Reverse individual words

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Given string str, we need to print the reverse of individual words.

Examples:

Input : Hello World

Output : olleH dlroW

Input : Geeks for Geeks

Output : skeeG rof skeeG

Method 1 (Simple): Generate all words separated by space. One by one reverse word and print them separated by space.

Method 2 (Space Efficient): We use a stack to push all words before space. As soon as we encounter a space, we empty the stack.

Implementation:

C++

// C++ program to reverse individual words in a given
// string using STL list
 
#include <bits/stdc++.h>
using namespace std;
 
// reverses individual words of a string
void reverseWords(string str)
{
    stack<char> st;
 
    // Traverse given string and push all characters
    // to stack until we see a space.
    for (int i = 0; i < str.length(); ++i) {
        if (str[i] != ' ')
            st.push(str[i]);
 
        // When we see a space, we print contents
        // of stack.
        else {
            while (st.empty() == false) {
                cout << st.top();
                st.pop();
            }
            cout << " ";
        }
    }
 
    // Since there may not be space after
    // last word.
    while (st.empty() == false) {
        cout << st.top();
        st.pop();
    }
}
 
// Driver program to test function
int main()
{
    string str = "Geeks for Geeks";
    reverseWords(str);
    return 0;
}

Java

// Java program to reverse individual
// words in a given string using STL list
import java.io.*;
import java.util.*;
 
class GFG {
 
    // reverses individual words of a string
    static void reverseWords(String str)
    {
        Stack<Character> st = new Stack<Character>();
 
        // Traverse given string and push all
        // characters to stack until we see a space.
        for (int i = 0; i < str.length(); ++i) {
            if (str.charAt(i) != ' ')
                st.push(str.charAt(i));
 
            // When we see a space, we print
            // contents of stack.
            else {
                while (st.empty() == false) {
                    System.out.print(st.pop());
                }
                System.out.print(" ");
            }
        }
 
        // Since there may not be space after
        // last word.
        while (st.empty() == false) {
            System.out.print(st.pop());
        }
    }
 
    // Driver program to test above function
    public static void main(String[] args)
    {
        String str = "Geeks for Geeks";
        reverseWords(str);
    }
}

Python3

# Python3 program to reverse individual words
# in a given string using STL list
 
# reverses individual words of a string
 
 
def reverseWords(string):
    st = list()
 
    # Traverse given string and push all characters
    # to stack until we see a space.
    for i in range(len(string)):
        if string[i] != " ":
            st.append(string[i])
 
        # When we see a space, we print
        # contents of stack.
        else:
            while len(st) > 0:
                print(st[-1], end="")
                st.pop()
            print(end=" ")
 
    # Since there may not be space after
    # last word.
    while len(st) > 0:
        print(st[-1], end="")
        st.pop()
 
 
# Driver Code
if __name__ == "__main__":
    string = "Geeks for Geeks"
    reverseWords(string)
 
# This code is contributed by
# sanjeev2552

C#

// C# program to reverse individual
// words in a given string using STL list
using System;
using System.Collections.Generic;
 
class GFG
{
 
// reverses individual words
// of a string
public static void reverseWords(string str)
{
    Stack<char> st = new Stack<char>();
 
    // Traverse given string and push
    // all characters to stack until
    // we see a space.
    for (int i = 0; i < str.Length; ++i)
    {
        if (str[i] != ' ')
        {
            st.Push(str[i]);
        }
 
        // When we see a space, we
        // print contents of stack.
        else
        {
            while (st.Count > 0)
            {
                Console.Write(st.Pop());
 
            }
            Console.Write(" ");
        }
    }
 
    // Since there may not be
    // space after last word.
    while (st.Count > 0)
    {
        Console.Write(st.Pop());
 
    }
}
 
// Driver Code
public static void Main(string[] args)
{
    string str = "Geeks for Geeks";
    reverseWords(str);
}
}
 
// This code is contributed
// by Shrikant13

Javascript

// JS program to reverse individual words in a given
// string using STL list
 
 
// reverses individual words of a string
function reverseWords(str) {
    let st = [];
 
