Skip to content
Related Articles

Related Articles

Improve Article
Save Article
Like Article

Length Of Last Word in a String

  • Difficulty Level : Easy
  • Last Updated : 09 Jun, 2021

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:  

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.

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 implementation of simple
// approach to find length of last word
#include<bits/stdc++.h>
#include <boost/algorithm/string.hpp>
using namespace std;
 
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 str(a);
    boost::trim_right(str);
    for (int i = 0; i < str.length(); i++)
    {
        if (str.at(i) == ' ')
            len = 0;
        else
            len++;
    }
    return len;
}
 
// Driver code
int main()
{
    string input = "Geeks For Geeks ";
    cout << "The length of last word is "
        << lengthOfLastWord(input);
}
 
// This code is contributed by Rajput-Ji

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: 

Length of the 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 length(str):
 
    count = 0;
    flag = False;
    length = len(str)-1;
    while(length != 0):
        if(str[length] == ' '):
            return count;
        else:
            count += 1;
        length -= 1;
    return count;
 
# Driver code
str = "Geeks for Geeks";
print("The length of last word is",
                      length(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

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:  

Length of the last word is 5

This article is contributed by Saloni Baweja. 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.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

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:

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

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:

Length of the last word is 5



My Personal Notes arrow_drop_up
Recommended Articles
Page :