added totalHammingDistance

README.md

@@ -8,6 +8,7 @@ LeetCode
 
 | # | Title | Solution | Difficulty |
 |---| ----- | -------- | ---------- |
+|477|[Total Hamming Distance](https://leetcode.com/problems/total-hamming-distance/) | [C++](./algorithms/cpp/totalHammingDistance/totalHammingDistance.cpp)|Medium|
 |415|[Add Strings](https://leetcode.com/problems/add-strings/) | [C++](./algorithms/cpp/addStrings/AddStrings.cpp)|Easy|
 |414|[Third Maximum Number](https://leetcode.com/problems/third-maximum-number/) | [C++](./algorithms/cpp/thirdMaximumNumber/ThirdMaximumNumber.cpp)|Easy|
 |413|[Arithmetic Slices](https://leetcode.com/problems/arithmetic-slices/) | [C++](./algorithms/cpp/arithmeticSlices/ArithmeticSlices.cpp)|Medium|

algorithms/cpp/totalHammingDistance/totalHammingDistance.cpp

@@ -0,0 +1,56 @@
+// Source : https://leetcode.com/problems/total-hamming-distance/
+// Author : Calinescu Valentin
+// Date   : 2017-01-09
+
+/*************************************************************************************** 
+ *
+ * The Hamming distance between two integers is the number of positions at which the 
+ * corresponding bits are different.
+ * 
+ * Now your job is to find the total Hamming distance between all pairs of the given 
+ * numbers.
+ * 
+ * Example:
+ * Input: 4, 14, 2
+ * 
+ * Output: 6
+ * 
+ * Explanation: In binary representation, the 4 is 0100, 14 is 1110, and 2 is 0010 (just
+ * showing the four bits relevant in this case). So the answer will be:
+ * HammingDistance(4, 14) + HammingDistance(4, 2) + HammingDistance(14, 2) = 2 + 2 + 2 = 6.
+ * 
+ * Note:
+ * Elements of the given array are in the range of 0 to 10^9
+ * Length of the array will not exceed 10^4.
+ ***************************************************************************************/
+
+/*
+*  Solution 1 - O(N)
+*
+* The total Hamming Distance is equal to the sum of all individual Hamming Distances
+* between every 2 numbers. However, given that this depends on the individual bits of
+* each number, we can see that we only need to compute the number of 1s and 0s for each
+* bit position. For example, we look at the least significant bit. Given that we need to
+* calculate the Hamming Distance for each pair of 2 numbers, we see that the answer is
+* equal to the number of 1s at this position * the number of 0s(which is the total number
+* of numbers - the number of 1s), because for each 1 we need to have a 0 to form a pair.
+* Thus, the solution is the sum of all these distances at every position.
+*/
+class Solution {
+public:
+    int totalHammingDistance(vector<int>& nums) {
+        long long solution = 0;
+        int ones[31];
+        for(int i = 0; i < 31; i++)
+            ones[i] = 0;
+        for(vector<int>::iterator it = nums.begin(); it != nums.end(); ++it)
+        {
+            for(int i = 0; (1 << i) <= *it; i++) //i is the position of the bit
+                if((1 << i) & *it)//to see if the bit at i-position is a 1
+                    ones[i]++;
+        }
+        for(int i = 0; i < 31; i++)
+            solution += ones[i] * (nums.size() - ones[i]);
+        return solution;
+    }
+};