Complete Leetcode 254

Author: yuchia0221

Backtracking/254-FactorCombinations/README.md

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+# Factor Combinations
+
+Problem can be found in [here](https://leetcode.com/problems/factor-combinations/)!
+
+### [Solution](/Backtracking/254-FactorCombinations/solution.py): Backtracking
+
+```python
+def getFactors(n: int) -> List[List[int]]:
+    def helper(factors: List[int], least_factor: int, target: int):
+        if len(factors) > 0:
+            result.append(factors + [target])
+        for candidate_factor in range(least_factor, int(sqrt(target))+1):
+            if target % candidate_factor == 0:
+                helper(factors+[candidate_factor], candidate_factor, target // candidate_factor)
+
+    result = []
+    if n == 1:
+        return result
+    helper([], 2, n)
+    return result
+```
+
+Explanation: We can use the backtracking technique to solve this problem. We will append valid factor into the factors list ascendingly to avoid duplicated factor combinations. If n = 12, we will only append [2, 2, 3] but not [2, 3, 2] with the help of a variable called least_factor.
+
+Time Complexity: ![O(logn*logn)](<https://latex.codecogs.com/svg.image?\inline&space;O(logn*logn)>), Space Complexity: ![O(n)](<https://latex.codecogs.com/svg.image?\inline&space;O(n)>)

Backtracking/254-FactorCombinations/solution.py

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+from math import sqrt
+from typing import List
+
+
+class Solution:
+    def getFactors(self, n: int) -> List[List[int]]:
+        def helper(factors: List[int], least_factor: int, target: int):
+            if len(factors) > 0:
+                result.append(factors + [target])
+            for candidate_factor in range(least_factor, int(sqrt(target))+1):
+                if target % candidate_factor == 0:
+                    helper(factors+[candidate_factor], candidate_factor, target // candidate_factor)
+
+        result = []
+        if n == 1:
+            return result
+        helper([], 2, n)
+        return result