Combinations

Medium Recursion & Backtracking Pick / Not-Pick

Combinations is a medium recursion & backtracking problem solved with the pick / not-pick pattern. The best approach, optimal (loop with pruning), runs in O(k · C(n, k)) time and O(k) space. Below are 2 approaches in Java, from pick / not-pick over 1..n up.

Problem

Given n and k, return all combinations of k numbers chosen from 1..n, in any order. [1,2] and [2,1] are the same combination.

Examples

Example 1

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

Example 2

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

Constraints

1 <= k <= n <= 20

Animated walkthrough

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Solutions

Try it yourself first. Then compare: each approach lists its idea, the steps, its time and space, and the Java code.

Approach	Time	Space
Pick / not-pick over 1..n	`O(k · C(n, k))`	`O(n)`
Optimal (loop with pruning)	`O(k · C(n, k))`	`O(k)`

1Pick / not-pick over 1..n

TimeO(k · C(n, k))

SpaceO(n)Recursion depth up to n.

For each number decide to take it or skip it, and record when the path has k numbers.

If path.size() == k, record.
If i > n, return.
Take i and recurse; skip i and recurse.

Java

class Solution {
    public List<List<Integer>> combine(int n, int k) {
        List<List<Integer>> out = new ArrayList<>();
        pick(1, n, k, new ArrayList<>(), out);
        return out;
    }

    private void pick(int i, int n, int k, List<Integer> path, List<List<Integer>> out) {
        if (path.size() == k) { out.add(new ArrayList<>(path)); return; }
        if (i > n) return;
        path.add(i);
        pick(i + 1, n, k, path, out);
        path.remove(path.size() - 1);
        pick(i + 1, n, k, path, out);
    }
}

2Optimal (loop with pruning)

TimeO(k · C(n, k))

SpaceO(k)

At each level loop over the next number from start. If fewer than k - path.size() numbers remain, no complete combination is possible, so stop the loop early.

need = k - path.size(); if need == 0, record.
for i from start to n - need + 1: add i, recurse with i + 1, remove.

Java

class Solution {
    public List<List<Integer>> combine(int n, int k) {
        List<List<Integer>> out = new ArrayList<>();
        build(1, n, k, new ArrayList<>(), out);
        return out;
    }

    private void build(int start, int n, int k, List<Integer> path, List<List<Integer>> out) {
        int need = k - path.size();
        if (need == 0) { out.add(new ArrayList<>(path)); return; }
        for (int i = start; i <= n - need + 1; i++) {
            path.add(i);
            build(i + 1, n, k, path, out);
            path.remove(path.size() - 1);
        }
    }
}

Edge cases to test

k = n (one combination)
k = 1

Hints

Hint 1

Only choose numbers larger than the last one so each set appears once, and stop early when too few numbers are left to reach k.

FAQ

What is the best time complexity for Combinations?

Optimal (loop with pruning) runs in O(k · C(n, k)) time and O(k) extra space.

Which pattern does Combinations use?

It is a recursion & backtracking problem that uses the pick / not-pick pattern. Other problems with the same pattern: Generate all binary strings without consecutive 1's, Subsets (2 choices per element), Subsets II.

Is there a brute force solution for Combinations?

Yes. Pick / not-pick over 1..n takes O(k · C(n, k)) time and O(n) space. For each number decide to take it or skip it, and record when the path has k numbers.

Which edge cases should I test for Combinations?

k = n (one combination); k = 1.