Added hamming distance program

Author: mandliya

README.md

@@ -226,7 +226,8 @@ Include contains single header implementation of data structures and some algori
 | Find all possible combinations of k numbers that add up to a number n, given that only numbers from 1 to 9 can be used and each combination should be a unique set of numbers. Ensure that numbers within the set are sorted in ascending order. Example : for k = 3, n = 9 result would be [[1,2,6], [1,3,5], [2,3,4]], similarly for k = 3, n = 7, result would be [[1,2,4]]. | [combinationSum3.cpp](leet_code_problems/combinationSum3.cpp) |
 | Given a non-negative integer num, repeatedly add all its digits until the result has only one digit. For example: Given num = 38, the process is like: 3 + 8 = 11, 1 + 1 = 2. Since 2 has only one digit, return it. Follow up: Could you do it without any loop/recursion in O(1) runtime?| [addDigits.cpp](leet_code_problems/addDigits.cpp) |
 | Given a matrix with cell values 0 or 1. Find the length of the shortest path from (a1, b1) to (a2, b2), such that path can only be constructed through cells which have value 1 and you can only travel in 4 possible directions, i.e. left, right, up and down.|[shortest_path_maze.cpp](leet_code_problems/shortest_path_maze.cpp) |
-
+| The Hamming distance between two integers is the number of positions at which the corresponding bits are different.
+Given two integers x and y, calculate the Hamming distance.| [hamming_distance.cpp](leet_code_problems/hamming_distance.cpp)|
 
 
 

leet_code_problems/hamming_distance.cpp

@@ -0,0 +1,42 @@
+/**
+ The Hamming distance between two integers is the number of positions at which
+ the corresponding bits are different.
+ Given two integers x and y, calculate the Hamming distance.
+
+ Input: x = 1, y = 4
+
+Output: 2
+
+Explanation:
+1   (0 0 0 1)
+4   (0 1 0 0)
+       ↑   ↑
+
+The above arrows point to positions where the corresponding bits are different
+*/
+
+#include <iostream>
+
+int hamming_distance(int x, int y)
+{
+    // XOR will set only those bits in z which are different in x and y
+    int z = x ^ y;
+
+    // Now just count set bits in z.
+    int count = 0;
+    while (z)
+    {
+        count += (z % 2);
+        z >>= 2;
+    }
+    return count;
+}
+
+
+int main()
+{
+    int x = 4;
+    int y = 1;
+    std::cout << "Hamming distance between " << x << " and " << y << " is : "
+        <<  hamming_distance(x, y) << std::endl;
+}