Given string str, we need to print the reverse of individual words.
Examples:
Input: Hello World
Output: olleH dlroW
Explanation: Each word in "Hello World" is reversed individually, preserving the original order, resulting in "olleH dlroW".Input: Geeks for Geeks
Output: skeeG rof skeeG
Try It Yourself
[Expected Approach] Stack-Based Word Reversal - O(n) Time and O(n) Space
We use a stack to store words character by character until we encounter a space. When a space is found, we pop and print the characters from the stack, effectively reversing the word. This process continues for all words in the string
#include <bits/stdc++.h>
using namespace std;
string reverseWords(string str) {
stack<char> st;
string result = "";
for (int i = 0; i < str.length(); ++i) {
if (str[i] != ' ') {
st.push(str[i]);
} else {
while (!st.empty()) {
result += st.top();
st.pop();
}
result += " "; // Add space after the word
}
}
// Reverse the last word (since it may not end with a space)
while (!st.empty()) {
result += st.top();
st.pop();
}
return result;
}
// Driver program to test function
int main() {
string str = "Geeks for Geeks";
string reversedString = reverseWords(str);
cout << reversedString << endl;
return 0;
}
import java.util.Stack;
class GfG {
static String reverseWords(String str) {
Stack<Character> st = new Stack<>();
StringBuilder result = new StringBuilder();
for (int i = 0; i < str.length(); ++i) {
if (str.charAt(i) != ' ')
st.push(str.charAt(i));
else {
while (!st.isEmpty()) {
result.append(st.pop());
}
result.append(" ");
}
}
// Reverse the last word (if there's no space after it)
while (!st.isEmpty()) {
result.append(st.pop());
}
return result.toString();
}
// Driver code
public static void main(String[] args) {
String str = "Geeks for Geeks";
String reversed = reverseWords(str);
System.out.println(reversed);
}
}
def reverseWords(string):
st = []
result = ""
for i in range(len(string)):
if string[i] != " ":
st.append(string[i])
else:
while st:
result += st.pop()
result += " "
# Reverse the last word (if any)
while st:
result += st.pop()
return result
# Driver Code
if __name__ == "__main__":
string = "Geeks for Geeks"
print(reverseWords(string))
using System;
using System.Collections.Generic;
class GfG {
static string ReverseWords(string str)
{
Stack<char> st = new Stack<char>();
string result = "";
for (int i = 0; i < str.Length; ++i) {
if (str[i] != ' ') {
st.Push(str[i]);
}
else {
while (st.Count > 0) {
result += st.Pop();
}
result += " ";
}
}
// Reverse the last word (since there may not be
// space at the end)
while (st.Count > 0) {
result += st.Pop();
}
return result;
}
// Driver Code
public static void Main(string[] args)
{
string str = "Geeks for Geeks";
string reversed = ReverseWords(str);
Console.WriteLine(reversed);
}
}
function reverseWords(str)
{
let st = [];
let result = "";
for (let i = 0; i < str.length; ++i) {
if (str[i] !== " ")
// Push character to stack
st.unshift(str[i]);
else {
while (st.length !== 0) {
// Pop from stack and append
result += st.shift();
}
// Add space after word is reversed
result += " ";
}
}
// Reverse the last word
while (st.length !== 0) {
result += st.shift();
}
return result;
}
// Test the function
let str = "Geeks for Geeks";
// Output: "skeeG rof skeeG"
console.log(reverseWords(str));
Output
skeeG rof skeeG
[Alternate Approach] Using Inbulit Functions - O(n) Time and O(n) Space
To reverse individual words in a string, we can use built-in functions like
stringstreamin C++, StringBuilder in Java,splitin Python and other languages. After splitting the string into words, we iterate over each word and reverse it using thereversefunction.
#include <bits/stdc++.h>
using namespace std;
string reverseWords(string str)
{
// word variable to store word
string word;
// making a string stream
stringstream iss(str);
// Read and print each word.
string s1 = "";
while (iss >> word) {
reverse(word.begin(), word.end());
s1 += word;
s1 += " ";
}
return s1;
}
// Driver code
int main()
{
string s = "Geeks for Geeks";
string s1 = reverseWords(s);
cout<<s1;
return 0;
}
public class GfG {
public static String reverseWords(String sentence)
{
// Split the string into words
String[] words = sentence.split(" ");
StringBuilder reverseSentence = new StringBuilder();
// Reverse each word and append to the result
for (String word : words) {
reverseSentence
.append(new StringBuilder(word)
.reverse()
.toString())
.append(" ");
}
// Remove the trailing space and return the result
return reverseSentence.toString().trim();
}
public static void main(String[] args)
{
String input = "Geeks for Geeks";
System.out.println(reverseWords(input));
}
}
# Function to make the reverse of the string
def reverseWords(string: 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(reverseWords(string))
using System;
using System.Linq;
public class GfG {
static string reverseWords(string s)
{
return string.Join(
" ", s.Split(' ').Select(
word => new string(
word.Reverse().ToArray())));
}
public static void Main(string[] args)
{
string str = "Geeks for Geeks";
// Splitting the string based on space and reverse
// each part and then join
string result = reverseWords(str);
Console.WriteLine(result);
}
}
function reverseWords(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 = "Geeks for Geeks";
const reversedStr = reverseWords(str);
console.log(reversedStr);
Output
skeeG rof skeeG