Length Of Last Word in a String

Given a string s consisting of upper/lower-case alphabets and empty space characters ‘ ‘, return the length of the last word in the string. If the last word does not exist, return 0.

Examples:

Input  : str = "Geeks For Geeks"
Output : 5
length(Geeks)= 5
Input : str = "Start Coding Here"
Output : 4
length(Here) = 4
Input  :  **
Output : 0

Approach 1: Iterate String from index 0
If we iterate the string from left to right, we would have to be careful about the spaces after the last word. The spaces before the first word can be ignored easily. However, it is difficult to detect the length of the last word if there are spaces at the end of the string. This can be handled by trimming the spaces before or at the end of the string. If modifying the given string is restricted, we need to create a copy of the string and trim spaces from that.  

C++

// c++ program for the above approach

#include <iostream>
#include <string>
using namespace std;

class GFG {
public:
    int lengthOfLastWord(const string& a) {
        int len = 0;
        string x = a;
        x.erase(0, x.find_first_not_of(" "));
        x.erase(x.find_last_not_of(" ") + 1);

        for (int i = 0; i < x.length(); i++) {
            if (x[i] == ' ')
                len = 0;
            else
                len++;
        }

        return len;
    }
};

int main() {
    string input = "Geeks For Geeks  ";
    GFG gfg;
    cout << "The length of last word is " << gfg.lengthOfLastWord(input) << endl;
    return 0;
}


// This code is contributed by prince

Java

// Java program for implementation of simple
// approach to find length of last word
public class GFG {
    public int lengthOfLastWord(final String a)
    {
        int len = 0;

        /* String a is 'final'-- can not be modified
           So, create a copy and trim the spaces from
           both sides */
        String x = a.trim();

        for (int i = 0; i < x.length(); i++) {
            if (x.charAt(i) == ' ')
                len = 0;
            else
                len++;
        }

        return len;
    }

    // Driver code
    public static void main(String[] args)
    {
        String input = "Geeks For Geeks  ";
        GFG gfg = new GFG();
        System.out.println("The length of last word is "
                           + gfg.lengthOfLastWord(input));
    }
}

Python3

# Python3 program for implementation of simple
# approach to find length of last word


def lengthOfLastWord(a):
    l = 0

    # String a is 'final'-- can not be modified
    # So, create a copy and trim the spaces from
    # both sides
    x = a.strip()

    for i in range(len(x)):
        if x[i] == " ":
            l = 0
        else:
            l += 1
    return l


# Driver code
if __name__ == "__main__":
    inp = "Geeks For Geeks "
    print("The length of last word is",
          lengthOfLastWord(inp))

# This code is contributed by
# sanjeev2552

C#

// C# program for implementation of simple
// approach to find length of last word
using System;

class GFG {

    public virtual int lengthOfLastWord(string a)
    {
        int len = 0;

        // String a is 'final'-- can
        // not be modified So, create
        // a copy and trim the
        // spaces from both sides
        string x = a.Trim();

        for (int i = 0; i < x.Length; i++) {
            if (x[i] == ' ') {
                len = 0;
            }
            else {
                len++;
            }
        }

        return len;
    }

    // Driver code
    public static void Main(string[] args)
    {
        string input = "Geeks For Geeks ";
        GFG gfg = new GFG();
        Console.WriteLine("The length of last word is "
                          + gfg.lengthOfLastWord(input));
    }
}

// This code is contributed by shrikanth13

Javascript

<script>

// js program for implementation of simple
// approach to find length of last word


function lengthOfLastWord(a)
{
    let len = 0;

    // String a is 'final'-- can
    // not be modified So, create
    // a copy and trim the
    // spaces from both sides
    x = a.trim();

    for (let i = 0; i < x.length; i++) {
        if (x[i] == ' ') {
            len = 0;
        }
        else {
            len++;
        }
    }

    return len;
}

// Driver code

input = "Geeks For Geeks ";
document.write("The length of last word is "+ lengthOfLastWord(input));



</script>

Output

The length of last word is 5

Approach 2: Iterate the string from the last index. This idea is more efficient since we can easily ignore the spaces from the last. The idea is to start incrementing the count when you encounter the first alphabet from the last and stop when you encounter a space after those alphabets.

