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 = “neveropen”
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++
// C++ program of the above approach#include <iostream>using namespace std;// Function to convert the given// string into its abbreviationvoid 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 Codeint main(){ string str = "internationalization"; abbreviateWord(str); return 0;} |
Java
// Java program for the above approachimport 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. |
Python3
# Python code for the above approach# Function to convert the given# string into its abbreviationdef 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 Codestr = "internationalization";abbreviateWord(str);# This code is contributed gfgking |
C#
// C# program of the above approachusing System;class GFG{ // Function to convert the given// string into its abbreviationstatic 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 Codepublic static void Main(){ string str = "internationalization"; abbreviateWord(str);}}// This code is contributed by Samim Hossain Mondal. |
Javascript
<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> |
Output
i18n
Time Complexity: O(1)
Auxiliary Space: O(1)
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