Reverse words in a string
Last Updated :
18 Mar, 2025
Given a string str, your task is to reverse the order of the words in the given string. Note that str may contain leading or trailing dots(.) or multiple trailing dots(.) between two words. The returned string should only have a single dot(.) separating the words.
Examples:
Input: str = "i.like.this.program.very.much"
Output: str = "much.very.program.this.like.i"
Explanation:The words in the input string are reversed while maintaining the dots as separators, resulting in "much.very.program.this.like.i".
Input: str = ”..geeks..for.geeks.”
Output: str = “geeks.for.geeks”
Input: str = "...home......"
Output: str = "home"
[Naive Approach] Using Stack - O(n) Time and O(n) Space
Push all words separated by dots into a stack, then pop each word one by one and append it back to form the reversed string.
C++
#include <bits/stdc++.h>
using namespace std;
string reverseWords(string str) {
stack<string> st;
string s = "";
for (int i = 0; i < str.length(); i++) {
if (str[i] != '.') {
s += str[i];
}
// If we see a dot, we push the
// previously seen word into the stack.
else if (!s.empty()) {
st.push(s);
s = "";
}
}
// Last word remaining, add it to stack
if (!s.empty()) {
st.push(s);
}
s = "";
// Now add from top to bottom of the stack
while (!st.empty()) {
s += st.top();
st.pop();
if (!st.empty()) {
s += ".";
}
}
return s;
}
int main() {
string s = "..geeks..for.geeks.";
cout << reverseWords(s) << endl;
return 0;
}
import java.util.Stack;
class GfG {
static String reverseWords(String str) {
Stack<String> stack = new Stack<>();
StringBuilder word = new StringBuilder();
// Iterate through the string
for (int i = 0; i < str.length(); i++) {
// If not a dot, build the current word
if (str.charAt(i) != '.') {
word.append(str.charAt(i));
}
// If we see a dot, push the word into the stack
else if (word.length() > 0) {
stack.push(word.toString());
word.setLength(0);
}
}
// Last word remaining, push it to stack
if (word.length() > 0) {
stack.push(word.toString());
}
// Rebuild the string from the stack
StringBuilder result = new StringBuilder();
while (!stack.isEmpty()) {
result.append(stack.pop());
if (!stack.isEmpty()) {
result.append(".");
}
}
return result.toString();
}
public static void main(String[] args) {
String str = "..geeks..for.geeks.";
System.out.println(reverseWords(str));
}
}
def reverse_words(str):
stack = []
word = ""
# Iterate through the string
for ch in str:
# If not a dot, build the current word
if ch != '.':
word += ch
# If we see a dot, push the word into the stack
elif word:
stack.append(word)
word = ""
# Last word remaining, push it to stack
if word:
stack.append(word)
# Rebuild the string from the stack
return ".".join(stack[::-1])
if __name__ == "__main__":
str = "..geeks..for.geeks."
print(reverse_words(str))
using System;
using System.Collections.Generic;
class GfG {
static string ReverseWords(string str) {
Stack<string> stack = new Stack<string>();
string word = "";
// Iterate through the string
for (int i = 0; i < str.Length; i++) {
// If not a dot, build the current word
if (str[i] != '.') {
word += str[i];
}
// If we see a dot, push the word into the stack
else if (word.Length > 0) {
stack.Push(word);
word = "";
}
}
// Last word remaining, push it to stack
if (word.Length > 0) {
stack.Push(word);
}
// Rebuild the string from the stack
string result = "";
while (stack.Count > 0) {
result += stack.Pop();
if (stack.Count > 0) {
result += ".";
}
}
return result;
}
static void Main() {
string str = "..geeks..for.geeks.";
Console.WriteLine(ReverseWords(str));
}
}
function reverseWords(str) {
let stack = [];
let word = "";
// Iterate through the string
for (let i = 0; i < str.length; i++) {
// If not a dot, build the current word
if (str[i] != '.') {
word += str[i];
}
// If we see a dot, push the word into the stack
else if (word.length > 0) {
stack.push(word);
word = "";
}
}
// Last word remaining, push it to stack
if (word.length > 0) {
stack.push(word);
}
// Rebuild the string from the stack
return stack.reverse().join('.');
}
let str = "..geeks..for.geeks.";
console.log(reverseWords(str));
[Expected Approach] Using Two Pointer - O(n) Time and O(1) Space
Reverse the entire string, then iterate through it to extract words separated by dots. Reverse each word individually and update the original string until the end is reached. Refer to the figure to understand better...
