216. Combination Sum III

Description

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.

Constraints

• 2 <= k <= 9

• 1 <= n <= 60

Approach

Links

• GeeksforGeeks

• Leetcode

• ProgramCreek

• YouTube

Examples

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. [1, 2, 1] is not valid because 1 is used twice.

Example 4

Input: k = 3, n = 2

Output: []

Explanation: There are no valid combinations.

Example 5

Input: k = 9, n = 45

Output: [[1, 2, 3, 4, 5, 6, 7, 8, 9]]

Explanation:

1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 45 ​​​​​​​ There are no other valid combinations.

Solutions

Solution 1

/**
 * Time complexity : 
 * Space complexity : 
 */

class Solution {
    public List<List<Integer>> combinationSum3(int k, int n) {
        List<List<Integer>> resultList = new ArrayList();
        helper(n, k, resultList, new LinkedList<Integer>(), 0, 1);
        return resultList;
    }
    
    private void helper(int n,
                        int k,
                        List<List<Integer>> resultList,
                        LinkedList<Integer> currList,
                        int sum,
                        int start) {
        if(currList.size() == k && sum == n) {
            resultList.add(new ArrayList(currList));
            return;
        }
        if(currList.size() < k) {
            for(int j = start; j <= 9; j++) {
                currList.add(j);
                helper(n, k, resultList, currList, sum+j, j+1);
                currList.removeLast();
            }
        }
    }
}

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