C++

// CPP program for implementation of efficient
// approach to find length of last word
#include <bits/stdc++.h>
#include <iostream>
using namespace std;

int length(string str)
{
    int count = 0;
    bool flag = false;
    for (int i = str.length() - 1; i >= 0; i--) {
        // Once the first character from last
        // is encountered, set char_flag to true.
        if ((str[i] >= 'a' && str[i] <= 'z')
            || (str[i] >= 'A' && str[i] <= 'Z')) {
            flag = true;
            count++;
        }
        // When the first space after the
        // characters (from the last) is
        // encountered, return the length
        // of the last word
        else {
            if (flag == true)
                return count;
        }
    }
    return count;
}

// Driver code
int main()
{
    string str = "Geeks for Geeks";
    cout << "The length of last word is " << length(str);
    return 0;
}

// This code is contributed by rahulkumawat2107

Java

// Java program for implementation of efficient
// approach to find length of last word
public class GFG {
    public int lengthOfLastWord(final String a)
    {
        boolean char_flag = false;
        int len = 0;
        for (int i = a.length() - 1; i >= 0; i--) {
            if (Character.isLetter(a.charAt(i))) {
                // Once the first character from last
                // is encountered, set char_flag to true.
                char_flag = true;
                len++;
            }
            else {
                // When the first space after the characters
                // (from the last) is encountered, return
                // the length of the last word
                if (char_flag == true)
                    return len;
            }
        }
        return len;
    }

    // Driver code
    public static void main(String[] args)
    {
        String input = "Geeks For Geeks  ";
        GFG gfg = new GFG();
        System.out.println("The length of last word is "
                           + gfg.lengthOfLastWord(input));
    }
}

Python3

# Python3 program for implementation of efficient
# approach to find length of last word


def findLength(str):
    count = 0
    flag = False
    for i in range(len(str) - 1, -1, -1):
        if ((str[i] >= 'a' and str[i] <= 'z') or (str[i] >= 'A' and str[i] <= 'Z')):
            flag = True
            count += 1
        elif (flag == True):
            return count
    return count


# Driver code
str = "Geeks for Geeks"
print("The length of last word is",
      findLength(str))

# This code is contributed by Rajput Ji

C#

// C# program for implementation of efficient
// approach to find length of last word
using System;

class GFG {

    public virtual int lengthOfLastWord(string a)
    {
        bool char_flag = false;
        int len = 0;
        for (int i = a.Length - 1; i >= 0; i--) {
            if (char.IsLetter(a[i])) {
                // Once the first character from last
                // is encountered, set char_flag to true.
                char_flag = true;
                len++;
            }
            else {
                // When the first space after the
                // characters (from the last) is
                // encountered, return the length
                // of the last word
                if (char_flag == true) {
                    return len;
                }
            }
        }
        return len;
    }

    // Driver code
    public static void Main(string[] args)
    {
        string input = "Geeks For Geeks ";
        GFG gfg = new GFG();
        Console.WriteLine("The length of last word is "
                          + gfg.lengthOfLastWord(input));
    }
}

// This code is contributed by Shrikant13

Javascript

// Function to find length of last word in a string
function length(str) {
    let count = 0;
    let flag = false;

    // Loop through the string backwards
    for (let i = str.length - 1; i >= 0; i--) {
        // Once the first character from last
        // is encountered, set char_flag to true.
        if ((str[i] >= 'a' && str[i] <= 'z')
            || (str[i] >= 'A' && str[i] <= 'Z')) {
            flag = true;
            count++;
        }
        // When the first space after the
        // characters (from the last) is
        // encountered, return the length
        // of the last word
        else {
            if (flag == true)
                return count;
        }
    }

    return count;
}

// Driver code
let str = "Geeks for Geeks";
console.log(`The length of last word is ${length(str)}`);

