Write a Program That Reads in Words and Prints Them Out in Reverse Order. Complete This Code.

Let there be a cord say "I AM A GEEK". And 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 than simpler fashion by using the property of the "%due south format specifier" .
Belongings: %s will become all the values until it gets Nil i.e. '\0'.

Instance: char String[] = "I AM A GEEK" is stored equally shown in the paradigm beneath :

Approach: Traverse the string from the terminal character, and motility towards the start grapheme. While traversing, if a space character is encountered, put a Nil in that position and print the remaining cord just after the NULL grapheme. Repeat this until the loop is over and when the loop ends, print the string, the %s will brand the press of characters until it encounters the kickoff NULL graphic symbol.

Let us see the approach with the help of diagrams:
pace i: Traverse from the last graphic symbol until it encounters a space graphic symbol .

Pace 2: Put a Zip character at the position of space grapheme and impress the string afterward it.

Step iii: At the finish, the loop ends when it reaches the kickoff graphic symbol, and then print the remaining characters, information technology will be printed the outset Cipher character, hence the kickoff word will exist printed.

C++

#include <iostream>

using namespace std;

string wordReverse(string str)

{

int i = str.length() - ane;

int start, end = i + i;

string event = "" ;

while (i >= 0) {

if (str[i] == ' ' ) {

start = i + ane;

while (start != end)

result += str[start++];

issue += ' ' ;

end = i;

}

i--;

}

start = 0;

while (beginning != end)

result += str[showtime++];

return result;

}

int main()

{

string str = "I AM A GEEK" ;

cout << wordReverse(str);

return 0;

}

C

#include <stdio.h>

#include <string.h>

void printReverse( char str[])

{

int length = strlen (str);

int i;

for (i = length - ane; i >= 0; i--) {

if (str[i] == ' ' ) {

str[i] = '\0' ;

printf ( "%s " , &(str[i]) + one);

}

}

printf ( "%s" , str);

}

int chief()

{

char str[] = "I AM A GEEK" ;

printReverse(str);

return 0;

}

Java

import java.io.*;

import java.lang.*;

import coffee.util.*;

course GFG {

static String wordReverse(String str)

{

int i = str.length() - ane ;

int first, end = i + 1 ;

String event = "" ;

while (i >= 0 ) {

if (str.charAt(i) == ' ' ) {

offset = i + 1 ;

while (start != end)

result += str.charAt(start++);

result += ' ' ;

end = i;

}

i--;

}

start = 0 ;

while (showtime != end)

issue += str.charAt(start++);

return result;

}

public static void main(Cord[] args)

{

Cord str = "I AM A GEEK" ;

System.out.print(wordReverse(str));

}

}

Python3

def wordReverse( str ):

i = len ( str ) - 1

start = end = i + 1

outcome = ''

while i > = 0 :

if str [i] = = ' ' :

start = i + i

while start ! = end:

result + = str [start]

commencement + = 1

result + = ' '

stop = i

i - = ane

start = 0

while start ! = end:

result + = str [start]

starting time + = ane

return result

str = 'I AM A GEEK'

print (wordReverse( str ))

C#

using Organization;

grade GFG {

static String wordReverse(String str)

{

int i = str.Length - 1;

int start, end = i + 1;

String consequence = "" ;

while (i >= 0) {

if (str[i] == ' ' ) {

start = i + 1;

while (get-go != end)

effect += str[start++];

effect += ' ' ;

cease = i;

}

i--;

}

start = 0;

while (start != end)

result += str[get-go++];

render result;

}

public static void Main()

{

String str = "I AM A GEEK" ;

Panel.Write(wordReverse(str));

}

}

PHP

<?php

function wordReverse( $str )

{

$i = strlen ( $str ) - 1;

$end = $i + ane;

$result = "" ;

while ( $i >= 0)

{

if ( $str [ $i ] == ' ' )

{

$get-go = $i + 1;

while ( $start != $end )

$result = $consequence . $str [ $get-go ++];

$result = $outcome . ' ' ;

$end = $i ;

}

$i --;

}

$start = 0;

while ( $starting time != $finish )

$result = $result . $str [ $start ++];

return $upshot ;

}

$str = "I AM A GEEK" ;

echo wordReverse( $str );

?>

Javascript

<script>

function wordReverse(str)

{

var i = str.length - i;

var offset, cease = i + 1;

var result = "" ;

while (i >= 0)

{

if (str[i] == ' ' )

{

first = i + 1;

while (beginning != end)

result += str[start++];

upshot += ' ' ;

stop = i;

}

i--;

}

start = 0;

while (start != end)

consequence += str[start++];

return result;

}

var str = "I AM A GEEK" ;

document.write(wordReverse(str));

</script>

Output:

GEEK A AM I

Fourth dimension Complexity: O(len(str))

Auxiliary Infinite: O(len(str))

Without using any actress infinite:

Get through the string and mirror each word in the string, so, at the end, mirror the whole string.

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 = due south.size();

while (s[i] == ' ' )

i++;

left = i;

while (i < n)

{

if (i + one == north || s[i] == ' ' )

{

int j = i - 1;

if (i + 1 == north)

j++;

while (left < j)

swap(s[left++], south[j--]);

left = i + i;

}

if (s[left] == ' ' && i > left)

left = i;

i++;

}

reverse(due south.begin(), southward.end());

return s;

}

int main()

{

cord str = "I AM A GEEK" ;

str = reverse_words(str);

cout << str;

return 0;

}

Output:

GEEK A AM I

Time Complexity: O(len(str))

Auxiliary Space: O(one)

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