problem: 0022 0040 0216 combinations

Author: cychiu8

0022 Generate Parentheses/0022 Generate Parentheses.md

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+# 0022. Generate Parentheses
+
+* Difficulty: medium
+* Link: https://leetcode.com/problems/generate-parentheses/
+* Topics: Backtracking
+
+# Clarification
+
+1. Check the inputs and outputs
+    - INPUT:
+        - n: int
+    - OUTPUT: List[String]
+
+# Naive Solution
+
+### Thought Process
+
+- Backtracking
+    - 停止條件
+        - open == n and close == n
+    - 非停止條件
+        - 當 open < n
+            - subset = subset + open
+        - 當 close < open
+            - subset = subest + close
+
+![Untitled](./Untitled.png)
+
+- Implement
+    
+    ```python
+    class Solution(object):
+        def generateParenthesis(self, n):
+            """
+            :type n: int
+            :rtype: List[str]
+            """
+            result = []
+            
+            def backtracking(substr, num_open, num_close):
+                if num_open == n and num_close == n:
+                    return result.append(substr)
+                if num_open < n:
+                    backtracking(substr + "(", num_open + 1, num_close)
+                if num_close < num_open:
+                    backtracking(substr + ")", num_open, num_close+1)
+            
+            backtracking("", 0,0)
+            return result
+    ```
+    
+
+# Note
+
+- **[Generate Parentheses - Stack - Leetcode 22](https://www.youtube.com/watch?v=s9fokUqJ76A)**

0022 Generate Parentheses/Untitled.png

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+# 40. Combination Sum II
+
+* Difficulty: medium
+* Link: https://leetcode.com/problems/combination-sum-ii/
+* Topics: Array-String, Backtracking
+
+# Clarification
+
+1. Check the inputs and outputs
+    - INPUT:
+        - canditates: List[int]
+        - target: int
+    - OUTPUT: List[List[int]
+
+# Naive Solution
+
+### Thought Process
+
+- Backtracking
+    - 停止條件
+        - sum of subset == target : add to result
+        - sum of subset > target : return
+    - 未達停止條件
+        - subset + res[i]
+        - res = res[i+1:]
+- 注意：當 res[i] == res[i+1] 時跳過，避免重複
+- Implement
+    
+    ```python
+    class Solution(object):
+        def combinationSum2(self, candidates, target):
+            """
+            :type candidates: List[int]
+            :type target: int
+            :rtype: List[List[int]]
+            """
+            candidates.sort()
+            result = []
+            def backtracking(subset, res):
+                if sum(subset) == target:
+                    return result.append(subset)
+                if sum(subset) > target:
+                    return
+                for idx in range(len(res)):
+                    if idx > 0 and res[idx] == res[idx-1]:
+                        continue
+                    backtracking(subset + [res[idx]], res[idx+1:])
+                
+            backtracking([], candidates)
+            return result
+    ```

0040 Combination Sum II/0040 Combination Sum II.md

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+# 216. Combination Sum III
+
+* Difficulty: medium
+* Link: https://leetcode.com/problems/combination-sum-iii/
+* Topics: Backtracking
+
+# Clarification
+
+1. Check the inputs and outputs
+    - INPUT:
+        - k: int
+        - n: int (target)
+    - OUTPUT:List[List[int]]
+
+# Naive Solution
+
+### Thought Process
+
+- Backtracking
+    - 停止條件
+        - len(subset) == k and sum(subset) == n ⇒ add to result
+        - sum(subset) > n ⇒ return
+        - len(subset) > k ⇒ return
+    - 非停止條件
+        - subset = subset + res[i]
+        - res = res[i+1:]
+- Implement
+    
+    ```python
+    class Solution(object):
+        def combinationSum3(self, k, n):
+            """
+            :type k: int
+            :type n: int
+            :rtype: List[List[int]]
+            """
+            result = []
+            candidates = range(1,10)
+            def backtracking(subset, res):
+                if len(subset) == k and sum(subset) == n:
+                    return result.append(subset)
+                if len(subset) > k:
+                    return
+                if sum(subset) > n:
+                    return
+                for idx, num in enumerate(res):
+                    backtracking(subset + [res[idx]], res[idx+1:])
+                    
+            backtracking([], candidates)
+            return result
+    ```