String str is given which contains lowercase English letters and spaces. It may contain multiple spaces. Get the first letter of every word and return the result as a string. The result should not contain any space.
Examples:
Input : str = "neveropen for neveropen" Output : gfg Input : str = "neveropen for neveropen"" Output : hc
Source: https://www.geeksforgeeks.org/amazon-interview-set-8-2/
The idea is to traverse each character of string str and maintain a boolean variable, which was initially set as true. Whenever we encounter space we set the boolean variable is true. And if we encounter any character other than space, we will check the boolean variable, if it was set as true as copy that charter to the output string and set the boolean variable as false. If the boolean variable is set false, do nothing.
Algorithm:
1. Traverse string str. And initialize a variable v as true. 2. If str[i] == ' '. Set v as true. 3. If str[i] != ' '. Check if v is true or not. a) If true, copy str[i] to output string and set v as false. b) If false, do nothing.
Implementation:
C++
// C++ program to find the string which contain// the first character of each word of another// string.#include<bits/stdc++.h>using namespace std;// Function to find string which has first// character of each word.string firstLetterWord(string str){ string result = ""; // Traverse the string. bool v = true; for (int i=0; i<str.length(); i++) { // If it is space, set v as true. if (str[i] == ' ') v = true; // Else check if v is true or not. // If true, copy character in output // string and set v as false. else if (str[i] != ' ' && v == true) { result.push_back(str[i]); v = false; } } return result;}// Driver codeint main(){ string str = "neveropen for neveropen"; cout << firstLetterWord(str); return 0;} |
Java
// Java program to find the string which // contain the first character of each word // of another string. class GFG { // Function to find string which has first // character of each word. static String firstLetterWord(String str) { String result = ""; // Traverse the string. boolean v = true; for (int i = 0; i < str.length(); i++) { // If it is space, set v as true. if (str.charAt(i) == ' ') { v = true; } // Else check if v is true or not. // If true, copy character in output // string and set v as false. else if (str.charAt(i) != ' ' && v == true) { result += (str.charAt(i)); v = false; } } return result; } // Driver code public static void main(String[] args) { String str = "neveropen for neveropen"; System.out.println(firstLetterWord(str)); }}// This code is contributed by // 29AjayKumar |
Python 3
# Python 3 program to find the string which # contain the first character of each word # of another string.# Function to find string which has first# character of each word.def firstLetterWord(str): result = "" # Traverse the string. v = True for i in range(len(str)): # If it is space, set v as true. if (str[i] == ' '): v = True # Else check if v is true or not. # If true, copy character in output # string and set v as false. elif (str[i] != ' ' and v == True): result += (str[i]) v = False return result# Driver Codeif __name__ == "__main__": str = "neveropen for neveropen" print(firstLetterWord(str))# This code is contributed by ita_c |
C#
// C# program to find the string which // contain the first character of each word // of another string. using System;class GFG { // Function to find string which has first // character of each word. static String firstLetterWord(String str) { String result = ""; // Traverse the string. bool v = true; for (int i = 0; i < str.Length; i++) { // If it is space, set v as true. if (str[i] == ' ') { v = true; } // Else check if v is true or not. // If true, copy character in output // string and set v as false. else if (str[i] != ' ' && v == true) { result += (str[i]); v = false; } } return result; } // Driver code public static void Main() { String str = "neveropen for neveropen"; Console.WriteLine(firstLetterWord(str)); } } // This code is contributed by PrinciRaj1992 |
Javascript
<script> // Javascript program to find the string which // contain the first character of each word // of another string. // Function to find string which has first // character of each word. function firstLetterWord(str) { let result = ""; // Traverse the string. let v = true; for (let i = 0; i < str.length; i++) { // If it is space, set v as true. if (str[i] == ' ') { v = true; } // Else check if v is true or not. // If true, copy character in output // string and set v as false. else if (str[i] != ' ' && v == true) { result += (str[i]); v = false; } } return result; } let str = "neveropen for neveropen"; document.write(firstLetterWord(str)); </script> |
Output
gfg
Time Complexity: O(n)
Auxiliary space: O(1).
Approach 1 : Reverse Iterative Approach
This is simplest approach to getting first letter of every word of the string. In this approach we are using reverse iterative loop to get letter of words. If particular letter ( i ) is 1st letter of word or not is can be determined by checking pervious character that is (i-1). If the pervious letter is space (‘ ‘) that means (i+1) is 1st letter then we simply add that letter to the string. Except character at 0th position. At the end we simply reverse the string and function will return string which contain 1st letter of word of the string.
