LintCode: Combination Sum

Given a set of candidate numbers (C) and a target number (T), find all unique combinations in C where the candidate numbers sums to T.

The same repeated number may be chosen from C unlimited number of times.

For example, given candidate set 2,3,6,7 and target 7,
A solution set is:
[7]
[2, 2, 3] 

Have you met this question in a real interview? 

Yes

Example

given candidate set 2,3,6,7 and target 7, 
A solution set is: 
[7] 
[2, 2, 3] 

Note

• All numbers (including target) will be positive integers.
• Elements in a combination (a1, a2, … , ak) must be in non-descending order. (ie, a1 ≤ a2 ≤ … ≤ ak).
• The solution set must not contain duplicate combinations.

Tags Expand 

Backtracking Array

Reference:
//codes
http://yucoding.blogspot.com/2012/12/leetcode-question-16-combination-sum.html
//complexity
https://gist.github.com/daifu/5844049

DFS recursively like problem combination
Note:
1. The candidates should be ordered first to make sure output an ordered output;
2. Same number can appear multiple times;
Time: O( (n + k)! )   n ---- size of candidates   k----max repeated times for each candidates
Space: O( m )   ----- solution size
stack space: the max length of the longest combination in result.

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30

class Solution { public: /** * @param candidates: A list of integers * @param target:An integer * @return: A list of lists of integers */ vector<vector<int> > combinationSum(vector<int> &candidates, int target) { // write your code here vector<vector<int> > res; vector<int> tempres; sort(candidates.begin(), candidates.end()); DFS(0, target, 0, candidates, res, tempres); return res; } void DFS(int start, int target, int sum, vector<int> &candidates, vector<vector<int> > &res, vector<int> &tempres){ if(sum > target) return; else if(sum == target) { res.push_back(tempres); return; } else{ for(int i = start; i < candidates.size(); i++){ tempres.push_back(candidates[i]); DFS(i, target, sum + candidates[i], candidates, res, tempres); tempres.pop_back(); } } } };

Lintcode second:
add line 24 to in case that there is 0 in candidates

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29

class Solution { public: /** * @param candidates: A list of integers * @param target:An integer * @return: A list of lists of integers */ vector<vector<int> > combinationSum(vector<int> &candidates, int target) { // write your code here sort(candidates.begin(), candidates.end()); vector<vector<int> > res; vector<int> temp; dfs(0, 0, target, candidates, temp, res); return res; } void dfs(int start, int sum, int target, vector<int> &candidates, vector<int> &temp, vector<vector<int> > &res){ if(sum == target){ res.push_back(temp); return; } if(sum > target) return; for(int i = start; i < candidates.size(); i++){ temp.push_back(candidates[i]); if(candidates[i] == 0) dfs(i + 1, sum, target, candidates, temp, res); else dfs(i, sum + candidates[i], target, candidates, temp, res); temp.pop_back(); } } };