Let there be a string say “I AM A GEEK”. So, the output should be “GEEK A AM I” . This can done in many ways. One of the solutions is given in Reverse words in a string .
Examples:
Input : I AM A GEEK Output : GEEK A AM I Input : GfG IS THE BEST Output : BEST THE IS GfG
This can be done in more simpler way by using the property of the “%s format specifier” .
Property: %s will get all the values until it gets NULL i.e. ‘\0’.
Example: char String[] = “I AM A GEEK” is stored as shown in the image below :
Approach: Traverse the string from the last character, and move towards the first character. While traversing, if a space character is encountered, put a NULL in that position and print the remaining string just after the NULL character. Repeat this until the loop is over and when the loop ends, print the string, the %s will make the printing of characters until it encounters the first NULL character.
Let us see the approach with the help of diagrams:
step 1: Traverse from the last character until it encounters a space character .
Step 2: Put a NULL character at the position of space character and print the string after it.
Step 3: At the end, the loop ends when it reaches the first character, so print the remaining characters, it will be printed the first NULL character, hence the first word will be printed.
Implementation:
C++
// C++ program to print reverse// of words in a string.#include <iostream>using namespace std;string wordReverse(string str){ int i = str.length() - 1; int start, end = i + 1; string result = ""; while (i >= 0) { if (str[i] == ' ') { start = i + 1; while (start != end) result += str[start++]; result += ' '; end = i; } i--; } start = 0; while (start != end) result += str[start++]; return result;}// Driver codeint main(){ string str = "I AM A GEEK"; cout << wordReverse(str); return 0;}// This code is contributed// by Imam |
C
// C program to print reverse of words in// a string.#include <stdio.h>#include <string.h>void printReverse(char str[]){ int length = strlen(str); // Traverse string from end int i; for (i = length - 1; i >= 0; i--) { if (str[i] == ' ') { // putting the NULL character at the // position of space characters for // next iteration. str[i] = '\0'; // Start from next character printf("%s ", &(str[i]) + 1); } } // printing the last word printf("%s", str);}// Driver codeint main(){ char str[] = "I AM A GEEK"; printReverse(str); return 0;} |
Java
// Java program to print reverse// of words in a string.import java.io.*;import java.lang.*;import java.util.*;class GFG { static String wordReverse(String str) { int i = str.length() - 1; int start, end = i + 1; String result = ""; while (i >= 0) { if (str.charAt(i) == ' ') { start = i + 1; while (start != end) result += str.charAt(start++); result += ' '; end = i; } i--; } start = 0; while (start != end) result += str.charAt(start++); return result; } // Driver code public static void main(String[] args) { String str = "I AM A GEEK"; System.out.print(wordReverse(str)); }}// This code is contributed// by Akanksha Rai(Abby_akku) |
Python3
# Python3 program to print reverse# of words in a string.def wordReverse(str): i = len(str)-1 start = end = i+1 result = '' while i >= 0: if str[i] == ' ': start = i+1 while start != end: result += str[start] start += 1 result += ' ' end = i i -= 1 start = 0 while start != end: result += str[start] start += 1 return result# Driver Codestr = 'I AM A GEEK'print(wordReverse(str))# This code is contributed# by SamyuktaSHegde |
C#
// C# program to print reverse// of words in a string.using System;class GFG { static String wordReverse(String str) { int i = str.Length - 1; int start, end = i + 1; String result = ""; while (i >= 0) { if (str[i] == ' ') { start = i + 1; while (start != end) result += str[start++]; result += ' '; end = i; } i--; } start = 0; while (start != end) result += str[start++]; return result; } // Driver code public static void Main() { String str = "I AM A GEEK"; Console.Write(wordReverse(str)); }}// This code is contributed// by Akanksha Rai(Abby_akku) |
PHP
<?php // PHP program to print reverse // of words in a stringfunction wordReverse($str){ $i = strlen($str) - 1; $end = $i + 1; $result = ""; while($i >= 0) { if($str[$i] == ' ') { $start = $i + 1; while($start != $end) $result = $result . $str[$start++]; $result = $result . ' '; $end = $i; } $i--; } $start = 0; while($start != $end) $result = $result . $str[$start++]; return $result;}// Driver code$str = "I AM A GEEK";echo wordReverse($str);// This code is contributed by ita_c?> |
Javascript
<script>// Javascript program to print reverse // of words in a string.function wordReverse(str){ var i = str.length - 1; var start, end = i + 1; var result = ""; while (i >= 0) { if (str[i] == ' ') { start = i + 1; while (start != end) result += str[start++]; result += ' '; end = i; } i--; } start = 0; while (start != end) result += str[start++]; return result;}// Driver codevar str = "I AM A GEEK";document.write(wordReverse(str));// This code is contributed by rutvik_56</script> |
Output
GEEK A AM I
Time Complexity: O(len(str))
Auxiliary Space: O(len(str))
Without using any extra space:
Go through the string and mirror each word in the string, then, at the end, mirror the whole string.
