Combination Sum, Combination Sum II

Author: idf

038 Combination Sum II py3.py

@@ -0,0 +1,67 @@
+#!/usr/bin/python3
+"""
+Given a collection of candidate numbers (candidates) and a target number
+(target), find all unique combinations in candidates where the candidate numbers
+sums to target.
+
+Each number in candidates may only be used once in the combination.
+
+Note:
+
+All numbers (including target) will be positive integers.
+The solution set must not contain duplicate combinations.
+Example 1:
+
+Input: candidates = [10,1,2,7,6,1,5], target = 8,
+A solution set is:
+[
+  [1, 7],
+  [1, 2, 5],
+  [2, 6],
+  [1, 1, 6]
+]
+Example 2:
+
+Input: candidates = [2,5,2,1,2], target = 5,
+A solution set is:
+[
+  [1,2,2],
+  [5]
+]
+"""
+from typing import List
+
+
+class Solution:
+    def combinationSum2(self, candidates: List[int], target: int) -> List[List[int]]:
+        ret = []
+        candidates.sort()
+        self.dfs(candidates, 0, [], 0, target, ret)
+        return ret
+
+    def dfs(self, candidates, i, cur, cur_sum, target, ret):
+        if cur_sum == target:
+            ret.append(list(cur))
+            return
+
+        if cur_sum > target or i >= len(candidates):
+            return
+
+        # not choose A_i
+        # to de-dup, need to jump
+        j = i + 1
+        while j < len(candidates) and candidates[j] == candidates[i]:
+            j += 1
+
+        self.dfs(candidates, j, cur, cur_sum, target, ret)
+
+        # choose A_i
+        cur.append(candidates[i])
+        cur_sum += candidates[i]
+        self.dfs(candidates, i + 1, cur, cur_sum, target, ret)
+        cur.pop()
+        cur_sum -= candidates[i]
+
+
+if __name__ == "__main__":
+    assert Solution().combinationSum2([2,5,2,1,2], 5) == [[5], [1,2,2]]

039 Combination Sum py3.py

@@ -0,0 +1,57 @@
+#!/usr/bin/python3
+"""
+Given a set of candidate numbers (candidates) (without duplicates) and a target
+number (target), find all unique combinations in candidates where the candidate
+numbers sums to target.
+
+The same repeated number may be chosen from candidates unlimited number of
+times.
+
+Note:
+
+All numbers (including target) will be positive integers.
+The solution set must not contain duplicate combinations.
+Example 1:
+
+Input: candidates = [2,3,6,7], target = 7,
+A solution set is:
+[
+  [7],
+  [2,2,3]
+]
+Example 2:
+
+Input: candidates = [2,3,5], target = 8,
+A solution set is:
+[
+  [2,2,2,2],
+  [2,3,3],
+  [3,5]
+]
+"""
+from typing import List
+
+
+class Solution:
+    def combinationSum(self, candidates: List[int], target: int) -> List[List[int]]:
+        ret = []
+        self.dfs(candidates, 0, [], 0, target, ret)
+        return ret
+
+    def dfs(self, candidates, i, cur, cur_sum, target, ret):
+        if cur_sum == target:
+            ret.append(list(cur))
+            return
+            
+        if cur_sum > target or i >= len(candidates):
+            return
+
+        # not choose A_i
+        self.dfs(candidates, i + 1, cur, cur_sum, target, ret)
+
+        # choose A_i
+        cur.append(candidates[i])
+        cur_sum += candidates[i]
+        self.dfs(candidates, i, cur, cur_sum, target, ret)
+        cur.pop()
+        cur_sum -= candidates[i]