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Lexicographically next string

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  • Difficulty Level : Easy
  • Last Updated : 20 Jul, 2022
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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++




// C++ program to find lexicographically next
// string
#include <bits/stdc++.h>
using namespace std;
 
string nextWord(string s)
{
    // If string is empty.
    if (s == "")
        return "a";
 
    // Find first character from right
    // which is not z.
     
    int i = s.length() - 1;
    while (s[i] == 'z' && i >= 0)
        i--;
 
    // If all characters are 'z', append
    // an 'a' at the end.
    if (i == -1)
        s = s + 'a';
 
    // If there are some non-z characters
    else
        s[i]++;
 
    return s;
}
 
// Driver code
int main()
{
    string str = "samez";
    cout << nextWord(str);
    return 0;
}


Java




// Java program to find
// lexicographically next string
import java.util.*;
 
class GFG
{
public static String nextWord(String str)
{
     
    // if string is empty
    if (str == "")
    return "a";
     
    // Find first character from
    // right which is not z.
    int i = str.length() - 1;
    while (str.charAt(i) == 'z' && i >= 0)
        i--;
         
    // If all characters are 'z',
    // append an 'a' at the end.
    if (i == -1)
        str = str + 'a';
 
// If there are some
// non-z characters
else
    str = str.substring(0, i) +
         (char)((int)(str.charAt(i)) + 1) +
                      str.substring(i + 1);
return str;
}
 
// Driver Code
public static void main (String[] args)
{
    String str = "samez";
    System.out.print(nextWord(str));
}
}
 
// This code is contributed
// by Kirti_Mangal


Python3




# Python 3 program to find lexicographically
# next string
 
def nextWord(s):
     
    # If string is empty.
    if (s == " "):
        return "a"
 
    # Find first character from right
    # which is not z.
    i = len(s) - 1
    while (s[i] == 'z' and i >= 0):
        i -= 1
 
    # If all characters are 'z', append
    # an 'a' at the end.
    if (i == -1):
        s = s + 'a'
 
    # If there are some non-z characters
    else:
        s = s.replace(s[i], chr(ord(s[i]) + 1), 1)
 
    return s
 
# Driver code
if __name__ == '__main__':
    str = "samez"
    print(nextWord(str))
     
# This code is contributed by
# Sanjit_Prasad


C#




// C# program to find
// lexicographically next string
using System;
 
class GFG
{
public static string nextWord(string str)
{
 
    // if string is empty
    if (str == "")
    {
    return "a";
    }
 
    // Find first character from
    // right which is not z.
    int i = str.Length - 1;
    while (str[i] == 'z' && i >= 0)
    {
        i--;
    }
 
    // If all characters are 'z',
    // append an 'a' at the end.
    if (i == -1)
    {
        str = str + 'a';
    }
 
    // If there are some
    // non-z characters
    else
    {
        str = str.Substring(0, i) +
             (char)((int)(str[i]) + 1) +
              str.Substring(i + 1);
    }
    return str;
}
 
// Driver Code
public static void Main(string[] args)
{
    string str = "samez";
    Console.Write(nextWord(str));
}
}
 
// This code is contributed by Shrikant13


Javascript




<script>
 
// Javascript program to find lexicographically next
// string
 
function nextWord(s)
{
    // If string is empty.
    if (s == "")
        return "a";
 
    // Find first character from right
    // which is not z.
     
    var i = s.length - 1;
    while (s[i] == 'z' && i >= 0)
        i--;
 
    // If all characters are 'z', append
    // an 'a' at the end.
    if (i == -1)
        s = s + 'a';
 
    // If there are some non-z characters
    else
        s[i] = String.fromCharCode(s[i].charCodeAt(0)+1);
 
    return s.join('');
}
 
// Driver code
var str = "samez".split('');
document.write( nextWord(str));
 
// This code is contributed by noob2000.
</script>


Output

samfz

Time Complexity: O(n)
Auxiliary Space: O(1)

This article is contributed by Pawan Asipu. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks. 


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