Merge branch 'master' of https://github.com/cychiu8/leetcode-practice

Author: cychiu8

0039 Combination Sum/0039 Combination Sum.md

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+# 39. Combination Sum
+
+* Difficulty: medium
+* Link: https://leetcode.com/problems/combination-sum/
+* Topics: Backtracking
+
+# Clarification
+
+1. Check the inputs and outputs
+    - INPUT:
+        - List[int] candiates
+        - int target
+    - OUTPUT: List[List[int]]
+
+# Naive Solution
+
+### Thought Process
+
+1. 停止條件：
+    1. 加總等於 target
+    2. 加總大於 target
+2. subset = 現有 + candiate[i]
+3. 新的 candidate 則是從 candidate[i] 之後開始 (包含 candidate[i], 因為 The **same**
+ number may be chosen from `candidates` an **unlimited number of times**.)
+
+![Untitled](./Untitled.png)
+
+- Implement
+    
+    ```python
+    class Solution(object):
+        def combinationSum(self, candidates, target):
+            """
+            :type candidates: List[int]
+            :type target: int
+            :rtype: List[List[int]]
+            """
+            result = []
+            def backtrack(subset, res):
+                if sum(subset) == target:
+                    return result.append(subset)
+                elif sum(subset) > target:
+                    return
+                for i in range(len(res)):
+                    backtrack(subset + [res[i]], res[i:])
+            backtrack([], candidates)
+            return result
+    ```
+    
+
+### Complexity
+
+- Time complexity:
+    - $O(n^n)$
+        
+        ![737308D1-A159-4D3A-A961-D71B98C17CB6.jpeg](./737308D1-A159-4D3A-A961-D71B98C17CB6.jpeg)
+        
+    - $O(k*2^n)$
+        - **[Deadbeef-ECE 039_Combination_Sum.java](https://github.com/Deadbeef-ECE/Interview/blob/master/Leetcode/BackTracking/039_Combination_Sum.java)**
+- Space complexity:
+
+### Problems & Improvement
+
+- 可以更早排除不須 backtrack 的數字
+    - 當 candidate > target 時就不用進去 backtrack 了
+        - Implement
+            
+            ```python
+            class Solution(object):
+                def combinationSum(self, candidates, target):
+                    """
+                    :type candidates: List[int]
+                    :type target: int
+                    :rtype: List[List[int]]
+                    """
+                    result = []
+                    def backtrack(subset, res):
+                        if sum(subset) == target:
+                            return result.append(subset)
+                        elif sum(subset) > target:
+                            return
+                        for i in range(len(res)):
+                            if res[i] > target:
+                                return
+                            backtrack(subset + [res[i]], res[i:])
+                    backtrack([], candidates)
+                    return result
+            ```

0039 Combination Sum/737308D1-A159-4D3A-A961-D71B98C17CB6.jpeg

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+# 77. Combinations
+
+* Difficulty: medium
+* Link: https://leetcode.com/problems/combinations/
+* Topics: Backtracking
+
+# Clarification
+
+1. Check the inputs and outputs
+    - INPUT:
+        - int n : the range [1,n]
+        - int k : pick k numbers
+    - OUTPUT: List[List[int]]
+
+# Naive Solution
+
+### Thought Process
+
+- [NOT working] 用於 k 為少數時
+    1. 使用兩個 pointer
+    2. 第一個放一開始，第二個從第一個的下一個開始移動
+    3. 直到第二個 pointer 指到 Null
+- 停止條件：當 subset 的長度 = k 時
+    
+    ![Untitled](./Untitled.png)
+    
+- Implement
+    
+    ```python
+    class Solution(object):
+        def combine(self, n, k):
+            """
+            :type n: int
+            :type k: int
+            :rtype: List[List[int]]
+            """
+            result = []
+            def combination(subset, res):
+                if len(subset) == k:
+                    return result.append(subset)
+                for i in range(len(res)):
+                    combination(subset + [res[i]], res[i+1:])
+            combination([], range(1,n+1))
+            return result
+    ```
+    
+
+### Complexity
+
+- Time complexity:
+    - O(C(n,k))
+    - $C^n_k=(n!)/k!(n-k)!$
+- Space complexity: O(k)
+    - subset 的大小