    // Traverse given string and push all characters
    // to stack until we see a space.
    for (let i = 0; i < str.length; ++i) {
        if (str[i] != ' ')
            st.unshift(str[i]);
 
        // When we see a space, we print contents
        // of stack.
        else {
            while (st.length != 0) {
                process.stdout.write(st[0]);
            st.shift();
            }
            process.stdout.write(' ');
        }
    }
 
    // Since there may not be space after
    // last word.
    while (st.length != 0) {
    process.stdout.write(st[0]);
        st.shift();
    }
}
 
// Driver program to test function
let str = "Geeks for Geeks";
reverseWords(str);
 
// This code is contributed by adityamaharshi21

Output

skeeG rof skeeG

Time Complexity: O(n), where n is the length of the string

Auxiliary Space: O(n), where n is the length of the string

Python | Reverse each word in a sentence

Using stringstream in C++ :

Implementation:

C++

#include <bits/stdc++.h>
using namespace std;
 
void printWords(string str)
{
    // word variable to store word
    string word;
 
    // making a string stream
    stringstream iss(str);
 
    // Read and print each word.
    while (iss >> word) {
        reverse(word.begin(), word.end());
        cout << word << " ";
    }
}
 
// Driver code
int main()
{
    string s = "zambiatek is good to learn";
    printWords(s);
    return 0;
}
// This code is contributed by Nikhil Rawat

Java

import java.io.*;
import java.lang.*;
import java.util.*;
 
class Main {
    public static void printWords(String str)
    {
        // word variable to store word
        String word;
 
        // making a string stream
        StringTokenizer iss = new StringTokenizer(str);
 
        // Read and print each word.
        while (iss.hasMoreTokens()) {
            word = iss.nextToken();
            System.out.print(
                new StringBuilder(word).reverse().toString()
                + " ");
        }
    }
 
    public static void main(String[] args)
        throws java.lang.Exception
    {
        String s = "zambiatek is good to learn";
        printWords(s);
    }
}
 
// Contributed by rishabmalhdijo

Python3

def print_words(s):
    # word variable to store word
    word = ""
 
    # making a string stream
    iss = s.split()
 
    # Read and print each word.
    for i in iss:
        word = i[::-1]
        print(word, end=" ")
 
 
# Driver code
if __name__ == '__main__':
    s = "zambiatek is good to learn"
    print_words(s)

C#

using System;
using System.Linq;
using System.Collections.Generic;
using System.Text;
using System.IO;
 
public class GFG {
    static void printWords(string str)
    {
        // word variable to store word
        string word;
 
        // making a string stream
        using(var iss = new StringReader(str))
        {
            string line;
            while ((line = iss.ReadLine()) != null) {
                // Read and print each word.
                foreach(string w in line.Split(' '))
                {
                    word
                        = new string(w.Reverse().ToArray());
                    Console.Write(word + " ");
                }
            }
        }
    }
 
    // Driver code
    static void Main()
    {
        string s = "zambiatek is good to learn";
        printWords(s);
    }
}
// ksam24000

Javascript

function printWords(str) {
 
    // word variable to store word
    let word;
     
    // making a string stream
    let lines = str.split("\n");
    for (let i = 0; i < lines.length; i++) {
        let words = lines[i].split(" ");
         
        for (let j = 0; j < words.length; j++) {
            word = words[j].split("").reverse().join("");
            process.stdout.write(word + " ");
        }
    }
}
 
// Driver Code
let s = "zambiatek is good to learn";
printWords(s);

Output

skeeGrofskeeG si doog ot nrael

Time complexity : O(n)

Auxiliary Space : O(n)

Using Java 8 Streams:

Implementation:

C++

#include <bits/stdc++.h>
using namespace std;
 
int main() {
string str = "Welcome to GFG";
    string result = "";
 
    // Splitting the string based on space
    istringstream ss(str);
    vector<string> words;
    do {
        string word;
        ss >> word;
        words.push_back(word);
    } while (ss);
 
