New Problem Solution - "Maximum Product of Word Lengths"

README.md

@@ -62,6 +62,7 @@ LeetCode
 |322|[Coin Change](https://leetcode.com/problems/coin-change/) | [C++](./algorithms/cpp/coinChange/coinChange.cpp)|Medium|
 |321|[Create Maximum Number](https://leetcode.com/problems/create-maximum-number/) | [C++](./algorithms/cpp/createMaximumNumber/CreateMaximumNumber.cpp)|Hard|
 |319|[Bulb Switcher](https://leetcode.com/problems/bulb-switcher/) | [C++](./algorithms/cpp/bulbSwitcher/bulbSwitcher.cpp)|Medium|
+|318|[Maximum Product of Word Lengths](https://leetcode.com/problems/maximum-product-of-word-lengths/) | [C++](./algorithms/cpp/maximumProductOfWordLengths/MaximumProductOfWordLengths.cpp)|Medium|
 |316|[Remove Duplicate Letters](https://leetcode.com/problems/remove-duplicate-letters/) | [C++](./algorithms/cpp/removeDuplicateLetters/RemoveDuplicateLetters.cpp)|Hard|
 |315|[Count of Smaller Numbers After Self](https://leetcode.com/problems/count-of-smaller-numbers-after-self/) | [C++](./algorithms/cpp/countOfSmallerNumbersAfterSelf/countOfSmallerNumbersAfterSelf.cpp)|Hard|
 |313|[Super Ugly Number](https://leetcode.com/problems/super-ugly-number/) | [C++](./algorithms/cpp/superUglyNumber/SuperUglyNumber.cpp)|Medium|

algorithms/cpp/maximumProductOfWordLengths/MaximumProductOfWordLengths.cpp

@@ -0,0 +1,72 @@
+// Source : https://leetcode.com/problems/maximum-product-of-word-lengths/
+// Author : Hao Chen
+// Date   : 2017-01-02
+
+/*************************************************************************************** 
+ *
+ * Given a string array words, find the maximum value of length(word[i]) * 
+ * length(word[j]) where the two words do not share common letters.
+ *     You may assume that each word will contain only lower case letters.
+ *     If no such two words exist, return 0.
+ * 
+ *     Example 1:
+ * 
+ *     Given ["abcw", "baz", "foo", "bar", "xtfn", "abcdef"]
+ *     Return 16
+ *     The two words can be "abcw", "xtfn".
+ * 
+ *     Example 2:
+ * 
+ *     Given ["a", "ab", "abc", "d", "cd", "bcd", "abcd"]
+ *     Return 4
+ *     The two words can be "ab", "cd".
+ * 
+ *     Example 3:
+ * 
+ *     Given ["a", "aa", "aaa", "aaaa"]
+ *     Return 0
+ *     No such pair of words.    
+ * 
+ * Credits:Special thanks to @dietpepsi for adding this problem and creating all test 
+ * cases.
+ ***************************************************************************************/
+
+class Solution {
+public:
+    //
+    // there are two algorithms:
+    //
+    // 1) compare two words is same or not 
+    //    - we can use bit-mask to solve that. 
+    //    - we need be careful about one word is subset of another one, such as: "abc" is subset of "abcabc"
+    //
+    // 2) find out the max product - that needs O(N^2) time complexity algorithm.
+    //
+
+    int maxProduct(vector<string>& words) {
+        //Key is the word's bitmask, and the value the max length of that bit mask
+        unordered_map<int, int> maxLens;
+        //constructing the bitmask.
+        for(auto& w: words) {
+            int mask = 0;
+            for (auto& c: w) {
+                mask = mask | ( 1 << (c-'a') );
+            }
+            if ( maxLens.find(mask) == maxLens.end() || maxLens[mask] < w.size() ) {
+                maxLens[mask] = w.size();
+            }
+        }
+        
+        //find out the max product
+        int result = 0;
+        for (auto a : maxLens) {
+            for (auto b: maxLens) {
+                // if `a` and `b` is same, then just simply continue
+                if (a.first & b.first) continue; // there are common letters
+                result = max( result, a.second * b.second );
+            }
+        }
+        
+        return result;
+    }
+};