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 x and 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.

Below is the implementation of the above approach: 

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;

// Function to return the modified string
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;
}

// Driver Program to test above function
int main()
{
    string s = "xxyyxyy";
    cout << printFinalString(s);
}

// Java implementation of the approach
class GFG
{
// Function to return the modified String
static 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.charAt(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;
}

// Driver Code
public static void main(String args[])
{
    String s = "xxyyxyy";
    System.out.println(printFinalString(s));
}
}

// This code is contributed 
// by 29AjayKumar

# Python 3 implementation of the approach

# Function to return the modified string
def prFinalString(s):
    i, n = 0, 0
    n = len(s)
    x, y = 0, 0
    for i in range(0, n):
        
        # Count number of 'x'
        if (s[i] == 'x'):
            x += 1

        # Count number of 'y'
        else:
            y += 1

    finalString = ""

    # min(x, y) number of 'x' and 
    # 'y' will be deleted
    if (x > y):
        for i in range(0, x - y):
            finalString += "x"
    else:
        for i in range(0, y - x):
            finalString += "y"

    return finalString

# Driver Code
if __name__ == '__main__':
    s = "xxyyxyy"
    print(prFinalString(s))

# This code contributed by 29AjayKumar

// C# implementation of the approach
using System;

class GFG
{
// Function to return the modified String
static 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;
}

// Driver Code
public static void Main()
{
    string s = "xxyyxyy";
    Console.WriteLine(printFinalString(s));
}
}

// This code is contributed 
// by Akanksha Rai

<?php
// PHP implementation of the approach

// Function to return the modified string
function printFinalString($s)
{
    $n = strlen($s);
    $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++;
    }

    $finalString = (string)null;

    // 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;
}

// Driver Code
$s = "xxyyxyy";
echo printFinalString($s);

// This code is contributed by ihritik
?>

<script>

// Javascript implementation of the approach

// Function to return the modified string
function printFinalString(s)
{
    var i, n;
    n = s.length;
    var 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++;
    }

    var 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;
}

// Driver Program to test above function
var s = "xxyyxyy";
document.write( printFinalString(s));

// This code is contributed by famously.
</script>

y

Time Complexity: O(n), where n is the length of the given string.
Auxiliary Space: O(n), where n is the length of the given string.