Reverse the Words of a String using Stack

Difficulty Level : Medium
Last Updated : 30 Jun, 2021

Given string str consisting of multiple words, the task is to reverse the entire string word by word.
Examples:

Input: str = “I Love To Code”
Output: Code To Love I
Input: str = “data structures and algorithms”
Output: algorithms and structures data

Approach: This problem can be solved not only with the help of the strtok() but also it can be solved by using Stack Container Class in STL C++ by following the given steps:

Create an empty stack.
Traverse the entire string, while traversing add the characters of the string into a temporary
variable until you get a space(‘ ‘) and push that temporary variable into the stack.
Repeat the above step until the end of the string.
Pop the words from the stack until the stack is not empty which will be in reverse order.

Below is the implementation of the above approach:

C++

//C++ implementation of the above approach
#include<bits/stdc++.h>
using namespace std;
 
//function to reverse the words
//of the given string without using strtok().
void reverse(string s)
{
  //create an empty string stack
  stack<string> stc;
 
  //create an empty temporary string
  string temp="";
 
  //traversing the entire string
  for(int i=0;i<s.length();i++)
  {
    if(s[i]==' ')
    {
       
      //push the temporary variable into the stack
      stc.push(temp);
       
      //assigning temporary variable as empty
      temp="";         
    }
    else
    {
      temp=temp+s[i];
    }
 
  }
 
  //for the last word of the string
  stc.push(temp);
 
  while(!stc.empty()) {
 
    // Get the words in reverse order
    temp=stc.top();
    cout<<temp<<" ";
    stc.pop();
  }
  cout<<endl;
}
 
//Driver code
int main()
{
  string s="I Love To Code";
  reverse(s);
  return 0;
}
// This code is contributed by Konderu Hrishikesh.

Java

//Java implementation of
// the above approach
import java.util.*;
class GFG{
 
// Function to reverse the words
// of the given String without
// using strtok().
static void reverse(String s)
{
  // Create an empty String stack
  Stack<String> stc = new Stack<>();
 
  // Create an empty temporary String
  String temp = "";
 
  // Traversing the entire String
  for(int i = 0; i < s.length(); i++)
  {
    if(s.charAt(i) == ' ')
    {
       
      // Push the temporary
      // variable into the stack
      stc.add(temp);
       
      // Assigning temporary
      // variable as empty
      temp = "";         
    }
    else
    {
      temp = temp + s.charAt(i);
    }
 
  }
 
  // For the last word
  // of the String
  stc.add(temp);
 
  while(!stc.isEmpty())
  {
    // Get the words in
    // reverse order
    temp = stc.peek();
    System.out.print(temp + " ");
    stc.pop();
  }
   
  System.out.println();
}
 
//Driver code
public static void main(String[] args)
{
  String s = "I Love To Code";
  reverse(s);
}
}
 
// This code is contributed by gauravrajput1

Python3

# Python3 implementation of
# the above approach
 
# function to reverse the
# words of the given string
# without using strtok().
def reverse(s):
 
  # create an empty string
  # stack
  stc = []
 
  # create an empty temporary
  # string
  temp = ""
 
  # traversing the entire string
  for i in range(len(s)):
  
    if s[i] == ' ':
       
      # push the temporary variable
      # into the stack
      stc.append(temp)
       
      # assigning temporary variable
      # as empty
      temp = ""         
 
    else:
      temp = temp + s[i]
 
  # for the last word of the string
  stc.append(temp)
 
  while len(stc) != 0:
 
    # Get the words in reverse order
    temp = stc[len(stc) - 1]
    print(temp, end = " ")
    stc.pop()
   
  print()
 
# Driver code
s = "I Love To Code"
reverse(s)
 
# This code is contributed by divyeshrabadiya07

C#

// C# implementation of
// the above approach
using System;
using System.Collections;
 
class GFG
{
     
    // Function to reverse the words
    // of the given String without
    // using strtok().
    static void reverse(string s)
    {
       
      // Create an empty String stack
      Stack stc = new Stack();
      
      // Create an empty temporary String
      string temp = "";
      
      // Traversing the entire String
      for(int i = 0; i < s.Length; i++)
      {
        if(s[i] == ' ')
        {
            
          // Push the temporary
          // variable into the stack
          stc.Push(temp);
            
          // Assigning temporary
          // variable as empty
          temp = "";         
        }
        else
        {
          temp = temp + s[i];
        }
      
      }
      
      // For the last word
      // of the String
      stc.Push(temp);
      
      while(stc.Count != 0)
      {
         
        // Get the words in
        // reverse order
        temp = (string)stc.Peek();
        Console.Write(temp + " ");
        stc.Pop();
      }
        