PHP

<?php
// PHP program for implementation of efficient
// approach to find length of last word

function length($str)
{
    $count = 0;
    $flag = false;
    for($i = strlen($str)-1 ; $i>=0 ; $i--)
    {
        // Once the first character from last
        // is encountered, set char_flag to true.
        if( ($str[$i] >='a' && $str[$i]<='z') ||
            ($str[$i] >='A' && $str[$i]<='Z'))
        {
            $flag = true;
            $count++;
        }

        // When the first space after the
        // characters (from the last) is
        // encountered, return the length
        // of the last word
        else
        {
            if($flag == true)
                return $count;
        }

    }
    return $count;
}

// Driver code
$str = "Geeks for Geeks";
echo "The length of last word is ", length($str);

// This code is contributed by ajit.
?>

Output

The length of last word is 5

Method #3 : Using split()  and list 

• As all the words in a sentence are separated by spaces.
• We have to split the sentence by spaces using split().
• We split all the words by spaces and store them in a list.
• Print the length of the last word of the list.

Below is the implementation:

C++

// C++ code for the above aprroach

#include <iostream>
#include <string>
#include <vector>
using namespace std;

int length(string str) {
  // Split by space and converting
  // String to list and
  vector<string> lis;
  string word = "";
  for (char c : str) {
    if (c == ' ') {
      lis.push_back(word);
      word = "";
    } else {
      word += c;
    }
  }
  lis.push_back(word);
  return lis.back().length();
}

// Driver code
int main() {
  string str = "Geeks for Geeks";
  cout << "The length of last word is " << length(str) << endl;
  return 0;
}

// This code is contributed adityashatmfh

Java

// Java program for implementation of efficient
// approach to find length of last word

import java.util.ArrayList;
import java.util.List;

public class Main {
    public static int length(String str) {
        // Split by space and converting String to list
        List<String> lis = new ArrayList<>();
        String word = "";
        for (char c : str.toCharArray()) {
            if (c == ' ') {
                lis.add(word);
                word = "";
            } else {
                word += c;
            }
        }
        lis.add(word);
        return lis.get(lis.size() - 1).length();
    }

    // Driver code
    public static void main(String[] args) {
        String str = "Geeks for Geeks";
        System.out.println("The length of last word is " + length(str));
    }
}
// Contributed by adityasharmadev01

Python3

# Python3 program for implementation of efficient
# approach to find length of last word


def length(str):
  # Split by space and converting
    # String to list and
    lis = list(str.split(" "))
    return len(lis[-1])


# Driver code
str = "Geeks for Geeks"
print("The length of last word is",
      length(str))

# This code is contributed by vikkycirus

C#

// C# code for the above approach
using System;

public class Program {
    static int length(string str)
    {
        // Split by space and converting
        // String to list and
        string[] lis = str.Split(' ');
        return lis[lis.Length - 1].Length;
    } // Driver code
    static void Main()
    {
        string str = "Geeks for Geeks";
        Console.WriteLine("The length of last word is "
                          + length(str));
    }
}

Javascript

<script>

// Javascript program for implementation
// of efficient approach to find length
// of last word
function length(str)
{

    // Split by space and converting
    // String to list and
    var lis = str.split(" ")
    return lis[lis.length - 1].length;
}

// Driver code
var str = "Geeks for Geeks"
document.write("The length of last word is " +
               length(str));

// This code is contributed by bunnyram19

</script>

Output

The length of last word is 5

METHOD 4:Using regular expressions

APPROACH:

