Given string str, the task is to convert the given string into its abbreviation of the form: first character, number of characters between first and last character, and the last character of the string.
Examples:
Input: str = “internationalization”
Output: i18n
Explanation: First letter ‘i’, followed by number of letters between ‘i’ and ‘n’ i.e. 18, and the last letter ‘n’.
Input: str = “geeksforgeeks”
Output: g11s
Approach: The given problem is an implementation based problem that can be solved by following the below steps:
- Print the 1st character of the given string str[0].
- Store the length of the string in a variable len and print len – 2.
- Print the last character of the string i.e, str[len -1].
Below is the implementation of the above approach:
C++
#include <iostream>
using namespace std;
void abbreviateWord(string str)
{
int len = str.size();
cout << str[0];
cout << len - 2;
cout << str[len - 1];
}
int main()
{
string str = "internationalization" ;
abbreviateWord(str);
return 0;
}
|
Java
import java.util.*;
public class GFG
{
static void abbreviateWord(String str)
{
int len = str.length();
System.out.print(str.charAt( 0 ));
System.out.print(len - 2 );
System.out.print(str.charAt(len - 1 ));
}
public static void main(String args[])
{
String str = "internationalization" ;
abbreviateWord(str);
}
}
|
Python3
def abbreviateWord( str ):
_len = len ( str );
print ( str [ 0 ], end = "");
print (_len - 2 , end = "");
print ( str [_len - 1 ], end = "");
str = "internationalization" ;
abbreviateWord( str );
|
C#
using System;
class GFG
{
static void abbreviateWord( string str)
{
int len = str.Length;
Console.Write(str[0]);
Console.Write(len - 2);
Console.Write(str[len - 1]);
}
public static void Main()
{
string str = "internationalization" ;
abbreviateWord(str);
}
}
|
Javascript
<script>
function abbreviateWord(str)
{
let len = str.length;
document.write(str[0]);
document.write(len - 2);
document.write(str[len - 1]);
}
let str = "internationalization" ;
abbreviateWord(str);
</script>
|
Time Complexity: O(1)
Auxiliary Space: O(1)