Combination Sum III LeetCode Solution

Problem – Combination Sum III LeetCode Solution

Find all valid combinations of k numbers that sum up to n such that the following conditions are true:

  • Only numbers 1 through 9 are used.
  • 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.

Example 1:

Input: k = 3, n = 7
Output: [[1,2,4]]
Explanation:
1 + 2 + 4 = 7
There are no other valid combinations.

Example 2:

Input: k = 3, n = 9
Output: [[1,2,6],[1,3,5],[2,3,4]]
Explanation:
1 + 2 + 6 = 9
1 + 3 + 5 = 9
2 + 3 + 4 = 9
There are no other valid combinations.

Example 3:

Input: k = 4, n = 1
Output: []
Explanation: There are no valid combinations.
Using 4 different numbers in the range [1,9], the smallest sum we can get is 1+2+3+4 = 10 and since 10 > 1, there are no valid combination.

Constraints:

  • 2 <= k <= 9
  • 1 <= n <= 60

Combination Sum III LeetCode Solution in Java

 public List<List<Integer>> combinationSum3(int k, int n) {
    List<List<Integer>> ans = new ArrayList<>();
    combination(ans, new ArrayList<Integer>(), k, 1, n);
    return ans;
}

private void combination(List<List<Integer>> ans, List<Integer> comb, int k,  int start, int n) {
	if (comb.size() == k && n == 0) {
		List<Integer> li = new ArrayList<Integer>(comb);
		ans.add(li);
		return;
	}
	for (int i = start; i <= 9; i++) {
		comb.add(i);
		combination(ans, comb, k, i+1, n-i);
		comb.remove(comb.size() - 1);
	}
}

Combination Sum III LeetCode Solution in Python

class Solution(object):
    def combinationSum3(self, k, n):
        ret = []
        self.dfs(list(range(1, 10)), k, n, [], ret)
        return ret
    
    def dfs(self, nums, k, n, path, ret):
        if k < 0 or n < 0:
            return 
        if k == 0 and n == 0:
            ret.append(path)
        for i in range(len(nums)):
            self.dfs(nums[i+1:], k-1, n-nums[i], path+[nums[i]], ret)

Combination Sum III LeetCode Solution in C++

class Solution {
public:
  void combination(vector<vector<int>>& result, vector<int> sol, int k, int n) {
    if (sol.size() == k && n == 0) { result.push_back(sol); return ; }
    if (sol.size() < k) {
      for (int i = sol.empty() ? 1 : sol.back() + 1; i <= 9; ++i) {
        if (n - i < 0) break;
        sol.push_back(i);
        combination(result, sol, k, n - i);
        sol.pop_back();
      }
    }
  }

  vector<vector<int>> combinationSum3(int k, int n) {
    vector<vector<int>> result;
    vector<int> sol;
    combination(result, sol, k, n);
    return result;
  }
};
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