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++

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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);
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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

chevron_right


Python3

filter_none

edit
close

play_arrow

link
brightness_4
code

# 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

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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

chevron_right


PHP

$y)
for ($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 ?>

Output:

y


My Personal Notes arrow_drop_up

Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.





Article Tags :
Practice Tags :


Be the First to upvote.


Please write to us at contribute@geeksforgeeks.org to report any issue with the above content.