Combination Sum II

Medium Recursion & Backtracking Pick / Not-Pick Original on LeetCode

Combination Sum II is a medium recursion & backtracking problem solved with the pick / not-pick pattern. The best approach, optimal (sort, skip equal siblings, break early), runs in O(n · 2ⁿ) time and O(n) space. Below are 2 approaches in Java, from generate and deduplicate up.

Problem

Given candidates (which may contain duplicates) and a target, return all unique combinations that sum to target. Each element can be used at most once, and the answer must not contain duplicate combinations.

Examples

Example 1

Input: candidates = [10, 1, 2, 7, 6, 1, 5], target = 8
Output: [[1,1,6],[1,2,5],[1,7],[2,6]]

Example 2

Input: candidates = [2, 5, 2, 1, 2], target = 5
Output: [[1,2,2],[5]]

Constraints

1 <= candidates.length <= 100, values from 1 to 50
1 <= target <= 30
Each element is used at most once; the input may have duplicates.

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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
Generate and deduplicate	`O(n · 2ⁿ)`	`O(n · 2ⁿ)`
Optimal (sort, skip equal siblings, break early)	`O(n · 2ⁿ)`	`O(n)`

1Generate and deduplicate

TimeO(n · 2ⁿ)

SpaceO(n · 2ⁿ)

Sort, run pick / not-pick with each element at most once, and store results in a set.

Sort; pick / not-pick; add successful paths to a HashSet.

Java

class Solution {
    public List<List<Integer>> combinationSum2(int[] candidates, int target) {
        Arrays.sort(candidates);
        Set<List<Integer>> seen = new HashSet<>();
        go(candidates, 0, target, new ArrayList<>(), seen);
        return new ArrayList<>(seen);
    }

    private void go(int[] c, int i, int target, List<Integer> path, Set<List<Integer>> seen) {
        if (target == 0) { seen.add(new ArrayList<>(path)); return; }
        if (i == c.length || target < 0) return;
        path.add(c[i]);
        go(c, i + 1, target - c[i], path, seen);
        path.remove(path.size() - 1);
        go(c, i + 1, target, path, seen);
    }
}

2Optimal (sort, skip equal siblings, break early)

TimeO(n · 2ⁿ)

SpaceO(n)

Sort. Loop from start; skip c[i] when it equals c[i - 1] at the same level (i > start); break once c[i] exceeds the remaining target; recurse with i + 1 so each element is used once.

for i from start: if i > start && c[i] == c[i - 1], continue.
if c[i] > target, break.
Add, recurse(i + 1, target - c[i]), remove.

Java

class Solution {
    public List<List<Integer>> combinationSum2(int[] candidates, int target) {
        Arrays.sort(candidates);
        List<List<Integer>> out = new ArrayList<>();
        go(candidates, 0, target, new ArrayList<>(), out);
        return out;
    }

    private void go(int[] c, int start, int target, List<Integer> path, List<List<Integer>> out) {
        if (target == 0) { out.add(new ArrayList<>(path)); return; }
        for (int i = start; i < c.length; i++) {
            if (i > start && c[i] == c[i - 1]) continue;
            if (c[i] > target) break;
            path.add(c[i]);
            go(c, i + 1, target - c[i], path, out);
            path.remove(path.size() - 1);
        }
    }
}

Edge cases to test

Many equal values
No combination possible

Hints

Hint 1

This is Subsets II plus a target sum: sort, skip equal siblings, recurse with i + 1.

FAQ

What is the best time complexity for Combination Sum II?

Optimal (sort, skip equal siblings, break early) runs in O(n · 2ⁿ) time and O(n) extra space.

Which pattern does Combination Sum II 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 Combination Sum II?

Yes. Generate and deduplicate takes O(n · 2ⁿ) time and O(n · 2ⁿ) space. Sort, run pick / not-pick with each element at most once, and store results in a set.

Which edge cases should I test for Combination Sum II?

Many equal values; No combination possible.