[20200705] Solve july challenge questions

Author: iamminji

README.md

@@ -227,4 +227,5 @@
 | [Arranging Coins](https://leetcode.com/explore/challenge/card/july-leetcoding-challenge/544/week-1-july-1st-july-7th/3377/)      | [python](challenge/2020/july/Arranging_Coins.py)  | [solutions](challenge/2020/july/Arranging_Coins.md)|
 | [Binary Tree Level Order Traversal II](https://leetcode.com/explore/challenge/card/july-leetcoding-challenge/544/week-1-july-1st-july-7th/3378/)      | [python](challenge/2020/july/Binary_Tree_Level_Order_Traversal_II.py)  | [solutions](challenge/2020/july/Binary_Tree_Level_Order_Traversal_II.md)|
 | [Prison Cells After N Days](https://leetcode.com/explore/challenge/card/july-leetcoding-challenge/544/week-1-july-1st-july-7th/3379/)      | [python](challenge/2020/july/Prison_Cells_After_N_Days.py)  | [solutions](challenge/2020/july/Prison_Cells_After_N_Days.md)|
-| [Ugly Number II](https://leetcode.com/explore/challenge/card/july-leetcoding-challenge/544/week-1-july-1st-july-7th/3380/)      | [python](challenge/2020/july/Ugly_Number_II.py)  | [solutions](challenge/2020/july/Ugly_Number_II.md)|
+| [Ugly Number II](https://leetcode.com/explore/challenge/card/july-leetcoding-challenge/544/week-1-july-1st-july-7th/3380/)      | [python](challenge/2020/july/Ugly_Number_II.py)  | [solutions](challenge/2020/july/Ugly_Number_II.md)|
+| [Hamming Distance](https://leetcode.com/explore/challenge/card/july-leetcoding-challenge/544/week-1-july-1st-july-7th/3381/)      | [python](challenge/2020/july/Hamming_Distance.py)  | [solutions](challenge/2020/july/Hamming_Distance.md)|

challenge/2020/july/Hamming_Distance.md

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+## 문제
+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.
+
+Note:
+```
+0 ≤ x, y < 2^31.
+```
+
+Example:
+```
+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.
+
+### 솔루션
+
+#### 해결 1
+integer 를 bin 메서드로 변경 후 최대 입력 값이 2^31 이므로 31개의 string 으로 만들어주었다.
+그리고 값 비교를 누적하여 리턴하였다.
+
+코드는 아래와 같다.
+
+```python3
+class Solution:
+    def hammingDistance(self, x: int, y: int) -> int:
+        p = bin(x)
+        q = bin(y)
+
+        res = 0
+        for a, b in zip(("%031d" % int(p[2:])), ("%031d" % int(q[2:]))):
+            if a != b:
+                res += 1
+        return res
+
+```
+
+#### 해결 2
+bit 값이 서로 다른 경우의 count 이므로 서로 다를 경우 1 로 해주는 exclusive or (xor) 연산을 해준다.
+그리고 1 카운트를 해주었다.
+
+```python3
+    def hammingDistanceUseXor(self, x: int, y: int) -> int:
+        p = x ^ y
+        res = 0
+        for q in bin(p)[2:]:
+            if q == "1":
+                res += 1
+        return res
+```

challenge/2020/july/Hamming_Distance.py

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+class Solution:
+    def hammingDistanceUseString(self, x: int, y: int) -> int:
+        p = bin(x)
+        q = bin(y)
+
+        res = 0
+        for a, b in zip(("%031d" % int(p[2:])), ("%031d" % int(q[2:]))):
+            if a != b:
+                res += 1
+        return res
+
+    def hammingDistanceUseXor(self, x: int, y: int) -> int:
+        p = x ^ y
+        res = 0
+        for q in bin(p)[2:]:
+            if q == "1":
+                res += 1
+        return res
+
+
+if __name__ == '__main__':
+    sol = Solution()
+    print(sol.hammingDistanceUseString(1, 4))
+    print(sol.hammingDistanceUseXor(1, 4))