C++
// C++ program to reverse words in a string
#include <bits/stdc++.h>
using namespace std;
string reverseWords(string str) {
// reverse the whole string
reverse(str.begin(), str.end());
int n = str.size();
int i = 0;
for (int l = 0; l < n; ++l) {
if (str[l] != '.') {
// go to the beginning of the word
if (i != 0) str[i++] = '.';
// go to the end of the word
int r = l;
while (r < n && str[r] != '.') str[i++] = str[r++];
// reverse the word
reverse(str.begin() + i - (r - l), str.begin() + i);
// move to the next word
l = r;
}
}
str.erase(str.begin() + i, str.end());
return str;
}
int main() {
string str = "..geeks..for.geeks.";
cout << reverseWords(str) << endl;
return 0;
}
// C program to reverse words in a string
#include <stdio.h>
#include <string.h>
// Function to reverse a string
void reverse(char* begin, char* end) {
char temp;
while (begin < end) {
temp = *begin;
*begin++ = *end;
*end-- = temp;
}
}
// Function to reverse words in a string
char* reverseWords(char* str) {
// reverse the whole string
reverse(str, str + strlen(str) - 1);
int n = strlen(str);
int i = 0;
for (int l = 0; l < n; ++l) {
if (str[l] != '.') {
// go to the beginning of the word
if (i != 0) str[i++] = '.';
// go to the end of the word
int r = l;
while (r < n && str[r] != '.') str[i++] = str[r++];
// reverse the word
reverse(str + i - (r - l), str + i - 1);
// move to the next word
l = r;
}
}
str[i] = '\0';
return str;
}
int main() {
char str[] = "..geeks..for.geeks.";
printf("%s\n", reverseWords(str));
return 0;
}
// Java program to reverse words in a string
import java.util.*;
class GfG {
static String reverseWords(String str) {
// Convert the string to mutable StringBuilder
StringBuilder s = new StringBuilder(str);
// Reverse the whole string
s.reverse();
int n = s.length();
int i = 0;
for (int l = 0; l < n; ++l) {
if (s.charAt(l) != '.') {
// go to the beginning of the word
if (i != 0) s.setCharAt(i++, '.');
// go to the end of the word
int r = l;
while (r < n && s.charAt(r) != '.') {
s.setCharAt(i++, s.charAt(r++));
}
// reverse the word
int start = i - (r - l);
int end = i - 1;
while (start < end) {
char temp = s.charAt(start);
s.setCharAt(start, s.charAt(end));
s.setCharAt(end, temp);
start++;
end--;
}
// move to the next word
l = r;
}
}
return s.substring(0, i);
}
public static void main(String[] args) {
String str = "..geeks..for.geeks.";
System.out.println(reverseWords(str));
}
}
# Python program to reverse words in a string
def reverseWords(s):
# reverse the whole string
s = s[::-1]
n = len(s)
i = 0
result = []
l = 0
while l < n:
if s[l] != '.':
# go to the beginning of the word
if i != 0:
result.append('.')
i += 1
# go to the end of the word
r = l
while r < n and s[r] != '.':
result.append(s[r])
i += 1
r += 1
# reverse the word
result[i - (r - l):i] = reversed(result[i - (r - l):i])
# move to the next word
l = r
l += 1
return ''.join(result)
if __name__ == "__main__":
s = "..geeks..for.geeks."
