In this Leetcode Combination Sum III problem solution Find all valid combinations of k numbers that sum up to n such that the following conditions are true:

  1. Only numbers 1 through 9 are used.
  2. Each number is used at most once.

Return a list of all possible valid combinations. The list must not contain the same combination twice, and the combinations may be returned in any order.

Leetcode Combination Sum III problem solution


Problem solution in Python.

class Solution:
    def combinationSum3(self, k: int, n: int) -> List[List[int]]:
         return [list for list in itertools.combinations(range(1, 10), k) if sum(list) == n]



Problem solution in Java.

class Solution {
    public List<List<Integer>> combinationSum3(int k, int n) {
        List<Integer> comb = new ArrayList<Integer>();
        List<List<Integer>> op = new ArrayList<List<Integer>>();
        
        helper(op, comb, k, 1, n);
        
        return op;
    }
    
    public static void helper(List<List<Integer>> op, List<Integer> comb, int k, int start, int n){
        if(comb.size() == k && n == 0){
            List<Integer> li = new ArrayList<Integer>(comb);
            op.add(li);
            return;
        }
        
        for(int i=start; i<10; i++){
            if (comb.size() < k) {
                comb.add(i);
                helper(op, comb, k, i+1, n-i);
                comb.remove(comb.size() - 1);
            }
        }
    }
}


Problem solution in C++.

class Solution {
public:
void dfs(vector<vector<int>>& res, vector<int> &pre, int start, int k, int n){
    if (k==0 && n==0){
            res.push_back(pre); return;
    }
    if (start>9 || start>n || k==0) return;
    pre.push_back(start);
    dfs(res,pre,start+1,k-1,n-start);
    pre.pop_back();
    dfs(res,pre,start+1,k,n);
}
vector<vector<int>> combinationSum3(int k, int n) {
    vector<vector<int>> res;
    vector<int> pre;
    dfs(res,pre,1,k,n);
    return res;
}
};


Problem solution in C.

void dfs(int **ret,int *candidate,int *rindex,int *cindex,int k,int n,int index){
    if(n==0 && k==0){
        ret[*rindex] = (int*)malloc(sizeof(int)*(*cindex));
        for(int i=0;i<(*cindex);i++){
            ret[*rindex][i]=candidate[i];
        }
        (*rindex)++;
        int **tmp = (int **)realloc(ret,sizeof(int*)*((*rindex)+1));
        if(tmp != NULL){
            ret=tmp;
        }
        return;
    }
    if(index>9 || n<=0 || k<=0){
        return;
    }
    for(int i=index;i<=9;i++){
        if(i>n){
            break;
        }
        *(candidate+(*cindex))=i;
        (*cindex)++;
        dfs(ret,candidate,rindex,cindex,k-1,n-i,i+1);
        (*cindex)--;
    }
}
  
int** combinationSum3(int k, int n, int** columnSizes, int* returnSize) {
    int *candidate = (int*)malloc(sizeof(int)*k);
    memset(candidate,0,sizeof(candidate));
    int **ret = (int **)malloc(sizeof(int*));
    int rindex=0;
    int cindex=0;
    dfs(ret,candidate,&rindex,&cindex,k,n,1);
    *columnSizes = (int *)malloc(sizeof(int)*(rindex));
    for(int i=0;i<rindex;i++){
        (*columnSizes)[i]=k;
    }
    *returnSize=rindex;
    free(candidate);
    return ret;
}