      Console.WriteLine();
    }
 
  // Driver code
  static void Main()
  {
    string s = "I Love To Code";
    reverse(s);
  }
}
 
// This code is contributed by divyesh072019

Javascript

<script>
 
    // Javascript implementation of
    // the above approach
     
      // Function to reverse the words
    // of the given String without
    // using strtok().
    function reverse(s)
    {
        
      // Create an empty String stack
      let stc = [];
       
      // Create an empty temporary String
      let temp = "";
       
      // Traversing the entire String
      for(let i = 0; i < s.length; i++)
      {
        if(s[i] == ' ')
        {
             
          // Push the temporary
          // variable into the stack
          stc.push(temp);
             
          // Assigning temporary
          // variable as empty
          temp = "";        
        }
        else
        {
          temp = temp + s[i];
        }
       
      }
       
      // For the last word
      // of the String
      stc.push(temp);
       
      while(stc.length != 0)
      {
          
        // Get the words in
        // reverse order
        temp = stc[stc.length - 1];
        document.write(temp + " ");
        stc.pop();
      }
    }
     
    let s = "I Love To Code";
    reverse(s);
     
</script>

Output:

Code To Love I

Time Complexity: O(N)

Another Approach: An approach without using stack is discussed here. This problem can also be solved using stack by following the below steps:

Create an empty stack.
Tokenize the input string into words using spaces as separator with the help of strtok()
Push the words into the stack.
Pop the words from the stack until the stack is not empty which will be in reverse order.

Below is the implementation of the above approach:

C++

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to reverse the words
// of the given sentence
void reverse(char k[])
{
 
    // Create an empty character array stack
    stack<char*> s;
    char* token = strtok(k, " ");
 
    // Push words into the stack
    while (token != NULL) {
        s.push(token);
        token = strtok(NULL, " ");
    }
 
    while (!s.empty()) {
 
        // Get the words in reverse order
        cout << s.top() << " ";
        s.pop();
    }
}
 
// Driver code
int main()
{
    char k[] = "geeks for geeks";
    reverse(k);
 
    return 0;
}

Java

// Java implementation of the approach
import java.util.Arrays;
import java.util.Stack;
 
class GFG {
 
    // Function to reverse the words
    // of the given sentence
    static void reverse(String k)
    {
 
        // Create an empty character array stack
        Stack<String> s = new Stack<>();
        String[] token = k.split(" ");
 
        // Push words into the stack
        for (int i = 0; i < token.length; i++) {
            s.push(token[i]);
        }
 
        while (!s.empty()) {
 
            // Get the words in reverse order
            System.out.print(s.peek() + " ");
            s.pop();
        }
    }
 
    // Driver code
    public static void main(String[] args)
    {
        String k = "geeks for geeks";
        reverse(k);
    }
}
 
// This code is contributed by Rajput-Ji

Python3

# Python3 implementation of the approach
 
# Function to reverse the words
# of the given sentence
def reverse(k):
    # Create an empty character array stack
    s = []
    token = k.split()
     
    # Push words into the stack
    for word in token :
        s.append(word);
         
    while (len(s)) :
        # Get the words in reverse order
        print(s.pop(), end = " ");
     
 
# Driver code
if __name__ == "__main__" :
 
    k = "geeks for geeks";
    reverse(k);
     
# This code is contributed by AnkitRai01

C#

// C# implementation of the approach
using System;
using System.Collections.Generic;
 
class GFG {
 
    // Function to reverse the words
    // of the given sentence
    static void reverse(String k)
    {
 
        // Create an empty character array stack
        Stack<String> s = new Stack<String>();
        String[] token = k.Split(' ');
 
        // Push words into the stack
        for (int i = 0; i < token.Length; i++) {
            s.Push(token[i]);
        }
 
        while (s.Count != 0) {
 
            // Get the words in reverse order
            Console.Write(s.Peek() + " ");
            s.Pop();
        }
    }
 
    // Driver code
    public static void Main(String[] args)
    {
        String k = "geeks for geeks";
        reverse(k);
    }
}
 
// This code is contributed by PrinciRaj1992

Javascript

<script>
 
    // JavaScript implementation of the approach
 
    // Function to reverse the words
    // of the given sentence
    function reverse(k)
    {
 
        // Create an empty character array stack
        let s = [];
          let token = k.split(" ");
 
        // Push words into the stack
        for (let i = 0; i < token.length; i++) {
            s.push(token[i]);
        }
  
        while (s.length > 0) {
  
            // Get the words in reverse order
            document.write(s[s.length - 1] + " ");
            s.pop();
        }
    }
 
    // Driver code
    let k = "geeks for geeks";
    reverse(k);
 
</script>

Output:

geeks for geeks

Time Complexity: O(N)

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