problem: 0078 subsets

Author: cychiu8

0078 Subsets/0078 Subsets.md

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+# 0078. Subsets
+
+* Difficulty: medium
+* Link: <https://leetcode.com/problems/subsets/>
+* Topics: Array-String, Backtracking
+
+# Clarification
+
+1. Check the inputs and outputs
+    * INPUT: List[int]
+    * OUTPUT: List[List[int]]
+
+# Naive Solution
+
+### Thought Process
+
+1. 一個一個決定取或不取
+2. 加到 result 前判斷是否存在於 result 了
+
+* Implement
+
+    ```python
+    class Solution:
+        def subsets(self, nums: List[int]) -> List[List[int]]:
+            result = []
+            subset = []
+            
+            def add_subset(subset, idx):
+                if subset not in result:
+                    result.append(subset)
+                if idx == len(nums):
+                    return
+                add_subset(subset + [nums[idx]], idx + 1)
+                add_subset(subset, idx + 1)
+            add_subset(subset,0)
+            return result
+    ```
+
+  * 應該是要在 idx == nums 再 append ，就不會有重複 append 的情況
+
+    ```jsx
+    class Solution:
+        def subsets(self, nums: List[int]) -> List[List[int]]:
+            result = []
+            subset = []
+            
+            def add_subset(subset, idx):   
+                if idx == len(nums):
+                    result.append(subset)
+                    return
+                add_subset(subset + [nums[idx]], idx + 1)
+                add_subset(subset, idx + 1)
+            add_subset(subset,0)
+            return result
+    ```
+
+### Complexity
+
+* Time complexity: $O(2^N)$
+* Space complexity:$O(N)$
+
+### Problems & Improvement
+
+* Time complexity 很高
+
+# Improvement
+
+### Thought Process
+
+1. 之前所有的組合 (沒有選當前數字)，再全部加入目前數字(選當前數字)
+
+```jsx
+Example : [1,2,3]
+result = [[]]
+==== idx = 0 ====
+[]
+----
+[1]
+==== idx = 1 ====
+[]
+[1]
+----
+[2]
+[1,2]
+==== idx = 2 ====
+[]
+[1]
+[2]
+[1,2]
+----
+[3]
+[1,3]
+[2,3]
+[1,2,3]
+```
+
+* Implement
+
+    ```python
+    class Solution:
+        def subsets(self, nums: List[int]) -> List[List[int]]:
+            result = [[]]
+            for num in nums:
+                for idx in range(len(result)):
+                    result.append(result[idx] + [num])
+            return result
+    ```
+
+    ```jsx
+    class Solution:
+        def subsets(self, nums: List[int]) -> List[List[int]]:
+            result = [[]]
+            for num in nums:
+                result+= [subset + [num] for subset in result]
+                    
+            return result
+    ```
+
+### Complexity
+
+* Time complexity: $O(N^2)$
+* Space complexity:$O(N)$