# Reverse individual words with O(1) extra space

Given a string str, the task is to reverse all the individual words.

Examples:

Input: str = “Hello World”
Output: olleH dlroW

Input: str = “Geeks for Geeks”
Output: skeeG rof skeeG

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach: A solution to the above problem has been discussed in this post. It has a time complexity of O(n) and uses O(n) extra space. In this post, we will discuss a solution which uses O(1) extra space.

• Traverse through the string until we encounter a space.
• After encountering the space, we use two variables ‘start’ and ‘end’ pointing to the first and last character of the word and we reverse that particular word.
• Repeat the above steps until the last word.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach ` `#include ` `using` `namespace` `std; ` ` `  `// Function to resturn the string after ` `// reversing the individual words ` `string reverseWords(string str) ` `{ ` ` `  `    ``// Pointer to the first character ` `    ``// of the first word ` `    ``int` `start = 0; ` `    ``for` `(``int` `i = 0; i <= str.size(); i++) { ` ` `  `        ``// If the current word has ended ` `        ``if` `(str[i] == ``' '` `|| i == str.size()) { ` ` `  `            ``// Pointer to the last character ` `            ``// of the current word ` `            ``int` `end = i - 1; ` ` `  `            ``// Reverse the current word ` `            ``while` `(start < end) { ` `                ``swap(str[start], str[end]); ` `                ``start++; ` `                ``end--; ` `            ``} ` ` `  `            ``// Pointer to the first character ` `            ``// of the next word ` `            ``start = i + 1; ` `        ``} ` `    ``} ` ` `  `    ``return` `str; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``string str = ``"Geeks for Geeks"``; ` ` `  `    ``cout << reverseWords(str); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java implementation of the approach  ` `class` `GFG ` `{ ` `    ``static` `String swap(String str, ``int` `i, ``int` `j)  ` `    ``{  ` `        ``StringBuilder sb = ``new` `StringBuilder(str);  ` `        ``sb.setCharAt(i, str.charAt(j));  ` `        ``sb.setCharAt(j, str.charAt(i));  ` `        ``return` `sb.toString();  ` `    ``}  ` `     `  `    ``// Function to resturn the string after  ` `    ``// reversing the individual words  ` `    ``static` `String reverseWords(String str)  ` `    ``{  ` `     `  `        ``// Pointer to the first character  ` `        ``// of the first word  ` `        ``int` `start = ``0``;  ` `        ``for` `(``int` `i = ``0``; i < str.length(); i++)  ` `        ``{  ` `     `  `            ``// If the current word has ended  ` `            ``if` `(str.charAt(i) == ``' '` `||  ` `                    ``i == str.length()-``1` `) ` `            ``{ ` `     `  `                ``// Pointer to the last character  ` `                ``// of the current word  ` `                ``int` `end; ` `                ``if` `(i == str.length()-``1``) ` `                    ``end = i ; ` `                ``else` `                    ``end = i - ``1` `;  ` `                     `  `                ``// Reverse the current word  ` `                ``while` `(start < end) ` `                ``{  ` `                    ``str = swap(str,start,end) ; ` `                    ``start++;  ` `                    ``end--;  ` `                ``}  ` `     `  `                ``// Pointer to the first character  ` `                ``// of the next word  ` `                ``start = i + ``1``;  ` `            ``}  ` `        ``}  ` `     `  `        ``return` `str ;  ` `    ``}  ` `     `  `    ``// Driver code  ` `    ``public` `static` `void` `main(String args[]) ` `    ``{ ` `        ``String str = ``"Geeks for Geeks"``;  ` `     `  `        ``System.out.println(reverseWords(str));  ` `    ``} ` `} ` ` `  `// This code is contributed by AnkitRai01 `

## Python3

 `# Python3 implementation of the approach ` ` `  `# Function to resturn the after ` `# reversing the individual words ` `def` `reverseWords(``Str``): ` ` `  `    ``# Pointer to the first character ` `    ``# of the first word ` `     `  `    ``start ``=` `0` `    ``for` `i ``in` `range``(``len``(``Str``)): ` `        ``# If the current word has ended ` `        ``if` `(``Str``[i] ``=``=` `' '` `or` `i ``=``=` `len``(``Str``)``-``1``): ` ` `  `            ``# Pointer to the last character ` `            ``# of the current word ` `            ``end ``=` `i ``-` `1` `            ``if``(i ``=``=` `len``(``Str``)``-``1``): ` `                ``end ``=` `i ` `             `  `            ``# Reverse the current word ` `            ``while` `(start < end): ` `                ``Str``[start], ``Str``[end]``=``Str``[end],``Str``[start] ` `                ``start``+``=``1` `                ``end``-``=``1` `             `  ` `  `            ``# Pointer to the first character ` `            ``# of the next word ` `            ``start ``=` `i ``+` `1` `         `  `    ``return` `"".join(``Str``) ` ` `  `# Driver code ` `Str` `=` `"Geeks for Geeks"` `Str``=``[i ``for` `i ``in` `Str``] ` ` `  `print``(reverseWords(``Str``)) ` ` `  `# This code is contributed by mohit kumar 29 `

## C#

 `// C# implementation of the approach  ` `using` `System; ` `     `  `class` `GFG ` `{ ` ` `  `    ``// Function to resturn the string after  ` `    ``// reversing the individual words  ` `    ``// Function to resturn the string after  ` `    ``// reversing the individual words  ` `    ``static` `String reverseWords(String str)  ` `    ``{  ` `     `  `        ``// Pointer to the first character  ` `        ``// of the first word  ` `        ``int` `start = 0;  ` `        ``for` `(``int` `i = 0; i < str.Length; i++)  ` `        ``{  ` `     `  `            ``// If the current word has ended  ` `            ``if` `(str[i] == ``' '` `||  ` `                    ``i == str.Length-1 ) ` `            ``{ ` `     `  `                ``// Pointer to the last character  ` `                ``// of the current word  ` `                ``int` `end; ` `                ``if` `(i == str.Length-1) ` `                    ``end = i ; ` `                ``else` `                    ``end = i - 1 ;  ` `                     `  `                ``// Reverse the current word  ` `                ``while` `(start < end) ` `                ``{  ` `                    ``str = swap(str,start,end) ; ` `                    ``start++;  ` `                    ``end--;  ` `                ``}  ` `     `  `                ``// Pointer to the first character  ` `                ``// of the next word  ` `                ``start = i + 1;  ` `            ``}  ` `        ``}  ` `     `  `        ``return` `str ;  ` `    ``} ` `     `  `    ``static` `String swap(String str, ``int` `i, ``int` `j)  ` `    ``{  ` `        ``char` `[]ch = str.ToCharArray();  ` `        ``char` `temp = ch[i];  ` `        ``ch[i] = ch[j];  ` `        ``ch[j] = temp;  ` `        ``return` `String.Join(``""``,ch);  ` `    ``} ` `     `  `    ``// Driver code  ` `    ``public` `static` `void` `Main(String []args) ` `    ``{ ` `        ``String str = ``"Geeks for Geeks"``;  ` `     `  `        ``Console.WriteLine(reverseWords(str));  ` `    ``} ` `} ` ` `  `/* This code is contributed by PrinciRaj1992 */`

Output:

```skeeG rof skeeG
```

Time Complexity: O(n)
Auxiliary Space: O(1)

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