C++
#include <iostream>using namespace std;void get(string s){ string str = "", temp = ""; for (int i = s.length() - 1; i > 0; i--) { if (isalpha(s[i]) && s[i - 1] == ' ') { temp += s[i]; } } str += s[0]; for (int i = temp.length() - 1; i >= 0; i--) { str += temp[i]; } cout << str << endl;}int main(){ string str = "neveropen for neveropen"; string str2 = "Code of the Day"; get(str); get(str2); return 0;}// This code is contributed by sarojmcy2e |
Java
public class GFG { public static void get(String s) { String str = "", temp = ""; // checking condition for (int i = s.length() - 1; i > 0; i--) { if (Character.isLetter(s.charAt(i)) && s.charAt(i - 1) == ' ') { temp += s.charAt(i); // if consition match // added it to the string } } // adding 1st letter of string str += s.charAt(0); // adding remaning letters for (int i = temp.length() - 1; i >= 0; i--) { str += temp.charAt(i); } System.out.println(str); } public static void main(String[] args) { String str = "neveropen for neveropen"; String str2 = "Code of the Day"; get(str); // function call get(str2); // function call }} |
Python3
def get(s): str = "" temp = "" for i in range(len(s)-1, 0, -1): if s[i].isalpha() and s[i-1] == ' ': temp += s[i] str += s[0] for i in range(len(temp)-1, -1, -1): str += temp[i] print(str)str = "neveropen for neveropen"str2 = "Code of the Day"get(str)get(str2) |
C#
using System;public class GFG { public static void Get(string s) { string str = "", temp = ""; // checking condition for (int i = s.Length - 1; i > 0; i--) { if (char.IsLetter(s[i]) && s[i - 1] == ' ') { temp += s[i]; // if consition match // added it to the string } } // adding 1st letter of string str += s[0]; // adding remaning letters for (int i = temp.Length - 1; i >= 0; i--) { str += temp[i]; } Console.WriteLine(str); } public static void Main(string[] args) { string str = "neveropen for neveropen"; string str2 = "Code of the Day"; Get(str); // function call Get(str2); // function call }} |
Javascript
function get(s) { let str = "", temp = ""; for (let i = s.length - 1; i > 0; i--) { if (s[i].match(/[a-zA-Z]/) && s[i - 1] === ' ') { temp += s[i]; } } str += s[0]; for (let i = temp.length - 1; i >= 0; i--) { str += temp[i]; } console.log(str);}const str = "neveropen for neveropen";const str2 = "Code of the Day";get(str);get(str2); |
Output
gfg CotD
Time Complexity: O(n)
Auxiliary space: O(1).
Approach 2: Using StringBuilder
This approach uses the StringBuilder class of Java. In this approach, we will first split the input string based on the spaces. The spaces in the strings can be matched using a regular expression. The split strings are stored in an array of strings. Then we can simply append the first character of each split string in the String Builder object.
Implementation:
C++
// C++ implementation of the above approach#include <bits/stdc++.h>using namespace std;string processWords(char *input){ /* we are splitting the input based on spaces (s)+ : this regular expression will handle scenarios where we have words separated by multiple spaces */ char *p; vector<string> s; p = strtok(input, " "); while (p != NULL) { s.push_back(p); p = strtok(NULL, " "); } string charBuffer; for (string values : s) /* charAt(0) will pick only the first character from the string and append to buffer */ charBuffer += values[0]; return charBuffer;}// Driver codeint main(){ char input[] = "neveropen for neveropen"; cout << processWords(input); return 0;}// This code is contributed by// sanjeev2552 |
Java
// Java implementation of the above approachclass GFG { private static StringBuilder charBuffer = new StringBuilder(); public static String processWords(String input) { /* we are splitting the input based on spaces (s)+ : this regular expression will handle scenarios where we have words separated by multiple spaces */ String s[] = input.split("(\\s)+"); for(String values : s) { /* charAt(0) will pick only the first character from the string and append to buffer */ charBuffer.append(values.charAt(0)); } return charBuffer.toString(); } // main function public static void main (String[] args) { String input = "neveropen for neveropen"; System.out.println(processWords(input)); }}// This code is contributed by Goutam Das |
Python3
# An efficient Python3 implementation# of above approachcharBuffer = []def processWords(input): """ we are splitting the input based on spaces (s)+ : this regular expression will handle scenarios where we have words separated by multiple spaces """ s = input.split(" ") for values in s: """ charAt(0) will pick only the first character from the string and append to buffer """ charBuffer.append(values[0]) return charBuffer# Driver Codeif __name__ == '__main__': input = "neveropen for neveropen" print(*processWords(input), sep = "")# This code is contributed# by SHUBHAMSINGH10 |
C#
// C# implementation of above approachusing System; using System.Text;class GFG {private static StringBuilder charBuffer = new StringBuilder(); public static String processWords(String input) { /* we are splitting the input based on spaces (s)+ : this regular expression will handle scenarios where we have words separated by multiple spaces */ String []s = input.Split(' '); foreach(String values in s) { /* charAt(0) will pick only the first character from the string and append to buffer */ charBuffer.Append(values[0]); } return charBuffer.ToString();} // Driver codepublic static void Main() { String input = "neveropen for neveropen"; Console.WriteLine(processWords(input));}}// This code is contributed by Rajput-Ji |
Javascript
<script>// Javascript implementation of the above approachvar charBuffer = "";function processWords(input){ /* we are splitting the input based on spaces (s)+ : this regular expression will handle scenarios where we have words separated by multiple spaces */ var s = input.split(' '); s.forEach(element => { /* charAt(0) will pick only the first character from the string and append to buffer */ charBuffer+=element[0]; }); return charBuffer;}// Driver codevar input = "neveropen for neveropen";document.write( processWords(input));// This code is contributed by rutvik_56.</script> |
Output
gfg
Time Complexity: O(n)
Auxiliary space: O(1).
Another Approach: Using boundary checker, refer https://www.geeksforgeeks.org/get-first-letter-word-string-using-regex-java/
This article is contributed by Aarti_Rathi and Anuj Chauhan. If you like neveropen and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the neveropen main page and help other Geeks.
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
