1+ // Source : https://leetcode.com/problems/combination-sum/
2+ // Author : henrytien
3+ // Date : 2022年02月17日
4+ 5+ /* ****************************************************************************************************
6+ *
7+ * Given an array of distinct integers candidates and a target integer target, return a list of all
8+ * unique combinations of candidates where the chosen numbers sum to target. You may return the
9+ * combinations in any order.
10+ *
11+ * The same number may be chosen from candidates an unlimited number of times. Two combinations are
12+ * unique if the frequency of at least one of the chosen numbers is different.
13+ *
14+ * It is guaranteed that the number of unique combinations that sum up to target is less than 150
15+ * combinations for the given input.
16+ *
17+ * Example 1:
18+ *
19+ * Input: candidates = [2,3,6,7], target = 7
20+ * Output: [[2,2,3],[7]]
21+ * Explanation:
22+ * 2 and 3 are candidates, and 2 + 2 + 3 = 7. Note that 2 can be used multiple times.
23+ * 7 is a candidate, and 7 = 7.
24+ * These are the only two combinations.
25+ *
26+ * Example 2:
27+ *
28+ * Input: candidates = [2,3,5], target = 8
29+ * Output: [[2,2,2,2],[2,3,3],[3,5]]
30+ *
31+ * Example 3:
32+ *
33+ * Input: candidates = [2], target = 1
34+ * Output: []
35+ *
36+ * Constraints:
37+ *
38+ * 1 <= candidates.length <= 30
39+ * 1 <= candidates[i] <= 200
40+ * All elements of candidates are distinct.
41+ * 1 <= target <= 500
42+ ******************************************************************************************************/
43+ 44+ #include " ../inc/ac.h"
45+ 46+ // Let N be the number of candidates, T be the target value, and M be the minimal value among the candidates.
47+ // Time Complexity: O(N^T/M)
48+ // Space Complexity: O(T/M)
49+ class Solution
50+ {
51+ public:
52+ vector<vector<int >> combinationSum (vector<int > &candidates, int target)
53+ {
54+ vector<vector<int >> res;
55+ vector<int > cur_res;
56+ backtrack (candidates,0 ,target,cur_res,res);
57+ return res;
58+ }
59+ 60+ private:
61+ void backtrack (const vector<int > &canditates, int index, int remain,
62+ vector<int > &cur_res, vector<vector<int >> &res)
63+ {
64+ if (remain == 0 )
65+ {
66+ res.push_back (cur_res);
67+ return ;
68+ }else if (remain < 0 ) {
69+ // exceed the scope, stop exploration.
70+ return ;
71+ }
72+ 73+ for (int i = index; i < canditates.size (); i++)
74+ {
75+ if (remain >= canditates[i])
76+ {
77+ // Add the number into the combination
78+ cur_res.push_back (canditates[i]);
79+ backtrack (canditates, i, remain - canditates[i], cur_res, res);
80+ // backtrack, remove the number from the combination
81+ cur_res.pop_back ();
82+ }
83+ }
84+ return ;
85+ }
86+ };
87+ 88+ int main ()
89+ {
90+ 91+ vector<int > candidates{2 , 3 , 6 , 7 };
92+ int target = 7 ;
93+ vector<vector<int >> res = Solution ().combinationSum (candidates, target);
94+ for (auto &&iter : res)
95+ {
96+ print_vec (iter);
97+ }
98+ 99+ return 0 ;
100+ }
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