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Program to reverse a word

Write a program to reverse a given word.

Examples:



Input: “Hello”
Output: “olleH”

Input: “Programming”
Output: “gnimmargorP”



Approach: To solve the problem, follow the below idea:

It can be observed that we can reverse a word by traversing over the original string in reverse order and then pushing the characters from the back to a new string. Finally, return the final string.

Step-by-step algorithm:

  1. Create a variable to store the reversed word.
  2. Traverse the original word from the last character to the first.
  3. Append each character to the reversed word.

Below is the implementation of the algorithm:




#include <bits/stdc++.h>;
using namespace std;
 
string reverseWord(string word) {
    string reversedWord = "";
      for(int i = word.length() - 1; i >= 0; i--) {
        reversedWord.push_back(word[i]);
    }
      return reversedWord;
}
 
int main() {
    string word = "Hello";
    cout<< "Original Word: " << word << endl;
    cout << "Reversed Word: " << reverseWord(word) << endl;
    return 0;
}




#include <stdio.h>
 
// Function to reverse a word and return the reversed word
char* reverseWord(const char* word)
{
    int length = strlen(word);
    // +1 for null terminator
    char* reversedWord
        = (char*)malloc((length + 1) * sizeof(char));
 
    for (int i = 0; i < length; i++) {
        reversedWord[i] = word[length - 1 - i];
    }
     
      // Null-terminate the reversed string
    reversedWord[length] = '\0';
    return reversedWord;
}
 
int main()
{
    const char* originalWord = "Hello";
    char* reversed = reverseWord(originalWord);
 
    printf("Original Word: %s\n", originalWord);
    printf("Reversed Word: %s\n", reversed);
 
    free(reversed); // Don't forget to free the allocated
                    // memory
    return 0;
}




public class ReverseWord {
    // Function to reverse a word and return the reversed
    // word
    static String reverseWord(String word)
    {
        int length = word.length();
        StringBuilder reversedWord
            = new StringBuilder(length);
 
        for (int i = length - 1; i >= 0; i--) {
            reversedWord.append(word.charAt(i));
        }
 
        return reversedWord.toString();
    }
 
    public static void main(String[] args)
    {
        String originalWord = "Hello";
        String reversed = reverseWord(originalWord);
 
        System.out.println("Original Word: "
                           + originalWord);
        System.out.println("Reversed Word: " + reversed);
    }
}




def reverse_word(word):
    return word[::-1]
 
 
def main():
    original_word = "Hello"
    reversed_word = reverse_word(original_word)
 
    print("Original Word:", original_word)
    print("Reversed Word:", reversed_word)
 
 
if __name__ == "__main__":
    main()




using System;
 
class Program {
    // Function to reverse a word and return the reversed
    // word
    static string ReverseWord(string word)
    {
        char[] charArray = word.ToCharArray();
        Array.Reverse(charArray);
        return new string(charArray);
    }
 
    static void Main()
    {
        string originalWord = "Hello";
        string reversedWord = ReverseWord(originalWord);
 
        Console.WriteLine("Original Word: " + originalWord);
        Console.WriteLine("Reversed Word: " + reversedWord);
    }
}




function reverseWord(word) {
    return word.split('').reverse().join('');
}
 
function main() {
    const originalWord = "Hello";
    const reversedWord = reverseWord(originalWord);
 
    console.log("Original Word:", originalWord);
    console.log("Reversed Word:", reversedWord);
}
 
main();

Output
Original Word: Hello
Reversed Word: olleH

Time Complexity: O(N) where N is the length of the word.
Auxiliary Space: O(N)


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