40. Combination Sum II - Detailed Explanation
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Problem Statement
Given a collection of candidate numbers candidates
(without duplicates) and a target integer target
, return a list of all unique combinations of candidates
where the chosen numbers sum to target
. You may only use each candidate once. The solution set 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
1 ≤ candidates[i] ≤ 50
- All elements of
candidates
are integers (may contain duplicates) 1 ≤ target ≤ 500
Hints
- Sort
candidates
first to help skip duplicates. - Use backtracking, advancing an index so each number is used at most once.
- When you see the same value at the same recursive depth, skip it to avoid duplicate sequences.
Approach 1 Backtracking with Pruning
Idea
We explore each number in sorted order, decide to include it or skip it, and recurse on the remaining target. By sorting we can easily detect and skip over duplicates at the same decision level.
Steps
- Sort
candidates
. - Define a recursive function
dfs(start, remaining, path)
:- If
remaining == 0
, record a copy ofpath
. - Otherwise, for
i
fromstart
tolen(candidates)-1
:- If
candidates[i] > remaining
, break (further ones are too big). - If
i > start
andcandidates[i] == candidates[i-1]
, continue (skip duplicate). - Append
candidates[i]
topath
, recursedfs(i+1, remaining - candidates[i], path)
, then pop.
- If
- If
- Call
dfs(0, target, [])
and return collected results.
Code (Python)
Python3
Python3
. . . .
Code (Java)
Java
Java
. . . .
Complexity Analysis
- Time ≈ O(2ⁿ) in the worst case due to exploring subsets, but pruning and duplicate‐skipping reduce the practical load.
- Space O(n) recursion depth plus O(#solutions × n) to store results.
Step‑by‑Step Walkthrough and Visual Example
For candidates = [1,1,2,5,6,7,10], target = 8
after sorting:
- Start at
1
(index 0): choose it → remaining 7 → recurse start 1 - At index 1 also
1
: choose second1
→ remaining 6 → recurse start 2 - At index 2
2
: choose → remaining 4 … eventually path [1,1,6] → record. - Backtrack to explore [1,2,…], skip duplicate
1
at index 1 when start 1, etc. - Explore [2,6], [1,7], [1,2,5], [1,7], [2,6], and so on, skipping repeated values at each level.
Common Mistakes
- Forgetting to sort, so duplicate‐skipping (
i>start && cand[i]==cand[i-1]
) fails. - Using
dfs(i,…)
instead ofdfs(i+1,…)
, allowing reuse of the same element. - Not breaking when
cand[i] > rem
, leading to unnecessary exploration.
Edge Cases
- Empty
candidates
→ return empty list. - All candidates larger than
target
→ return empty list. - Many duplicates in
candidates
but no combination sums totarget
.
Alternative Variations
- Combination Sum (LeetCode 39) allows unlimited use of each candidate.
- Combination Sum III (LeetCode 216) fixes the number of elements to choose.
- Subset Sum Count: count the number of ways instead of enumerating them.
Related Problems
TAGS
leetcode
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