Reverse Words in a Given String (gfg)

Problem Description

Given a string S, you need to reverse the order of words in the string. Words are separated by dots.

Examples

Example 1:

Input: S = "i.like.this.program.very.much"
Output: "much.very.program.this.like.i"
Explanation: After reversing the words, the new string is "much.very.program.this.like.i".

Example 2:

Input: S = "pqr.mno"
Output: "mno.pqr"
Explanation: After reversing the words, the new string is "mno.pqr".

Your Task

You don't need to read input or print anything. Your task is to complete the function reverseWords() which takes the string S as input and returns the string with the words reversed.

Expected Time Complexity: $O(|S|)$

Expected Auxiliary Space: $O(|S|)$

Constraints

• 1 ≤ |S| ≤ 2000

Problem Explanation

The problem is to reverse the order of words in a string where words are separated by dots.

Code Implementation

Python

Written by @arunimad6yuq

class Solution:
    def reverseWords(self, S):
        # Split the string by dots
        words = S.split('.')
        # Reverse the list of words
        words.reverse()
        # Join the reversed list back into a string with dots
        return '.'.join(words)

# Example usage
if __name__ == "__main__":
    solution = Solution()
    print(solution.reverseWords("i.like.this.program.very.much"))  # Expected output: "much.very.program.this.like.i"
    print(solution.reverseWords("pqr.mno"))  # Expected output: "mno.pqr"

C++

Written by @arunimad6yuq

//{ Driver Code Starts
#include <bits/stdc++.h>
using namespace std;

// } Driver Code Ends

class Solution {
public:
    // Function to reverse words in a given string.
    string reverseWords(string S) {
        // Split the string by dots
        vector<string> words;
        string word;
        for (char c : S) {
            if (c == '.') {
                words.push_back(word);
                word.clear();
            } else {
                word += c;
            }
        }
        words.push_back(word); // Add the last word

        // Reverse the list of words
        reverse(words.begin(), words.end());

        // Join the reversed list back into a string with dots
        string result;
        for (size_t i = 0; i < words.size(); ++i) {
            if (i != 0) {
                result += '.';
            }
            result += words[i];
        }

        return result;
    }
};

//{ Driver Code Starts.
int main() {
    int t;
    cin >> t;
    while (t--) {
        string s;
        cin >> s;
        Solution obj;
        cout << obj.reverseWords(s) << endl;
    }
    return 0;
}
// } Driver Code Ends

Example Walkthrough

For the string S = "i.like.this.program.very.much":

1. Split the string by dots into ["i", "like", "this", "program", "very", "much"].
2. Reverse the list to ["much", "very", "program", "this", "like", "i"].
3. Join the reversed list with dots to get "much.very.program.this.like.i".

For the string S = "pqr.mno":

1. Split the string by dots into ["pqr", "mno"].
2. Reverse the list to ["mno", "pqr"].
3. Join the reversed list with dots to get "mno.pqr".

Solution Logic:

1. Split the string into words using the dot as a delimiter.
2. Reverse the list of words.
3. Join the reversed list back into a string with dots.

Time Complexity

• The time complexity is $O(|S|)$, where |S| is the length of the input string.

Space Complexity

• The auxiliary space complexity is $O(|S|)$ due to the storage required for the list of words.

References

• gfg Problem: gfg Problem
• Solution Author: arunimad6yuq/