    // Reverse each part and then join
    for (int i = 0; i < words.size() - 1; i++) {
        reverse(words[i].begin(), words[i].end());
        result += words[i] + ' ';
    }
    reverse(words.back().begin(), words.back().end());
    result += words.back();
 
    cout << result << endl;
    return 0;
}
//This code is contributed by Shivam Tiwari

Java

import java.util.Arrays;
import java.util.stream.Collectors;
 
// This code is contributed by Mayank Sharma
public class reverseIndividual {
 
    public static void main(String[] args) {
 
        String str = "Welcome to GFG";
         
        // Splitting the string based on space and reverse each part
        // and then join
        String result = Arrays.asList(str.split(" "))
                .stream()
                .map(s -> new StringBuilder(s).reverse())
                .collect(Collectors.joining(" "));
 
        System.out.println(result);
 
    }
 
}

Python3

# python code
 
import re
 
def reverseIndividual(str):
    words = re.findall(r'\b\w+\b', str)
    result = " ".join(word[::-1] for word in words)
    return result
 
str = "Welcome to GFG"
print(reverseIndividual(str))
 
#code by ksam24000

C#

using System;
using System.Linq;
 
public class ReverseIndividual
{
    public static void Main(string[] args)
    {
        string str = "Welcome to GFG";
 
        // Splitting the string based on space and reverse each part
        // and then join
        string result = string.Join(" ", str.Split(' ')
            .Select(word => new string(word.Reverse().ToArray())));
 
        Console.WriteLine(result);
    }
}

Javascript

function reverseIndividualWords(str) {
    // Split the string based on space
    const words = str.split(' ');
 
    // Reverse each word and join them back with space
    const reversedWords = words.map(word => word.split('').reverse().join(''));
 
    // Join the reversed words to form the final result
    const result = reversedWords.join(' ');
 
    return result;
}
 
const str = 'Welcome to GFG';
const reversedStr = reverseIndividualWords(str);
console.log(reversedStr);

Output

emocleW ot GFG

Time complexity : O(n)

Auxiliary Space: O(n)

Way 3: Using StringBuffer Class.

Approach:

O(n) First, convert the string object into a StringBuffer object.
By using the reverse method of the StringBuffer class reverse the string.
Now, store the reverse sequence in a String array.
Run a loop that will create a new String by using these reverse words
Finally, return the new string.

Implementation:

C++

#include <algorithm>
#include <iostream>
#include <sstream>
#include <string>
 
std::string makeReverse(std::string str)
{
    std::reverse(str.begin(), str.end());
    std::istringstream iss(str);
    std::string word;
    std::string reverse;
    while (iss >> word) {
        reverse = word + " " + reverse;
    }
    return reverse;
}
 
int main()
{
    std::string str = "Geeks for Geeks";
    std::cout << makeReverse(str) << std::endl;
    return 0;
}

Java

/*package whatever //do not write package name here */
 
import java.io.*;
 
class GFG {
    static String makeReverse(String str)
    {
        StringBuffer s = new StringBuffer(str);
        str = s.reverse().toString();
        String[] rev = str.split(" ");
        StringBuffer reverse = new StringBuffer();
        for (int i = rev.length - 1; i >= 0; i--) {
            reverse.append(rev[i]).append(" ");
        }
        return reverse.toString();
    }
    public static void main(String[] args)
    {
        String str = "Geeks for Geeks";
        System.out.println(makeReverse(str));
    }
}
 
// This code is contributed by Adarsh Kumar

Python3

# Function to make the reverse of the string
def make_reverse(string: str) -> str:
    # Reversing the string
    string = string[::-1]
    # Splitting the string by space
    rev = string.split(" ")
    # Reversing the list of words
    rev = rev[::-1]
    # Joining the words to form a new string
    reversed_string = " ".join(rev)
    return reversed_string
 
 
# Driver code
if __name__ == "__main__":
    string = "Geeks for Geeks"
    print(make_reverse(string))
 
# This code is contributed by Shivam Tiwari

C#

using System;
 
class GFG {
    static string MakeReverse(string str)
    {
        char[] charArray = str.ToCharArray();
        Array.Reverse(charArray);
        str = new string(charArray);
        string[] rev = str.Split(' ');
        Array.Reverse(rev);
        return string.Join(" ", rev);
    }
 
    public static void Main()
    {
        string str = "Geeks for Geeks";
        Console.WriteLine(MakeReverse(str));
    }
}

