How do you solve Combination Sum II on LeetCode?

Combination Sum II (LeetCode #40) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n * 2^n) time complexity and O(n) space complexity. Key insight: Sorting the candidates helps in skipping duplicates efficiently.

What is the brute force solution for Combination Sum II?

The brute force approach for Combination Sum II is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves generating all possible combinations of the candidate numbers and checking if their sums equal the target. This method is straightforward but inefficient due to the exponential number of combinations. The optimal Optimal Solution approach improves this to O(n * 2^n).

What algorithm or data structure is used to solve Combination Sum II?

Combination Sum II uses the following concepts: Array, Backtracking. The recommended patterns to study are: Backtracking, Sorting.

What are the interview tips for Combination Sum II?

Always clarify if duplicates are allowed in combinations. Explain your thought process while coding to show your understanding. Practice backtracking problems to become familiar with the recursive approach.

What are common mistakes when solving Combination Sum II?

Failing to sort the array, leading to duplicate combinations. Not handling the case where a candidate exceeds the remaining target.

#40

Combination Sum II

Medium

ArrayBacktrackingBacktrackingSorting

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n * 2^n)
Space	O(1)	O(n)

💡

Intuition

Time O(n * 2^n)Space O(n)

The optimal solution uses backtracking to build combinations incrementally. By sorting the candidates and skipping duplicates, we efficiently explore valid combinations without generating all possible ones.

⚙️

Algorithm

4 steps

1Step 1: Sort the candidates array to facilitate skipping duplicates.
2Step 2: Use a backtracking function to explore combinations, adding elements to the current combination.
3Step 3: If the sum of the current combination equals the target, add it to the result.
4Step 4: Skip duplicates by checking if the current element is the same as the previous one.

solution.py15 lines

1def combinationSum2(candidates, target):
2    candidates.sort()
3    result = []
4    def backtrack(start, path, target):
5        if target == 0:
6            result.append(path)
7            return
8        for i in range(start, len(candidates)):
9            if i > start and candidates[i] == candidates[i - 1]:
10                continue
11            if candidates[i] > target:
12                break
13            backtrack(i + 1, path + [candidates[i]], target - candidates[i])
14    backtrack(0, [], target)
15    return result

ℹ

Complexity note: The time complexity is O(n * 2^n) due to the nature of backtracking, where each candidate can either be included or excluded. The space complexity is O(n) for the recursion stack and the result storage.

1Sorting the candidates helps in skipping duplicates efficiently.
2Backtracking allows us to explore combinations without generating all possible ones.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.