Implementation: The following C++ code can handle multiple contiguous spaces.
C++
#include <algorithm>#include <iostream>#include <string>using namespace std;string reverse_words(string s){ int left = 0, i = 0, n = s.length(); while (s[i] == ' '){ i++; } left = i; while (i < n) { if (i + 1 == n || s[i] == ' ') { int j = i - 1; if (i + 1 == n) j++; reverse(s.begin()+left, s.begin()+j+1); left = i + 1; } if (left < n && s[left] == ' ' && i > left) left = i; i++; } // reversing the string reverse(s.begin(), s.end()); return s;}int main(){ string str = "I AM A GEEK"; str = reverse_words(str); cout << str; return 0;// This code is contributed// by Gatea David} |
Java
/*package whatever //do not write package name here */import java.io.*;class GFG { static String reverse_words(String s) { int left = 0, i = 0, n = s.length(); while (s.charAt(i) == ' '){ i++; } left = i; while (i < n) { if (i + 1 == n || s.charAt(i) == ' ') { int j = i - 1; if (i + 1 == n) j++; while (left<n && j<n && left < j){ char ch[] = s.toCharArray(); char temp = ch[left]; ch[left] = ch[j]; ch[j] = temp; s = String.valueOf(ch); left++; j--; } left = i + 1; } if (left < n && s.charAt(left) == ' ' && i > left) left = i; i++; } // reversing the string char ch[] = s.toCharArray(); int len = s.length(); for (i=0; i < (len/2); i++) { char temp = ch[i]; ch[i] = ch[len - i - 1]; ch[len - i- 1] = temp; } s = String.valueOf(ch); return s; } public static void main(String args[]) { String str = "I AM A GEEK"; str = reverse_words(str); System.out.println(str); }}// This code is contributed by shinjanpatra. |
Python3
# Python code for the same approachdef reverse_words(s): left, i, n = 0, 0, len(s) while (s[i] == ' '): i += 1 left = i while (i < n): if (i + 1 == n or s[i] == ' '): j = i - 1 if (i + 1 == n): j += 1 while (left < j): s = s[0:left] + s[j] + s[left+1:j] + s[left] + s[j+1:] left += 1 j -= 1 left = i + 1 if (i > left and s[left] == ' '): left = i i += 1 s = s[::-1] return s# driver codeStr = "I AM A GEEK"Str = reverse_words(Str)print(Str)# This code is contributed by shinjanpatra |
C#
// C# program to print reverse// of words in a string.using System;class GFG { static String reverse_words(String s) { var left = 0; var i = 0; var n = s.Length; while (s[i] == ' ') { i++; } left = i; while (i < n) { if (i + 1 == n || s[i] == ' ') { var j = i - 1; if (i + 1 == n) { j++; } while (left < n && j < n && left < j) { char[] ch = s.ToCharArray(); var temp = ch[left]; ch[left] = ch[j]; ch[j] = temp; s = new string(ch); left++; j--; } left = i + 1; } if (left < n && s[left] == ' ' && i > left) { left = i; } i++; } // reversing the string int len = s.Length; char[] chh = s.ToCharArray(); for (i = 0; i < (len / 2); i++) { char temp = chh[i]; chh[i] = chh[len - i - 1]; chh[len - i - 1] = temp; } s = new string(chh); return s; } public static void Main(String[] args) { string str = "I AM A GEEK"; str = reverse_words(str); Console.WriteLine(str); }}// This code is contributed by Aarti_Rathi |
Javascript
<script>// JavaScript code for the approachfunction reverse_words(s){ let left = 0, i = 0, n = s.length; while (s[i] == ' ') i++; left = i; while (i < n) { if (i + 1 == n || s[i] == ' ') { let j = i - 1; if (i + 1 == n) j++; let temp; let a = s.split(""); while (left < j){ temp = a[left]; a[left] = a[j]; a[j] = temp; left++; j--; } s = a.join(""); left = i + 1; } if (s[left] == ' ' && i > left) left = i; i++; } s = s.split('').reverse().join(''); return s;}// driver codelet str = "I AM A GEEK";str = reverse_words(str);document.write(str);// This code is contributed by shinjanpatra </script> |
Output
GEEK A AM I
Time Complexity: O(len(str))
Auxiliary Space: O(1)
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