Skip to content
Related Articles
Lexicographically smallest string formed by appending a character from the first K characters of a given string
• Difficulty Level : Easy
• Last Updated : 14 May, 2021

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 ``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;` `        ``// 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;` `        ``// 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

 ` 0)``    ``{``        ``\$temp` `= ``\$s``;` `        ``// 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

 ``
Output:
`agangru`

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with industry experts, please refer DSA Live Classes

My Personal Notes arrow_drop_up