Lexicographically smallest string formed by appending a character from the first K characters of a given string

Given a string S consisting of lowercase alphabets. The task is to find the lexicographically smallest string X of the same length only that can be formed using the operation given below:

In a single operation, select any one character among the at most first K characters of string S, remove it from string S and append it to string X. Apply this operation as many times as he wants.

Examples:

Input: str = “gaurang”, k=3 
Output: agangru 
Remove ‘a’ in the first step and append to X. 
Remove ‘g’ in the second step and append to X. 
Remove ‘a’ in the third step and append to X. 
Remove ‘n’ in the third step and append to X. 
Pick the lexicographically smallest character at every step from the first K characters to get the 
string “agangru”

Input: str = “geeksforgeeks”, k=5 
Output: eefggeekkorss

Approach: 

• Find the smallest character in the first k characters in the string S.
• Delete the smallest character found from the string.
• Append the smallest character found to the new string X.
• Repeat the above steps till the string s is empty.

Below is the implementation of the above approach: 

C++

// C++ program to find the new string
// after performing deletions and append
// operation in the string s
#include <bits/stdc++.h>
using namespace std;

// Function to find the new string thus
// formed by removing characters
string newString(string s, int k)
{
    // new string
    string X = "";

    // Remove characters until
    // the string  is empty
    while (s.length() > 0) {

        char temp = s[0];

        // Traverse to find the smallest character in the
        // first k characters
        for (long long i = 1; i < k and i < s.length(); i++) {
            if (s[i] < temp) {
                temp = s[i];
            }
        }

        // append the smallest character
        X = X + temp;

        // removing the lexicographically smallest
        // character from the string
        for (long long i = 0; i < k; i++) {
            if (s[i] == temp) {

                s.erase(s.begin() + i);
                break;
            }
        }
    }

    return X;
}

// Driver Code
int main()
{

    string s = "gaurang";
    int k = 3;

    cout << newString(s, k);
}

Java

// Java program to find the new string
// after performing deletions and append
// operation in the string s

class GFG {

// Function to find the new string thus
// formed by removing characters
    static String newString(String s, int k) {
        // new string
        String X = "";

        // Remove characters until
        // the string  is empty
        while (s.length() > 0) {

            char temp = s.charAt(0);

            // Traverse to find the smallest character in the
            // first k characters
            for (int i = 1; i < k && i < s.length(); i++) {
                if (s.charAt(i) < temp) {
                    temp = s.charAt(i);
                }
            }

            // append the smallest character
            X = X + temp;

            // removing the lexicographically smallest
            // character from the string
            for (int i = 0; i < k; i++) {
                if (s.charAt(i) == temp) {

                    s = s.substring(0, i) + s.substring(i + 1);
                    //s.erase(s.begin() + i);
                    break;
                }
            }
        }

        return X;
    }
// Driver code

    public static void main(String[] args) {
        String s = "gaurang";
        int k = 3;

        System.out.println(newString(s, k));

    }
}

// This code contributed by Jajput-Ji

Python3

# Python 3 program to find the new string
# after performing deletions and append
# operation in the string s

# Function to find the new string thus
# formed by removing characters
def newString(s, k):

    # new string
    X = ""

    # Remove characters until
    # the string is empty
    while (len(s) > 0):
        temp = s[0]

        # Traverse to find the smallest
        # character in the first k characters
        i = 1
        while(i < k and i < len(s)):
            if (s[i] < temp):
                temp = s[i]

            i += 1

        # append the smallest character
        X = X + temp

        # removing the lexicographically
        # smallest character from the string
        for i in range(k):
            if (s[i] == temp):
                s = s[0:i] + s[i + 1:]
                break

    return X

# Driver Code
if __name__ == '__main__':
    s = "gaurang"
    k = 3
    print(newString(s, k))

# This code is contributed by
# Shashank_Sharma

C#

// C# program to find the new string
// after performing deletions and
// append operation in the string s
using System;

class GFG
{

// Function to find the new string thus
// formed by removing characters
static String newString(String s, int k)
{
    // new string
    String X = "";

    // Remove characters until
    // the string is empty
    while (s.Length > 0)
    {
        char temp = s[0];

        // Traverse to find the smallest
        // character in the first k characters
        for (int i = 1; i < k && i < s.Length; i++)
        {
            if (s[i] < temp)
            {
                temp = s[i];
            }
        }

        // append the smallest character
        X = X + temp;

        // removing the lexicographically smallest
        // character from the string
        for (int i = 0; i < k; i++)
        {
            if (s[i] == temp)
            {

                s = s.Substring(0, i) + s.Substring(i + 1);
                //s.erase(s.begin() + i);
                break;
            }
        }
    }

    return X;
}

// Driver code
public static void Main(String[] args)
{
    String s = "gaurang";
    int k = 3;

    Console.Write(newString(s, k));
}
}

// This code contributed by Rajput-Ji

PHP

<?php
// PHP program to find the new string
// after performing deletions and
// append operation in the string s

// Function to find the new string thus
// formed by removing characters
function newString($s, $k)
{
    // new string
    $X = "";

    // Remove characters until
    // the string is empty
    while (strlen($s) > 0)
    {
        $temp = $s[0];

        // Traverse to find the smallest
        // character in the first k characters
        for ($i = 1; $i < $k &&
             $i < strlen($s); $i++)
        {
            if ($s[$i] < $temp)
            {
                $temp = $s[$i];
            }
        }

        // append the smallest character
        $X = $X . $temp;

        // removing the lexicographically smallest
        // character from the string
        for ($i = 0; $i < $k; $i++)
        {
            if ($s[$i] == $temp)
            {

                $s = substr($s, 0, $i) .
                     substr($s, $i + 1, strlen($s));

                //s.erase(s.begin() + i);
                break;
            }
        }
    }

    return $X;
}

// Driver code
$s = "gaurang";
$k = 3;

echo(newString($s, $k));

// This code contributed by mits
?>

Javascript

<script>

      // JavaScript program to find the new string
      // after performing deletions and
      // append operation in the string s

      // Function to find the new string thus
      // formed by removing characters
      function newString(s, k) {
        // new string
        var X = "";

        // Remove characters until
        // the string is empty
        while (s.length > 0) {
          var temp = s[0];

          // Traverse to find the smallest
          // character in the first k characters
          for (var i = 1; i < k && i < s.length; i++)
          {
            if (s[i] < temp) {
              temp = s[i];
            }
          }

          // append the smallest character
          X = X + temp;

          // removing the lexicographically smallest
          // character from the string
          for (var i = 0; i < k; i++) {
            if (s[i] === temp) {
              s = s.substring(0, i) + s.substring(i + 1);

              break;
            }
          }
        }

        return X;
      }

      // Driver code
      var s = "gaurang";
      var k = 3;

      document.write(newString(s, k));

</script>

Output

agangru

Complexity Analysis:

• Time Complexity: O(n*n), as nested loops are used
• Auxiliary Space: O(1), as no extra space is used