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++ 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 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 |
# 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# 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 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 ?> |
<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> |
agangru
Complexity Analysis:
- Time Complexity: O(n*n), as nested loops are used
- Auxiliary Space: O(1), as no extra space is used