string would have min(x, y) number of x and y characters deleted.

string printFinalString(string s)

{

    int i, n;

    n = s.length();

    int x = 0, y = 0;

    for (i = 0; i < n; i++) {

  

        // Count number of 'x'

        if (s[i] == 'x')

            x++;

  

        // Count number of 'y'

        else

            y++;

    }

  

    string finalString = "";

  

    // min(x, y) number of 'x' and 'y' will be deleted

    if (x > y)

        for (i = 0; i < x - y; i++)

            finalString += "x";

    else

        for (i = 0; i < y - x; i++)

            finalString += "y";

  

    return finalString;

}




https://www.geeksforgeeks.org/final-state-of-the-string-after-modification/