Lexicographically next string
Given a string, find lexicographically next string.
Examples:
Input : Beginner
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.push( '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)
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