Javascript

// Javascript code addition
 
function makeReverse(string) {
// Reversing the string
string = string.split("").reverse().join("");
// Splitting the string by space
let rev = string.split(" ");
// Reversing the list of words
rev = rev.reverse();
// Joining the words to form a new string
let reversedString = rev.join(" ");
return reversedString;
}
 
// Driver code
let string = "Geeks for Geeks";
console.log(makeReverse(string));
 
// The code is contributed by Nidhi goel.

Output

skeeG rof skeeG

Time complexity : O(n)

Auxiliary Space : O(n)

Way 4 :

Approach: To store the reversed string instead of just printing

Steps:

Create an empty stack to hold characters from the input string. Create two empty strings: ‘rev’ (for the reversed string) and ‘temp’ (for temporary storage).
Iterate through each character in the input string.
If the current character is a letter (an alphabet character) Add it to the ‘temp’ string to form a word .
If the current character is a space Append a space to the ‘rev’ string to separate words. Append the content of ‘temp’ (a word) to ‘rev’. Clear ‘temp’ to prepare for the next word.
After processing all characters. If ‘temp’ is not empty then add the content of ‘temp’ to the beginning of the ‘rev’ string.
Finally, return the ‘rev’ string.

C++

#include <iostream>
#include <stack>
 
class Reverse {
public:
    // function to reverse the individual words
    std::string reverse(std::string str) {
        // create a stack to access the string from end
        std::stack<char> s;
 
        // push all the characters of the stack
        for (int i = 0; i < str.length(); i++)
            s.push(str[i]);
 
        // rev string to store the required output
        std::string rev = "";
        std::string temp = "";
 
        // till the stack becomes empty
        while (!s.empty()) {
            // if top of the stack is a letter,
            // then append it to temp;
            if (isalpha(s.top()))
                temp += s.top();
            // if it is a space, then append space to rev
            // and also temp to rev
            else {
                rev = " " + temp + rev;
                // make the temp empty
                temp = "";
            }
            s.pop();
        }
        // if temp is not empty, add temp to rev at the front
        if (!temp.empty())
            rev = temp + rev;
 
        // return the output string
        return rev;
    }
};
 
int main() {
    std::string str = "Geeks for Geeks";
 
    Reverse obj;
    std::cout << obj.reverse(str) << std::endl;
     
    return 0;
}
 
// Siddhesh

Java

// Java program to reverse the individual words
 
import java.util.Stack;
 
public class Reverse {
 
    // function to reverse the individual words
    String reverse(String str)
    {
        // create a stack to access the string from end
        Stack<Character> s = new Stack<>();
 
        // push all the characters of the stack
        for (int i = 0; i < str.length(); i++)
            s.push(str.charAt(i));
 
        // rev string to store the required output
        String rev = "";
        String temp = "";
 
        // till the stack becomes empty
        while (!s.isEmpty()) {
            // if top of the stack is a letter,
            // then append it to temp;
            if (Character.isLetter(s.peek()))
                temp = temp + s.pop();
            // if it is a space, the append space to rev
            // and also temp to rev
            else {
                rev = " " + temp + rev;
                // make the temp empty
                temp = "";
                s.pop();
            }
        }
        // if temp is not empty, add temp to rev at the
        // front
        if (temp != "")
            rev = temp + rev;
 
        // return the output string
        return rev;
    }
 
    public static void main(String[] args)
    {
        String str = "Geeks for Geeks";
 
        Reverse obj = new Reverse();
        System.out.println(obj.reverse(str));
    }
}

Python3

class Reverse:
 
    # function to reverse the individual words
    def reverse(self, string):
        # create a stack to access the string from end
        stack = []
 
        # push all the characters of the string onto the stack
        for i in range(len(string)):
            stack.append(string[i])
 
