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++ program of the above approach #include <iostream> using namespace std;
// Function to convert the given // string into its abbreviation void abbreviateWord(string str)
{ // Stores the length of the string
int len = str.size();
// Print 1st character
cout << str[0];
// Print count of characters
// in between
cout << len - 2;
// Print last character
cout << str[len - 1];
} // Driver Code int main()
{ string str = "internationalization" ;
abbreviateWord(str);
return 0;
} |
// Java program for the above approach import java.util.*;
public class GFG
{ // Function to convert the given
// string into its abbreviation
static void abbreviateWord(String str)
{
// Stores the length of the string
int len = str.length();
// Print 1st character
System.out.print(str.charAt( 0 ));
// Print count of characters
// in between
System.out.print(len - 2 );
// Print last character
System.out.print(str.charAt(len - 1 ));
}
// Driver Code
public static void main(String args[])
{
String str = "internationalization" ;
abbreviateWord(str);
}
} // This code is contributed by Samim Hossain Mondal. |
# Python code for the above approach # Function to convert the given # string into its abbreviation def abbreviateWord( str ):
# Stores the length of the string
_len = len ( str );
# Print 1st character
print ( str [ 0 ], end = "");
# Print count of characters
# in between
print (_len - 2 , end = "");
# Print last character
print ( str [_len - 1 ], end = "");
# Driver Code str = "internationalization" ;
abbreviateWord( str );
# This code is contributed gfgking |
// C# program of the above approach using System;
class GFG
{ // Function to convert the given // string into its abbreviation static void abbreviateWord( string str)
{ // Stores the length of the string
int len = str.Length;
// Print 1st character
Console.Write(str[0]);
// Print count of characters
// in between
Console.Write(len - 2);
// Print last character
Console.Write(str[len - 1]);
} // Driver Code public static void Main()
{ string str = "internationalization" ;
abbreviateWord(str);
} } // This code is contributed by Samim Hossain Mondal. |
<script> // JavaScript code for the above approach
// Function to convert the given
// string into its abbreviation
function abbreviateWord(str)
{
// Stores the length of the string
let len = str.length;
// Print 1st character
document.write(str[0]);
// Print count of characters
// in between
document.write(len - 2);
// Print last character
document.write(str[len - 1]);
}
// Driver Code
let str = "internationalization" ;
abbreviateWord(str);
// This code is contributed by Potta Lokesh </script>
|
i18n
Time Complexity: O(1)
Auxiliary Space: O(1)