LeetCode 1258: Synonymous Sentences Solution

Master LeetCode problem 1258 (Synonymous Sentences), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

1258. Synonymous Sentences

Medium

Array Hash Table String Backtracking Union Find

Problem Explanation

Explanation:

To solve this problem, we can use the concept of backtracking along with a depth-first search (DFS) approach. We will build a graph representing the synonyms relationships given in the input. Then, we can use backtracking to generate all possible sentences by replacing words with their synonyms.

Algorithm:

Build a graph where each word is a node and the synonyms relationships are edges.
Perform a Depth First Search (DFS) backtracking on the graph to generate all possible sentences by replacing words with their synonyms.
Keep track of the current sentence being built.
When we reach a word that has no synonyms (i.e., not present in the graph), add the current sentence to the result.
Continue the backtracking process until all words are processed.

Time Complexity:

The time complexity of this approach is O(n * s), where n is the number of words in the input sentence and s is the average number of synonyms for each word.

Space Complexity:

The space complexity is O(n), where n is the number of words in the input sentence.

: :

Solution Code

import java.util.*;

class Solution {
    public List<String> generateSentences(List<List<String>> synonyms, String text) {
        Map<String, List<String>> graph = new HashMap<>();
        for (List<String> synonymPair : synonyms) {
            String word1 = synonymPair.get(0);
            String word2 = synonymPair.get(1);
            graph.computeIfAbsent(word1, k -> new ArrayList<>()).add(word2);
            graph.computeIfAbsent(word2, k -> new ArrayList<>()).add(word1);
        }
        
        List<String> result = new ArrayList<>();
        dfs(graph, text.split(" "), 0, new StringBuilder(), result);
        Collections.sort(result);
        return result;
    }
    
    private void dfs(Map<String, List<String>> graph, String[] words, int index, StringBuilder current, List<String> result) {
        if (index == words.length) {
            result.add(current.toString().trim());
            return;
        }
        
        String word = words[index];
        if (!graph.containsKey(word)) {
            dfs(graph, words, index + 1, current.append(word).append(" "), result);
            current.setLength(current.length() - word.length() - 1);
        } else {
            for (String synonym : graph.get(word)) {
                dfs(graph, words, index + 1, current.append(synonym).append(" "), result);
                current.setLength(current.length() - synonym.length() - 1);
            }
        }
    }
}

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