# Solving Leetcode Interviews in Seconds with AI: Combination Sum II


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "40" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like [Chatmagic](https://www.chatmagic.app), we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

	# 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. Note: The solution set must not contain duplicate combinations.   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 1 <= target <= 30  

	# Explanation
	Here's a breakdown of the approach, complexity, and the Python code for solving the combination sum II problem:

*   **Key Approach:**
    *   **Sorting:** Sort the candidates to easily skip duplicate combinations.
    *   **Backtracking:** Use a recursive backtracking approach to explore all possible combinations.
    *   **Skipping Duplicates:** Within each recursive call, skip over consecutive duplicate numbers in the `candidates` array to avoid generating duplicate combinations in the result.

*   **Complexity:**
    *   **Runtime Complexity:** O(2<sup>n</sup>) in the worst case, where 'n' is the number of candidates. This is because, in the worst-case scenario, each number could either be included or excluded in a combination.
    *   **Storage Complexity:** O(n) due to the recursion depth and the space used to store the intermediate combinations. The sorting operation might add O(log n) space complexity.

	
	# Code
	```python
	def combinationSum2(candidates, target):
    result = []
    candidates.sort()  # Sort to handle duplicates efficiently

    def backtrack(combination, remaining, start):
        if remaining == 0:
            result.append(combination.copy())
            return
        elif remaining < 0:
            return

        for i in range(start, len(candidates)):
            # Skip duplicate numbers to avoid duplicate combinations
            if i > start and candidates[i] == candidates[i - 1]:
                continue

            combination.append(candidates[i])
            backtrack(combination, remaining - candidates[i], i + 1)  # Use each number only once
            combination.pop()  # Backtrack

    backtrack([], target, 0)
    return result
	```