Lexicographically next string
Given a string, find lexicographically next string.
Examples:
Input : geeks
Output : geekt
The last character 's' is changed to 't'.
Input : raavz
Output : raawz
Since we can't increase last character,
we increment previous character.
Input : zzz
Output : zzza
If string is empty, we return ‘a’. If string contains all characters as ‘z’, we append ‘a’ at the end. Otherwise we find first character from end which is not z and increment it.
Implementation:
C++
#include <bits/stdc++.h>
using namespace std;
string nextWord(string s)
{
if (s == "" )
return "a" ;
int i = s.length() - 1;
while (s[i] == 'z' && i >= 0)
i--;
if (i == -1)
s = s + 'a' ;
else
s[i]++;
return s;
}
int main()
{
string str = "samez" ;
cout << nextWord(str);
return 0;
}
|
Java
import java.util.*;
class GFG
{
public static String nextWord(String str)
{
if (str == "" )
return "a" ;
int i = str.length() - 1 ;
while (str.charAt(i) == 'z' && i >= 0 )
i--;
if (i == - 1 )
str = str + 'a' ;
else
str = str.substring( 0 , i) +
( char )(( int )(str.charAt(i)) + 1 ) +
str.substring(i + 1 );
return str;
}
public static void main (String[] args)
{
String str = "samez" ;
System.out.print(nextWord(str));
}
}
|
Python3
def nextWord(s):
if (s = = " " ):
return "a"
i = len (s) - 1
while (s[i] = = 'z' and i > = 0 ):
i - = 1
if (i = = - 1 ):
s = s + 'a'
else :
s = s.replace(s[i], chr ( ord (s[i]) + 1 ), 1 )
return s
if __name__ = = '__main__' :
str = "samez"
print (nextWord( str ))
|
C#
using System;
class GFG
{
public static string nextWord( string str)
{
if (str == "" )
{
return "a" ;
}
int i = str.Length - 1;
while (str[i] == 'z' && i >= 0)
{
i--;
}
if (i == -1)
{
str = str + 'a' ;
}
else
{
str = str.Substring(0, i) +
( char )(( int )(str[i]) + 1) +
str.Substring(i + 1);
}
return str;
}
public static void Main( string [] args)
{
string str = "samez" ;
Console.Write(nextWord(str));
}
}
|
Javascript
<script>
function nextWord(s)
{
if (s == "" )
return "a" ;
var i = s.length - 1;
while (s[i] == 'z' && i >= 0)
i--;
if (i == -1)
s.push( 'a' );
else
s[i] = String.fromCharCode(s[i].charCodeAt(0) + 1);
return s.join( '' );
}
var str = "samez" .split( '' );
document.write( nextWord(str));
</script>
|
Time Complexity: O(n)
Auxiliary Space: O(1)
Last Updated :
27 Dec, 2023
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