print(reverseWords(s))
// C# program to reverse words in a string
using System;
class GfG {
static void Reverse(char[] str, int start, int end) {
while (start < end) {
char temp = str[start];
str[start] = str[end];
str[end] = temp;
start++;
end--;
}
}
static string ReverseWords(string str) {
// Convert the string to a char array
char[] s = str.ToCharArray();
// reverse the whole string
Reverse(s, 0, s.Length - 1);
int n = s.Length;
int i = 0;
for (int l = 0; l < n; ++l) {
if (s[l] != '.') {
// go to the beginning of the word
if (i != 0) s[i++] = '.';
// go to the end of the word
int r = l;
while (r < n && s[r] != '.') s[i++] = s[r++];
// reverse the word
Reverse(s, i - (r - l), i - 1);
// move to the next word
l = r;
}
}
return new String(s, 0, i);
}
static void Main(string[] args) {
string str = "..geeks..for.geeks.";
Console.WriteLine(ReverseWords(str));
}
}
// JavaScript program to reverse words in a string
function reverse(str, start, end) {
while (start < end) {
let temp = str[start];
str[start] = str[end];
str[end] = temp;
start++;
end--;
}
}
function reverseWords(str) {
// Convert the string to an array of characters
str = str.split('');
// reverse the whole string
reverse(str, 0, str.length - 1);
let n = str.length;
let i = 0;
for (let l = 0; l < n; ++l) {
if (str[l] !== '.') {
// go to the beginning of the word
if (i !== 0) str[i++] = '.';
// go to the end of the word
let r = l;
while (r < n && str[r] !== '.') str[i++] = str[r++];
// reverse the word
reverse(str, i - (r - l), i - 1);
// move to the next word
l = r;
}
}
return str.slice(0, i).join('');
}
let str = "..geeks..for.geeks.";
console.log(reverseWords(str));
[Alternate Approach] Using inbuilt library functions - O(n) Time and O(n) Space
We can efficiently solve this problem using built-in library functions like split to break the string into words, reverse to reverse the order of words, and stringstream (or equivalent functions) to reconstruct the final reversed string. This approach simplifies implementation and improves readability.
C++
#include <bits/stdc++.h>
using namespace std;
string reverseWords(string str) {
// Split the input string by '.' while
// ignoring multiple consecutive dots
vector<string> words;
stringstream ss(str);
string word;
while (getline(ss, word, '.')) {
if (!word.empty()) {
// Ignore empty words caused by multiple dots
words.push_back(word);
}
}
// Reverse the words
reverse(words.begin(), words.end());
// Join the reversed words back into a string
string result;
for (int i = 0; i < words.size(); ++i) {
if (i > 0) {
result += '.';
}
result += words[i];
}
return result;
}
int main() {
string str = "..geeks..for.geeks.";
cout << reverseWords(str) << endl;
return 0;
}
import java.util.*;
class GfG {
static String reverseWords(String str) {
// Split the input string by '.' while
// ignoring multiple consecutive dots
List<String> words = new ArrayList<>();
String[] parts = str.split("\\.");
for (String word : parts) {
if (!word.isEmpty()) {
// Ignore empty words caused by multiple dots
words.add(word);
}
}
// Reverse the words
Collections.reverse(words);
// Join the reversed words back into a string
return String.join(".", words);
}
public static void main(String[] args) {
String str = "..geeks..for.geeks.";
System.out.println(reverseWords(str));
}
}
def reverse_words(str):
# Split the input string by '.' while
# ignoring multiple consecutive dots
words = [word for word in str.split('.') if word]
# Reverse the words
words.reverse()
# Join the reversed words back into a string
return '.'.join(words)
if __name__ == "__main__":
str = "..geeks..for.geeks."
print(reverse_words(str))
using System;
using System.Collections.Generic;
class GfG {
static string ReverseWords(string str) {
// Split the input string by '.' while
// ignoring multiple consecutive dots
List<string> words = new List<string>();
string[] parts = str.Split('.');
foreach (string word in parts) {
if (!string.IsNullOrEmpty(word)) {
// Ignore empty words caused by multiple dots
words.Add(word);
}
}
// Reverse the words
words.Reverse();
// Join the reversed words back into a string
return string.Join(".", words);
}
static void Main(string[] args) {
string str = "..geeks..for.geeks.";
Console.WriteLine(ReverseWords(str));
}
}
function reverseWords(str) {
// Split the input string by '.' while
// ignoring multiple consecutive dots
let words = str.split('.').filter(word => word);
// Reverse the words
words.reverse();
// Join the reversed words back into a string
return words.join('.');
}
let str = "..geeks..for.geeks.";
console.log(reverseWords(str));
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