        # rev string to store the required output
        rev = ""
        temp = ""
 
        # till the stack becomes empty
        while len(stack) > 0:
            # if top of the stack is a letter, then append it to temp
            if stack[-1].isalpha():
                temp = temp + stack.pop()
            # if it is a space, append space to rev and also temp to rev
            else:
                rev = " " + temp + rev
                # make the temp empty
                temp = ""
                stack.pop()
 
        # if temp is not empty, add temp to rev at the front
        if temp != "":
            rev = temp + rev
 
        # return the output string
        return rev
 
 
# main driver function
if __name__ == '__main__':
    str = "Geeks for Geeks"
 
    obj = Reverse()
    print(obj.reverse(str))

C#

using System;
using System.Collections.Generic;
 
class Reverse
{
    // Renamed the method to ReverseWords
    public string ReverseWords(string str)
    {
        // create a stack to access the string from end
        Stack<char> stack = new Stack<char>();
 
        // push all the characters onto the stack
        for (int i = 0; i < str.Length; i++)
        {
            stack.Push(str[i]);
        }
 
        // rev string to store the required output
        string rev = "";
        string temp = "";
 
        // until the stack becomes empty
        while (stack.Count > 0)
        {
            // if the top of the stack is a letter,
            // then append it to temp;
            if (char.IsLetter(stack.Peek()))
            {
                temp += stack.Pop();
            }
            // if it is a space, then append space to rev
            // and also temp to rev
            else
            {
                rev = " " + temp + rev;
                // make the temp empty
                temp = "";
                stack.Pop();
            }
        }
        // if temp is not empty, add temp to rev at the front
        if (!string.IsNullOrEmpty(temp))
        {
            rev = temp + rev;
        }
 
        // return the output string
        return rev;
    }
}
 
class Program
{
    static void Main()
    {
        string str = "Geeks for Geeks";
 
        Reverse obj = new Reverse();
        Console.WriteLine(obj.ReverseWords(str));
 
        // Pause execution to see the result
        Console.ReadLine();
    }
}
// contributed Siddhesh

Javascript

class Reverse {
    // function to reverse the individual words
    reverse(str) {
        // create an array (stack) to access the string from end
        const stack = [];
 
        // push all the characters onto the stack
        for (let i = 0; i < str.length; i++) {
            stack.push(str.charAt(i));
        }
 
        // rev string to store the required output
        let rev = '';
        let temp = '';
 
        // until the stack becomes empty
        while (stack.length > 0) {
            // if top of the stack is a letter,
            // then append it to temp;
            if (/[a-zA-Z]/.test(stack[stack.length - 1])) {
                temp += stack.pop();
            }
            // if it is a space, then append space to rev
            // and also temp to rev
            else {
                rev = ' ' + temp + rev;
                // make the temp empty
                temp = '';
                stack.pop();
            }
        }
        // if temp is not empty, add temp to rev at the front
        if (temp !== '') {
            rev = temp + rev;
        }
 
        // return the output string
        return rev;
    }
}
 
// Testing the Reverse class
const str = 'Geeks for Geeks';
const obj = new Reverse();
console.log(obj.reverse(str));

Output

skeeG rof skeeG

Time Complexity : O(N)

Auxiliary Space : O(N) for using stack .

Using Split and iterator in Python

Implementation:

Python3

def rev_sentence(sentence):
 
    # first split the string into words
    words = sentence.split(' ')
    reverse_sentence = ""
    for i in words:
        reverse_sentence += i[::-1]+' '
    # then reverse the split string list and join using space
 
    # finally return the joined string
    return reverse_sentence
 
 
if __name__ == "__main__":
    input = 'zambiatek quiz practice code'
    print(rev_sentence(input))

Output

skeeg ziuq ecitcarp edoc

Time Complexity: O(n)

Auxiliary Space:: O(n)

Method: Using join and split functions

Implementation:

Python3

# Python code to reverse words
 
s = " Geeks for Geeks"
l = []
# splitting the string
s = s.split()
for i in s:
    # reversing each word
    l.append(i[::-1])
# printing string using join
 
# function after reversing the
# words
 
 
print(" ".join(l))

Output

skeeG rof skeeG

Time Complexity: O(n)

Auxiliary Space: O(n)

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