Leetcode #40: Combination Sum II
In this guide, we solve Leetcode #40 Combination Sum II in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
Given a collection of candidate numbers (candidates) and a target number (target), find all unique combinations in candidates where the candidate numbers sum to target. Each number in candidates may only be used once in the combination.
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: Array, Backtracking
Intuition
We must explore combinations of choices, but many branches can be pruned early.
Backtracking enumerates valid candidates while keeping the search space under control.
Approach
Use DFS to build candidates step by step, and backtrack when constraints are violated.
Pruning keeps the exploration practical for typical constraints.
Steps:
- Define the decision tree.
- DFS through choices and backtrack.
- Prune invalid paths early.
Example
Input: candidates = [10,1,2,7,6,1,5], target = 8
Output:
[
[1,1,6],
[1,2,5],
[1,7],
[2,6]
]
Python Solution
class Solution:
def combinationSum2(self, candidates: List[int], target: int) -> List[List[int]]:
def dfs(i: int, s: int):
if s == 0:
ans.append(t[:])
return
if i >= len(candidates) or s < candidates[i]:
return
for j in range(i, len(candidates)):
if j > i and candidates[j] == candidates[j - 1]:
continue
t.append(candidates[j])
dfs(j + 1, s - candidates[j])
t.pop()
candidates.sort()
ans = []
t = []
dfs(0, target)
return ans
Complexity
The time complexity is , and the space complexity is . The space complexity is .
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.