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 that may satisfy the above condition, we convert string “a” into a string that is lexicographically next to “a”.
To find lexicographically next, we start traversing the string from backward and convert all the letters “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”.
Implementation:
C++
#include <iostream>
using namespace std;
void lexMiddle(string a, string b)
{
for ( int i = a.length() - 1; i >= 0; i--) {
if (a[i] == 'z' )
a[i] = 'a' ;
else {
a[i]++;
break ;
}
}
if (a < b)
cout << a;
else
cout << -1;
}
int main()
{
string a = "geeks" , b = "heeks" ;
lexMiddle(a, b);
return 0;
}
|
Java
class GFG
{
static void lexMiddle(String a, String b)
{
String new_String = "" ;
for ( int i = a.length() - 1 ; i >= 0 ; i--)
{
if (a.charAt(i) == 'z' )
new_String = 'a' + new_String;
else
{
new_String = ( char )(a.charAt(i) + 1 ) +
new_String;
for ( int j = i - 1 ; j >= 0 ; j--)
new_String = a.charAt(j) + new_String;
break ;
}
}
if (new_String.compareTo(b) < 0 )
System.out.println(new_String);
else
System.out.println(- 1 );
}
public static void main(String args[])
{
String a = "geeks" , b = "heeks" ;
lexMiddle(a, b);
}
}
|
Python3
def lexMiddle( a, b):
for i in range ( len (a) - 1 , - 1 , - 1 ):
ans = []
if (a[i] = = 'z' ):
a[i] = 'a'
else :
a[i] = chr ( ord (a[i]) + 1 )
break
if (a < b):
return a
else :
return - 1
if __name__ = = '__main__' :
a = list ( "geeks" )
b = list ( "heeks" )
ans = lexMiddle(a, b)
ans = ''.join( map ( str , ans))
print (ans)
|
C#
using System;
class GFG
{
static void lexMiddle( string a, string b)
{
string new_String = "" ;
for ( int i = a.Length - 1; i >= 0; i--)
{
if (a[i] == 'z' )
new_String = 'a' + new_String;
else
{
new_String = ( char )(a[i] + 1) +
new_String;
for ( int j = i - 1; j >= 0; j--)
new_String = a[j] + new_String;
break ;
}
}
if (new_String.CompareTo(b) < 0)
Console.Write(new_String);
else
Console.Write(-1);
}
public static void Main()
{
string a = "geeks" , b = "heeks" ;
lexMiddle(a, b);
}
}
|
Javascript
<script>
function lexMiddle(a, b)
{
let new_String = "" ;
for (let i = a.length - 1; i >= 0; i--)
{
if (a[i] == 'z' )
new_String = 'a' + new_String;
else
{
new_String = String.fromCharCode(
a[i].charCodeAt(0) + 1) + new_String;
for (let j = i - 1; j >= 0; j--)
new_String = a[j] + new_String;
break ;
}
}
if (new_String < (b))
document.write(new_String);
else
document.write(-1);
}
let a = "geeks" , b = "heeks" ;
lexMiddle(a, b);
</script>
|
Time Complexity: O(n) where n is length of string ‘a’.
Auxiliary Space: O(1), no extra space is required, so it is a constant.
Last Updated :
29 Nov, 2022
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