New Problem Solution "982. Triples with Bitwise AND Equal To Zero"

Author: haoel

README.md

@@ -25,6 +25,7 @@ LeetCode
 |985|[Sum of Even Numbers After Queries](https://leetcode.com/problems/sum-of-even-numbers-after-queries/) | [C++](./algorithms/cpp/sumOfEvenNumbersAfterQueries/SumOfEvenNumbersAfterQueries.cpp)|Easy|
 |984|[String Without AAA or BBB](https://leetcode.com/problems/string-without-aaa-or-bbb/) | [C++](./algorithms/cpp/stringWithoutAAAOrBBB/StringWithoutAAAOrBBB.cpp)|Easy|
 |983|[Minimum Cost For Tickets](https://leetcode.com/problems/minimum-cost-for-tickets/) | [C++](./algorithms/cpp/minimumCostForTickets/MinimumCostForTickets.cpp)|Medium|
+|982|[Triples with Bitwise AND Equal To Zero](https://leetcode.com/problems/triples-with-bitwise-and-equal-to-zero/) | [C++](./algorithms/cpp/triplesWithBitwiseANDEqualToZero/TriplesWithBitwiseAndEqualToZero.cpp)|Hard|
 |981|[Time Based Key-Value Store](https://leetcode.com/problems/time-based-key-value-store/) | [C++](./algorithms/cpp/timeBasedKeyValueStore/TimeBasedKeyValueStore.cpp)|Medium|
 |980|[Unique Paths III](https://leetcode.com/problems/unique-paths-iii/) | [C++](./algorithms/cpp/uniquePaths/UniquePaths.III.cpp)|Hard|
 |979|[Distribute Coins in Binary Tree](https://leetcode.com/problems/distribute-coins-in-binary-tree/) | [C++](./algorithms/cpp/distributeCoinsInBinaryTree/DistributeCoinsInBinaryTree.cpp)|Medium|

algorithms/cpp/triplesWithBitwiseANDEqualToZero/TriplesWithBitwiseAndEqualToZero.cpp

@@ -0,0 +1,61 @@
+// Source : https://leetcode.com/problems/triples-with-bitwise-and-equal-to-zero/
+// Author : Hao Chen
+// Date   : 2020-07-26
+
+/***************************************************************************************************** 
+ *
+ * Given an array of integers A, find the number of triples of indices (i, j, k) such that:
+ * 
+ * 	0 <= i < A.length
+ * 	0 <= j < A.length
+ * 	0 <= k < A.length
+ * 	A[i] & A[j] & A[k] == 0, where & represents the bitwise-AND operator.
+ * 
+ * Example 1:
+ * 
+ * Input: [2,1,3]
+ * Output: 12
+ * Explanation: We could choose the following i, j, k triples:
+ * (i=0, j=0, k=1) : 2 & 2 & 1
+ * (i=0, j=1, k=0) : 2 & 1 & 2
+ * (i=0, j=1, k=1) : 2 & 1 & 1
+ * (i=0, j=1, k=2) : 2 & 1 & 3
+ * (i=0, j=2, k=1) : 2 & 3 & 1
+ * (i=1, j=0, k=0) : 1 & 2 & 2
+ * (i=1, j=0, k=1) : 1 & 2 & 1
+ * (i=1, j=0, k=2) : 1 & 2 & 3
+ * (i=1, j=1, k=0) : 1 & 1 & 2
+ * (i=1, j=2, k=0) : 1 & 3 & 2
+ * (i=2, j=0, k=1) : 3 & 2 & 1
+ * (i=2, j=1, k=0) : 3 & 1 & 2
+ * 
+ * Note:
+ * 
+ * 	1 <= A.length <= 1000
+ * 	0 <= A[i] < 2^16
+ * 
+ ******************************************************************************************************/
+
+class Solution {
+public:
+    int countTriplets(vector<int>& A) {
+        int n = A.size();
+
+        //using a map to aggregate the duplication
+        unordered_map<int, int> rec;
+        for (int i=0; i<n; i++) {
+            for (int j=0; j<n; j++) {
+                rec[A[i] & A[j]]++;
+            }
+        }
+
+        int result = 0;
+        for (auto &r : rec ) {
+            for (int k=0; k<n; k++) {
+                if ((r.first & A[k]) == 0) result+=r.second;
+            }
+        }
+        return result;
+    }
+
+};