problem: 0090 subsets ii

Author: cychiu8

0090 Subsets II/0090 Subsets II.md

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+# 0090. Subsets II
+
+* Difficulty: medium
+* Link: https://leetcode.com/problems/subsets-ii/
+* Topics: Array-String, Backtracking
+* highlight: 只有在不 duplicate 的情況加上新的 num
+
+# Clarification
+
+1. Check the inputs and outputs
+    - INPUT:List[int]
+    - OUTPUT:List[List[int]]
+
+# Naive Solution
+
+<aside>
+💡 從最簡單的方法開始 easy solution → only speak out
+
+</aside>
+
+### Thought Process
+
+1. 算出每個 number 的數量
+2. 真對每個 subset 加上不同數量的 number
+- Implement
+    
+    ```python
+    class Solution:
+        def subsetsWithDup(self, nums: List[int]) -> List[List[int]]:
+            result = [[]]
+            num_map = {}
+            for num in nums:
+                num_map[num] = num_map.get(num, 0) + 1
+            for key, value in num_map.items():
+                for idx in range(len(result)):
+                    for val in range(1,value + 1):
+                        result.append(result[idx] + [key] * val)
+            return result
+    ```
+    
+
+### Complexity
+
+- Time complexity:$O(N+N^3)$
+- Space complexity:$O(N)$
+
+### Problems & Improvement
+
+- 有使用額外的 space 去存個數
+- time complexity 是否可降到 $O(N^2)$
+
+# Improvement
+
+### Thought Process
+
+1. 將 nums 排序
+2. 當數字相同時，只新增於相同數字的 subset
+    - 因為如果新增到其他的 subset 就會重複
+3.  當數字不同時，則可以對 result 內的所有 subset 加上新的數字
+- Implement
+    
+    ```python
+    class Solution:
+        def subsetsWithDup(self, nums: List[int]) -> List[List[int]]:
+            result = [[]]
+            curr = []
+            nums.sort()
+            for idx, num in enumerate(nums):
+                if idx > 0 and nums[idx - 1] == num:
+                    curr = [subset + [num] for subset in curr]
+                else:
+                    curr = [subset + [num] for subset in result]
+                result += curr
+                    
+            return result
+    ```
+    
+    - curr: 還可以再增長的 subset
+        - 當數字相同時，curr 就只考慮由前一輪的 curr 加上新的 num
+        - 當數字不相同時，curr 就可以為全部的 result 加上新的 num
+    
+    ```jsx
+    Example: [1, 2, 2, 3, 3]
+    result: [[]]
+    ==== idx: 0 ====
+    []
+    ----
+    [1]
+    ==== idx: 1 ====
+    []
+    [1]
+    ----
+    [2]
+    [1, 2]
+    ==== idx: 2 ====
+    []
+    [1]
+    [2]
+    [1, 2]
+    ----
+    [2, 2]
+    [1, 2, 2]
+    ==== idx: 3 ====
+    []
+    [1]
+    [2]
+    [1, 2]
+    [2, 2]
+    [1, 2, 2]
+    ----
+    [3]
+    [1, 3]
+    [2, 3]
+    [1, 2, 3]
+    [2, 2, 3]
+    [1, 2, 2, 3]
+    ==== idx: 4 ====
+    []
+    [1]
+    [2]
+    [1, 2]
+    [2, 2]
+    [1, 2, 2]
+    ----
+    [3]
+    [1, 3]
+    [2, 3]
+    [1, 2, 3]
+    [2, 2, 3]
+    [1, 2, 2, 3]
+    ----
+    [3, 3]
+    [1, 3, 3]
+    [2, 3, 3]
+    [1, 2, 3, 3]
+    [2, 2, 3, 3]
+    [1, 2, 2, 3, 3]
+    ```
+    
+
+### Complexity
+
+- Time complexity: $O(N^2)$
+- Space complexity:$O(1)$