String containing first letter of every word in a given string with spaces

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 = "geeks for geeks"
Output : gfg

Input : str = "happy   coding"
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 code
int main()
{
    string str = "geeks for geeks";
    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 = "geeks for geeks";
        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 Code
if __name__ == "__main__":

    str = "geeks for geeks"
    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 = "geeks for geeks";
        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 = "geeks for geeks";
      document.write(firstLetterWord(str));

</script>

Output

gfg

Time Complexity: O(n)
Auxiliary space: O(n). 

Another Approach: 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 code
int main()
{
    char input[] = "geeks for geeks";
    cout << processWords(input);
    return 0;
}

// This code is contributed by
// sanjeev2552

Java

// Java implementation of the above approach

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("(\\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 = "geeks for geeks";
      System.out.println(processWords(input));
   }
}

// This code is contributed by Goutam Das

Python3

# An efficient Python3 implementation
# of above approach
charBuffer = []
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 Code
if __name__ == '__main__':
    input = "geeks for geeks"
    print(*processWords(input), sep = "")

# This code is contributed
# by SHUBHAMSINGH10

C#

// C# implementation of above approach
using 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 code
public static void Main()
{
    String input = "geeks for geeks";
    Console.WriteLine(processWords(input));
}
}

// This code is contributed by Rajput-Ji

Javascript

<script>
// Javascript implementation of the above approach
var 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 code
var input = "geeks for geeks";
document.write( processWords(input));

// This code is contributed by rutvik_56.
</script>

Output

gfg

Time Complexity: O(n²)
Auxiliary space: O(n). 

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 GeeksforGeeks 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 GeeksforGeeks main page and help other Geeks.