Problem Description

Given two integers n and k, return all possible comibnations of k numbers out of the range [1, n].

You may return the answer in any order.

Example 1:

Input: n = 4, k = 2
Output:
[
  [2,4],
  [3,4],
  [2,3],
  [1,2],
  [1,3],
  [1,4],
]

Example 2:

Input: n = 1, k = 1
Output: [[1]]

Constraints:

• 1 <= n <= 20
• 1 <= k <= n

Solution

We can use backtracking to generate all combinations. The width of the recursion tree is n, and the depth is k.

Apporach 1: Without Pruning

class Solution {
    List<List<Integer>> result = new ArrayList<>();
    LinkedList<Integer> current = new LinkedList<>();
    
    public List<List<Integer>> combine(int n, int k) {
        backtrack(n, k, 1);
        return result;
    }
    
    private void backtrack(int n, int k, int startIndex) {
        if (current.size() == k) {
            result.add(new ArrayList<>(current));
            return;
        }
        for (int i = startIndex; i <= n; i++) {
            current.add(i);
            backtrack(n, k, i + 1);
            current.removeLast();
        }
    }
}

Approach 2: With Pruning

The number of remaining elements needed is k - current.size(). Therefore, the last starting index should be at most n - (k - current.size()) + 1 to have enough elements.

class Solution {
    List<List<Integer>> result = new ArrayList<>();
    LinkedList<Integer> path = new LinkedList<>();
    
    public List<List<Integer>> combine(int n, int k) {
        combineHelper(n, k, 1);
        return result;
    }
    
    private void combineHelper(int n, int k, int startIndex) {
        if (path.size() == k) {
            result.add(new ArrayList<>(path));
            return;
        }
        for (int i = startIndex; i <= n - (k - path.size()) + 1; i++) {
            path.add(i);
            combineHelper(n, k, i + 1);
            path.removeLast();
        }
    }
}