The regular expression essentially matches all non-space characters at the end of the string. The re.findall() function then returns a list of all non-overlapping matches in the input string, which in this case is just a single match consisting of the last word. Finally, the join() function is used to join the list of characters into a single string, and the length of this string (i.e., the length of the last word) is returned by the functio

ALGORITHM:

1.Import the re module for regular expressions
2.Use the findall() function to find all non-space characters at the end of the string
3.Join the resulting list of characters into a string
4.Return the length of the resulting string

C++

#include <iostream>
#include <regex>

using namespace std;

int length_of_last_word_4(string s) {
    // Use regular expressions to find the last word
    regex re("[^ ]*$");
    smatch match;
    regex_search(s, match, re);
    string last_word = match.str();
    // Return the length of the last word
    return last_word.length();
}

int main() {
    string s = "Geeks For Geeks";
    cout << length_of_last_word_4(s) << endl;

    s = "Start Coding Here";
    cout << length_of_last_word_4(s) << endl;

    s = "";
    cout << length_of_last_word_4(s) << endl;

    return 0;
}

Java

import java.util.regex.*;

public class Main {
    public static int lengthOfLastWord(String s) {
        // Use regular expressions to find the last word
        Pattern pattern = Pattern.compile("[^ ]*$");
        Matcher matcher = pattern.matcher(s);
        if (matcher.find()) {
            String lastWord = matcher.group();
            // Return the length of the last word
            return lastWord.length();
        }
        // If no last word is found, return 0
        return 0;
    }

    public static void main(String[] args) {
        String s = "Geeks For Geeks";
        System.out.println(lengthOfLastWord(s));

        s = "Start Coding Here";
        System.out.println(lengthOfLastWord(s));

        s = "";
        System.out.println(lengthOfLastWord(s));
    }
}

Python3

import re

def length_of_last_word_4(s):
    # use regular expressions to find the last word
    last_word = ''.join(re.findall('[^ ]*$', s))
    # return the length of the last word
    return len(last_word)
s = "Geeks For Geeks"
print(length_of_last_word_4(s))


s = "Start Coding Here"
print(length_of_last_word_4(s))

s = ""
print(length_of_last_word_4(s))

C#

using System;
using System.Text.RegularExpressions;

class Program
{
    // Function to calculate the length of the last word in a string
    static int LengthOfLastWord(string s)
    {
        // Use regular expressions to find the last word
        Regex re = new Regex(@"[^ ]*$"); // Define a regular expression to match non-space characters at the end of the string.
        Match match = re.Match(s); // Search for the regular expression in the input string.
        string lastWord = match.Value; // Get the matched portion as the last word.

        // Return the length of the last word
        return lastWord.Length; // Return the length of the last word.
    }

    static void Main()
    {
        string s = "Geeks For Geeks";
        Console.WriteLine(LengthOfLastWord(s)); // Call the function and print the result.

        s = "Start Coding Here";
        Console.WriteLine(LengthOfLastWord(s)); // Call the function and print the result.

        s = "";
        Console.WriteLine(LengthOfLastWord(s)); // Call the function and print the result.
    }
}

Javascript

function lengthOfLastWord(s) {
    // Use regular expressions to find the last word
    const re = /[^ ]*$/;
    const match = s.match(re);

    if (match) {
        const lastWord = match[0];
        // Return the length of the last word
        return lastWord.length;
    } else {
        // If no last word is found, return 0
        return 0;
    }
}

// Main function
const s1 = "Geeks For Geeks";
console.log("Length of the last word in '" + s1 + "': " + lengthOfLastWord(s1));

const s2 = "Start Coding Here";
console.log("Length of the last word in '" + s2 + "': " + lengthOfLastWord(s2));

const s3 = "";
console.log("Length of the last word in an empty string: " + lengthOfLastWord(s3));

Output

5
4
0

Time complexity: O(n), where n is the length of the input string
Space complexity: O(n), where n is the length of the input string

This article is contributed by Saloni Baweja.