Reverse words in a given string | Set 2
Last Updated :
30 Apr, 2023
Given string str, the task is to reverse the string by considering each word of the string, str as a single unit.
Examples:
Input: str = “geeks quiz practice code”
Output: code practice quiz geeks
Explanation:
The words in the given string are [“geeks”, “quiz”, “practice”, “code”].
Therefore, after reversing the order of the words, the required output is“code practice quiz geeks”.
Input: str = “getting good at coding needs a lot of practice”
Output: practice of lot a needs coding at good getting
In-place Reversal Approach: Refer to the article Reverse words in a given string for the in-place reversal of words followed by a reversal of the entire string.
Time Complexity: O(N)
Auxiliary Space: O(1)
Stack-based Approach: In this article, the approach to solving the problem using Stack is going to be discussed. The idea here is to push all the words of str into the Stack and then print all the elements of the Stack. Follow the steps below to solve the problem:
- Create a Stack to store each word of the string str.
- Iterate over string str, and separate each word of str by a space delimiter.
- Push all the words of str into the stack.
- Print all the elements of the stack one by one.
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
void printRev(string str)
{
stack<string> st;
stringstream ss(str);
string temp;
while (getline(ss, temp, ' ')) {
st.push(temp);
}
while (!st.empty()) {
cout << st.top() << " ";
st.pop();
}
}
int main()
{
string str;
str = "geeks quiz practice code";
printRev(str);
return 0;
}
|
Java
import java.util.*;
class GFG{
static void printRev(String str)
{
Stack<String> st = new Stack<String>();
String[] ss = str.split(" ");
for (String temp : ss)
{
st.add(temp);
}
while (!st.isEmpty())
{
System.out.print(st.peek() + " ");
st.pop();
}
}
public static void main(String[] args)
{
String str;
str = "geeks quiz practice code";
printRev(str);
}
}
|
Python3
def printRev(strr):
strr = strr.split(" ")
st = []
for i in strr:
st.append(i)
while len(st) > 0:
print(st[-1], end = " ")
del st[-1]
if __name__ == '__main__':
strr = "geeks quiz practice code"
printRev(strr)
|
C#
using System;
using System.Collections;
class GFG{
static void printRev(string str)
{
Stack st = new Stack();
String[] separator = {" "};
string[] ss = str.Split(separator,
int.MaxValue,
StringSplitOptions.RemoveEmptyEntries);
foreach(string temp in ss)
{
st.Push(temp);
}
while (st.Count > 0)
{
Console.Write(st.Peek() + " ");
st.Pop();
}
}
public static void Main(string[] args)
{
string str;
str = "geeks quiz practice code";
printRev(str);
}
}
|
Javascript
<script>
function printRev(str)
{
let st = [];
let ss = str.split(" ");
for (let temp=0;temp< ss.length;temp++)
{
st.push(ss[temp]);
}
while (st.length!=0)
{
document.write(st.pop() + " ");
}
}
let str;
str = "geeks quiz practice code";
printRev(str);
</script>
|
Output
code practice quiz geeks
Time Complexity: O(N), where N denotes the length of the string.
Auxiliary Space: O(N)
Method #2: Using built-in python functions
- As all the words in a sentence are separated by spaces.
- We have to split the sentence by spaces using split().
- We split all the words by spaces and store them in a list.
- Reverse this list and print it.
C++
#include <bits/stdc++.h>
using namespace std;
void printRev(string lis[])
{
reverse(lis, lis + 4);
for(int i = 0; i < 4; i++)
{
cout << lis[i] << " ";
}
}
int main()
{
string strr[] = {"geeks", "quiz", "practice", "code"};
printRev(strr);
return 0;
}
|
Java
import java.util.*;
class GFG{
static void printRev(String string)
{
String[] lis = string.split(" ", 0);
Collections.reverse(Arrays.asList(lis));
for(String li : lis)
{
System.out.print(li + " ");
}
}
public static void main(String[] args)
{
String strr = "geeks quiz practice code";
printRev(strr);
}
}
|
Python3
def printRev(string):
lis = list(string.split())
lis.reverse()
print(*lis)
if __name__ == '__main__':
strr = "geeks quiz practice code"
printRev(strr)
|
C#
using System;
class GFG {
static void printRev(string String)
{
string[] lis = String.Split(' ');
Array.Reverse(lis);
foreach(string li in lis)
{
Console.Write(li + " ");
}
}
static void Main() {
string strr = "geeks quiz practice code";
printRev(strr);
}
}
|
Javascript
<script>
function printRev(string){
var lis = string.split(' ');
console.log(lis);
lis.reverse();
console.log(lis);
document.write(lis.join(' '));
}
var strr = "geeks quiz practice code"
printRev(strr)
</script>
|
Output
code practice quiz geeks
Time Complexity : O(n), n is the number of strings.
Auxiliary Space : O(1)
Using a loop and stack in python:
Approach:
The function starts by importing the time module, which is used to measure the time taken to perform the operation.
The function defines a variable stack as an empty list, which will be used to store the words in reverse order.
The function also defines a variable word as an empty string, which will be used to store each word in the input string as it is being read.
The function then iterates over each character c in the input string s.
If the character is a space, the word variable is appended to the stack list, and the word variable is reset to an empty string to start building the next word.
If the character is not a space, it is added to the current word being built.
After all characters have been processed, the last word variable is appended to the stack list.
The function then defines a variable result as an empty string, which will be used to store the reversed string.
The function then enters a loop that pops words from the stack list in reverse order, concatenates them with a space, and adds the resulting string to the result variable.
The loop continues until all words have been popped from the stack list.
The resulting string in result is stripped of any trailing spaces and returned along with the time taken to perform the operation.
Finally, the function is called twice with two different input strings (s1 and s2) to demonstrate how the function works.
Python3
import time
def reverse_words(s):
start_time = time.time()
stack = []
word = ""
for c in s:
if c == " ":
stack.append(word)
word = ""
else:
word += c
stack.append(word)
result = ""
while stack:
result += stack.pop() + " "
result = result.strip()
end_time = time.time()
return result, end_time - start_time
s1 = "geeks quiz practice code"
s2 = "getting good at coding needs a lot of practice"
result, time_taken = reverse_words(s1)
print(result)
print(f"Time taken: {time_taken} seconds")
result, time_taken = reverse_words(s2)
print(result)
print(f"Time taken: {time_taken} seconds")
|
Output
code practice quiz geeks
Time taken: 1.049041748046875e-05 seconds
practice of lot a needs coding at good getting
Time taken: 6.9141387939453125e-06 seconds
Time complexity: O(n), where n is the length of the input string. This is because we iterate over each character in the string once, which takes O(n) time. We also append each word to a stack, which takes O(1) time per operation.
Space complexity: O(n), where n is the length of the input string. This is because we create a stack to store the words, which can take up to O(n) space in the worst case.
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