Lexicographically middle string

Given two strings a and b. Our task is to print any string which is greater than a(lexicographically) but smaller than b(lexicographically). If it is impossible to get such string, print -1;

Examples:

Input : a = "abg", b = "abj"
Output : abh
The string "abh" is lexicographically 
greater than "abg" and smaller than
"abj"        

Input : a = "abc", b = "abd"
Output :-1
There is no string which is lexicographically
greater than a but smaller than b/

Since, there can be multiple strings which may satisfy the above condition, we convert string “a” into a string which is lexicographically next to “a”.
To find lexicographically next, we start traversing the string from backward and convert all the letter “z” to letter”a”. If we encounter any letter which is not “z”, then we increment it by one and further traversal will not be carried out.If this string is not smaller than “b”, then we will print -1 as no string can satisfy the above condition.
For example, string a=”ddzzz” and string b=”deaao”.So, starting from backward, we will convert all letter “z” to letter “a” until we reach to letter “d”(in this case).Increment “d” by one (to “e”) and break out from the loop.So, string a will become “deaaa” which is lexicographically greater than “ddzzz” and smaller than “deaao”.



C++

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// CPP program to implement above approach
#include <iostream>
using namespace std;
  
// function to find lexicographically mid
// string.
void lexMiddle(string a, string b)
{
    // converting string "a" into its 
    // lexicographically next string
    for (int i = a.length() - 1; i >= 0; i--) {
  
        // converting all letter "z" to letter "a"
        if (a[i] == 'z'
            a[i] = 'a';
        else {
  
            // if letter other than "z" is
            // encountered, increment it by one 
            // and break
            a[i]++; 
            break;
        }
    }
  
    // if this new string "a" is lexicographically 
    // smaller than b
    if (a < b)
        cout << a;
    else
        cout << -1;
}
  
// Driver function
int main()
{
    string a = "geeks", b = "heeks";
    lexMiddle(a, b);
    return 0;
}

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Java

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// Java program to implement
// above approach
class GFG
{
  
// function to find lexicographically 
// mid String.
static void lexMiddle(String a, String b)
{
    String new_String = "";
      
    // converting String "a" into its 
    // lexicographically next String
    for (int i = a.length() - 1; i >= 0; i--) 
    {
  
        // converting all letter 
        // "z" to letter "a"
        if (a.charAt(i) == 'z'
            new_String = 'a' + new_String;
        else
        {
              
            // if letter other than "z" is
            // encountered, increment it by 
            // one and break
            new_String = (char)(a.charAt(i) + 1) +
                                       new_String;
              
            //compose the remaining string
            for(int j = i - 1; j >= 0; j--)
            new_String = a.charAt(j) + new_String;
              
            break;
        }
    }
  
    // if this new String new_String is 
    // lexicographically smaller than b
    if (new_String.compareTo(b) < 0)
    System.out.println(new_String);
    else
        System.out.println(-1);
}
  
// Driver Code
public static void main(String args[])
{
    String a = "geeks", b = "heeks";
    lexMiddle(a, b);
}
}
  
// This code is contributed
// by Arnab Kundu

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Python3

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# Python3 program to implement above approach 
  
# function to find lexicographically mid 
# string. 
def lexMiddle( a, b): 
  
    # converting string "a" into its 
    # lexicographically next string 
    for i in range(len(a)-1,-1,-1):
      
        ans=[]
        # converting all letter "z" to letter "a" 
        if (a[i] == 'z'): 
            a[i] = 'a'
        else
  
            # if letter other than "z" is 
            # encountered, increment it by one 
            # and break 
            a[i]=chr(ord(a[i])+1
            break
      
      
  
    # if this new string "a" is lexicographically 
    # smaller than b 
    if (a < b):
        return a
    else:
        return -1
  
  
  
# Driver function 
if __name__=='__main__':
    a = list("geeks")
    b = list("heeks"
    ans=lexMiddle(a, b)
    ans = ''.join(map(str, ans))
    print(ans)
  
# this code is contributed by ash264

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C#

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// C# program to implement above approach
using System;
  
class GFG
{
  
// function to find lexicographically 
// mid String.
static void lexMiddle(string a, string b)
{
    string new_String = "";
      
    // converting String "a" into its 
    // lexicographically next String
    for (int i = a.Length - 1; i >= 0; i--) 
    {
  
        // converting all letter 
        // "z" to letter "a"
        if (a[i] == 'z'
            new_String = 'a' + new_String;
        else
        {
              
            // if letter other than "z" is
            // encountered, increment it by 
            // one and break
            new_String = (char)(a[i] + 1) +
                                new_String;
              
            //compose the remaining string
            for(int j = i - 1; j >= 0; j--)
                new_String = a[j] + new_String;
              
            break;
        }
    }
  
    // if this new String new_String is 
    // lexicographically smaller than b
    if (new_String.CompareTo(b) < 0)
    Console.Write(new_String);
    else
        Console.Write(-1);
}
  
// Driver Code
public static void Main()
{
    string a = "geeks", b = "heeks";
    lexMiddle(a, b);
}
}
  
// This code is contributed by ita_c

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Output:

geekt

Time Complexity: O(n) where n is length of string ‘a’



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