Add Word Break II solution with backtracking in C++

Implement solution to generate all possible sentences from a string using words from a dictionary. Key features:

- Use backtracking with memoization for efficient word segmentation
- Convert dictionary to hash set for O(1) lookups
- Handle edge cases (empty string, no valid segmentation)
- Add comprehensive test cases covering various scenarios
- Include helper function for result visualization

LeetCode: #140 Word Break II

cpp/Word_Break_II.cpp

@@ -0,0 +1,122 @@
+/*
+  PROBLEM
+    Word Break II
+  
+    Given a string s and a dictionary of strings wordDict, add spaces in s to construct a sentence where each word is a valid dictionary word. Return all such possible sentences in any order.
+
+    Note that the same word in the dictionary may be reused multiple times in the segmentation.
+
+    LeetCode: https://leetcode.com/problems/word-break-ii
+*/
+
+#include <iostream>
+#include <vector>
+#include <string>
+#include <unordered_set>
+#include <unordered_map>
+using namespace std;
+
+class Solution {
+public:
+    vector<string> wordBreak(string s, vector<string>& wordDict) {
+        vector<string> ans;  // Store all possible sentences
+
+        // Convert wordDict to set for O(1) lookup
+        unordered_set<string> st(wordDict.begin(), wordDict.end());
+
+        // Memoization map: index -> vector of possible sentences from that index
+        unordered_map<int, vector<string>> memo;
+
+        // Start recursive solution with empty sentence
+        solve(ans, st, memo, s, "", 0);
+        return ans;
+    }
+
+    void solve(vector<string>& ans, unordered_set<string>& st, 
+               unordered_map<int,vector<string>>& memo, 
+               string s, string sent, int idx) {
+        // Base case: if input string is empty, we found a valid sentence
+        if(s.size() == 0) {
+            ans.push_back(sent);
+            return;
+        }
+
+        // Try all possible prefixes of current string
+        for(int i = 0; i < s.size(); i++) {
+            // Extract current prefix
+            string w = s.substr(0, i + 1);
+
+            // If prefix is a valid dictionary word
+            if(st.find(w) != st.end()) {
+                string new_sent;
+
+                // Construct new sentence by appending current word
+                if(sent == "") {
+                    new_sent = w;  // First word
+                } else {
+                    new_sent = sent + " " + w;  // Add space between words
+                }
+
+                // Remaining string to process
+                string new_s = s.substr(i + 1);
+
+                // Recursive call with remaining string and updated sentence
+                solve(ans, st, memo, new_s, new_sent, idx + i);
+            }
+        }
+    }
+};
+
+// Helper function to print vector of strings
+void printVector(const vector<string>& vec) {
+    cout << "[";
+    for(int i = 0; i < vec.size(); i++) {
+        cout << "\"" << vec[i] << "\"";
+        if(i < vec.size() - 1) cout << ", ";
+    }
+    cout << "]" << endl;
+}
+
+int main() {
+    // Test cases
+    vector<tuple<string, vector<string>>> testCases = {
+        // Format: {input string, dictionary words}
+        {"catsanddog", {"cat","cats","and","sand","dog"}},
+        {"pineapplepenapple", {"apple","pen","applepen","pine","pineapple"}},
+        {"catsandog", {"cats","dog","sand","and","cat"}},
+        {"", {"cat","dog"}},  // Empty string case
+        {"a", {"b"}},         // No valid segmentation case
+        {"aaaa", {"a","aa"}}, // Multiple valid segmentations case
+    };
+
+    Solution solution;
+
+    // Process each test case
+    for(int i = 0; i < testCases.size(); i++) {
+        cout << "\nTest Case " << i + 1 << ":\n";
+
+        // Extract input string and dictionary
+        string s = get<0>(testCases[i]);
+        vector<string>& dict = get<1>(testCases[i]);
+
+        // Print input
+        cout << "Input String: \"" << s << "\"\n";
+        cout << "Dictionary: ";
+        printVector(dict);
+
+        // Calculate and print result
+        vector<string> result = solution.wordBreak(s, dict);
+        cout << "Possible Sentences: ";
+        printVector(result);
+
+        // Print explanation
+        cout << "Explanation: ";
+        if(result.empty()) {
+            cout << "No valid sentence can be formed.\n";
+        } else {
+            cout << "Found " << result.size() << " valid sentence(s) using dictionary words.\n";
+        }
+    }
+
+    return 0;
+}