Combination Sum

1.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.

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.

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

class Solution {
public:
    vector<vector<int>> combinationSum(vector<int>& candidates, int target) {
        sort(candidates.begin(),candidates.end());
        vector<vector<int>> result;
        vector<int> path;
        dfs(0,0,target,candidates,path,result);
        return result;
    }
private:
    void dfs(int curr,int k,int target,vector<int>& candiates,
        vector<int>& path,vector<vector<int>>& result){
        if(curr == target){
            result.push_back(path);
            return;
        }

        for(int i=k;i<candiates.size();i++){
            curr = curr+candiates[i];
            if(curr > target) break;
            else{
                path.push_back(candiates[i]);
                dfs(curr,i,target, candiates, path, result);
                path.pop_back();
                curr = curr - candiates[i];
            }
        }
    }
};

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

Each number in C may only be used once in the combination.

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.

For example, given candidate set 10,1,2,7,6,1,5 and target 8, 
A solution set is: 
[1, 7] 
[1, 2, 5] 
[2, 6] 
[1, 1, 6]

class Solution {
public:
     vector<vector<int>> combinationSum2(vector<int>& candidates, int target) {
        sort(candidates.begin(),candidates.end());
        vector<vector<int>> result;
        vector<int> path;
        dfs(0,0,target,candidates,path,result);
        return result;
    }
private:
    void dfs(int curr,int k,int target,vector<int>& candiates,
        vector<int>& path,vector<vector<int>>& result){
        if(curr == target){
            if(find(result.begin(),result.end(),path)==result.end())
                result.push_back(path);
            return;
        }

        for(int i=k;i<candiates.size();i++){
            curr = curr+candiates[i];
            if(curr > target) break;
            else{
                path.push_back(candiates[i]);
                dfs(curr,i+1,target, candiates, path, result);
                path.pop_back();
                curr = curr - candiates[i];
            }
        }
    }
};

3.Find all possible combinations of k numbers that add up to a number n, given that only numbers from 1 to 9 can be used and each combination should be a unique set of numbers.

Ensure that numbers within the set are sorted in ascending order.

Example 1:

Input: k = 3, n = 7

Output:

 [[1,2,4]]

 Example 2:

Input: k = 3, n = 9

Output:

 [[1,2,6], [1,3,5], [2,3,4]]

class Solution {
public:
    vector<vector<int>> combinationSum3(int k, int n) {
        vector<vector<int>> result;
        vector<int> path;
        dfs(0,1,k,n,path,result);
        return result;
    }
private:
    void dfs(int curr,int j,int k,int n,vector<int>& path,
        vector<vector<int>>& result){
        if(path.size()==k){
            if(curr == n){
                result.push_back(path);
            }
            return;
        }
        for(int i=j;i<10;i++){
            curr = curr + i;
            if(curr > n) break;
            else{
                path.push_back(i);
                dfs(curr, i+1, k, n, path, result);
                path.pop_back();
                curr = curr - i;
            }
        }
    }
};