From 19330b48d06db69c36a98319dfe3254c2a936f3d Mon Sep 17 00:00:00 2001
From: axayjha <akshayjha@live.in>
Date: Thu, 15 Oct 2020 21:30:11 +0530
Subject: [PATCH] adding documentation and other enhancements

---
 dynamic_programming/word_break.cpp | 84 +++++++++++++++++++++++++++---
 1 file changed, 76 insertions(+), 8 deletions(-)

diff --git a/dynamic_programming/word_break.cpp b/dynamic_programming/word_break.cpp
index 6de8eaf8d..e13dfff55 100644
--- a/dynamic_programming/word_break.cpp
+++ b/dynamic_programming/word_break.cpp
@@ -36,47 +36,115 @@ using std::string;
 using std::unordered_set;
 using std::vector;
 
+/**
+ * @brief Solution class
+ */
 class Solution {
  public:
-    bool exists(const string &s, const unordered_set<string> &strSet) {
-        return strSet.find(s) != strSet.end();
+    /**
+     * @brief Function that checks if the string passed in param is present in
+     * the the unordered_set passed
+     *
+     * @param str the string to be searched
+     * @param strSet unordered set of string, that is to be looked into
+     * @returns bool, true if str is present in strSet
+     */
+    bool exists(const string &str, const unordered_set<string> &strSet) {
+        return strSet.find(str) != strSet.end();
     }
 
+    /**
+     * @brief Function that checks if the string passed in param can be
+     * segmented from position 'pos', and then correctly go on to segment the
+     * rest of the string correctly as well to reach a solution
+     *
+     * @param s the complete string to be segmented
+     * @param strSet unordered set of string, that is to be used as the
+     * reference dictionary
+     * @param pos the index value at which we will segment string and test
+     * further if it is correctly segmented at pos
+     * @param dp the vector to memoize solution for each position
+     * @returns bool, true if str is present in strSet
+     */
     bool check(const string &s, const unordered_set<string> &strSet, int pos,
                vector<int> *dp) {
         if (pos == s.length()) {
+            // if we have reached till the end of the string, means we have
+            // segmented throughout correctly hence we have a solution, thus
+            // returning true
             return true;
         }
 
         if (dp->at(pos) != INT_MAX) {
+            // if dp[pos] is not INT_MAX, means we must have saved a solution
+            // for the position pos; then return if the solution at pos is true
+            // or not
             return dp->at(pos) == 1;
         }
 
-        string wordTillNow = "";
+        string wordTillNow =
+            "";  // string to save the prefixes of word till different positons
+
         for (int i = pos; i < s.length(); i++) {
-            wordTillNow += string(1, s[i]);
+            // Loop starting from pos to end, to check valid set of
+            // segmentations if any
+            wordTillNow +=
+                string(1, s[i]);  // storing the prefix till the position i
+
+            // if the prefix till current position is present in the dictionary
+            // and the remaining substring can also be segmented legally, then
+            // set solution at position pos in the memo, and return true
             if (exists(wordTillNow, strSet) and check(s, strSet, i + 1, dp)) {
                 dp->at(pos) = 1;
                 return true;
             }
         }
-        dp->at(pos) = 0;
-        return false;
+        // if function has still not returned, then there must be no legal
+        // segmentation possible after segmenting at pos
+        dp->at(pos) = 0;  // so set solution at pos as false
+        return false;     // and return no solution at position pos
     }
 
+    /**
+     * @brief Function that checks if the string passed in param can be
+     * segmented into the strings present in the vector.
+     * In others words, it checks if any permutation of strings in
+     * the vector can be concatenated to form the final string.
+     *
+     * @param s the complete string to be segmented
+     * @param wordDict a vector of words to be used as dictionary to look into
+     * @returns bool, true if s can be
+     */
     bool wordBreak(const string &s, const vector<string> &wordDict) {
+        // unordered set to store words in the dictionary for contant time
+        // search
         unordered_set<string> strSet;
         for (const auto &s : wordDict) {
             strSet.insert(s);
         }
-
+        // a vector to be used for memoization, whose value at index i will
+        // tell if the string s can be segmented (correctly) at position i.
+        // initializing it with INT_MAX (which will denote no solution)
         vector<int> dp(s.length(), INT_MAX);
+
+        // calling check method with position = 0, to check from left
+        // from where can be start segmenting the complete string in correct
+        // manner
         return check(s, strSet, 0, &dp);
     }
 };
 
+/**
+ * @brief Main function
+ * @returns 0 on exit
+ */
 int main() {
+    // the complete string
     const string s = "applepenapple";
+    // the dictionary to be used
     const vector<string> wordDict = {"apple", "pen"};
+
+    // should return true, as applepenapple can be segmented as apple + pen +
+    // apple
     cout << Solution().wordBreak(s, wordDict) << endl;
-}
\ No newline at end of file
+}