Given the string the task is to remove the first and last character of each word in a string.
Input: Geeks for geeks Output: eek o eek Input: Geeksforgeeks is best Output: eeksforgeek es
- Split the String based on the space
- Run a loop from the first letter to the last letter.
- Check if the character is the starting or end of the word
- Remove this character from the String.
Below is the implementation of the above approach.
Geeks for Geeks eek o eek
- Print the first and last character of each word in a String
- Print last character of each word in a string
- Capitalize the first and last character of each word in a string
- Remove all occurrences of a character in a string
- Remove a character from a string to make it a palindrome
- Find a string such that every character is lexicographically greater than its immediate next character
- Replace every character of string by character whose ASCII value is K times more than it
- Check if frequency of character in one string is a factor or multiple of frequency of same character in other string
- Map every character of one string to another such that all occurrences are mapped to the same character
- Most frequent word in first String which is not present in second String
- String containing first letter of every word in a given string with spaces
- Replace every character of a string by a different character
- Lexicographically smallest string formed by appending a character from the first K characters of a given string
- Lexicographically smallest string formed by appending a character from first K characters of a string | Set 2
- Find the character in first string that is present at minimum index in second string
If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to firstname.lastname@example.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.