Combination Sum II Leetcode Problem 40 [Python Solution]

The Combination Sum II LeetCode problem is a classic backtracking problem, specifically categorized under “Backtracking” with medium-difficulty.

It can be found under problem number 40. This problem is often associated with companies like Amazon.

Problem Statement:
Given a collection of candidate numbers (candidates) and a target number (target), the task is to find all unique combinations in candidates where the candidate numbers sum to the target.

However, there’s a crucial constraint to keep in mind: each number in the candidates may only be used once in the combination.

Additionally, the solution set must not contain duplicate combinations.

Example:

Let’s consider an example to better understand the problem:

Input:

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

Output:

[
    [1, 1, 6],
    [1, 2, 5],
    [1, 7],
    [2, 6]
]

In this example, the task is to find all combinations of numbers from the candidates list that sum up to the target value of 8. You can observe that there are multiple valid combinations, but duplicate combinations should be avoided.

Constraints:

• 1 <= candidates.length <= 100
• 1 <= candidates[i] <= 50
• 1 <= target <= 30

Now that we have a clear understanding of the problem statement, let’s dive into the solutions.

Efficient Python Code Solution

def combinationSum2(self, candidates: List[int], target: int) -> List[List[int]]:
        candidates.sort()  # Sort the candidates list
        result = []

        def backtrack(current, position, target):
            if target == 0:
                result.append(current.copy())
                return
            if target < 0:
                return

            prev = -1
            for i in range(position, len(candidates)):
                if candidates[i] == prev:
                    continue
                current.append(candidates[i])
                backtrack(current, i + 1, target - candidates[i])
                current.pop()
                prev = candidates[i]

        backtrack([], 0, target)
        return result

Now, let’s break down this solution and understand the reasoning behind it.

Reasoning Behind Our Approach

The key idea behind solving this problem efficiently is to use backtracking, which is a common technique for generating all possible solutions to a problem.

In this case, we want to find all combinations of numbers from the candidates list that sum up to the target value without duplicates.

Here’s the step-by-step explanation of our approach:

1. We start by sorting the candidates list.

Sorting helps us avoid duplicates and ensures that we process candidates in a specific order.

2. We define a result list to store the valid combinations that sum up to the target.
3. The heart of the solution is the backtrack function.

This function takes three parameters:

• current: This list keeps track of the current combination being explored.
• position: This is the starting index in the candidates list where we begin exploring candidates.
• target: This parameter represents the current target value we aim to achieve.

1. Inside the backtrack function, we have three main cases:

• If target becomes zero, we have found a valid combination.

We append a copy of the current combination to the result list.

• If target becomes negative, we have exceeded the target value, so we backtrack without considering this path.
• For all other cases, we iterate through the remaining candidates, starting from the position index.

We avoid duplicates by checking if the current candidate is the same as the previous candidate (prev) and continue to the next candidate if they are the same.

5. We then recursively call the backtrack function with updated parameters:

• We append the current candidate to the current combination.
• The position is updated to i + 1 to ensure that we don’t reuse the same candidate in the same combination.
• We subtract the current candidate from the target.

5. After the recursive call, we remove the last added candidate using current.pop() to backtrack and explore other possibilities.
6. Finally, we initiate the backtrack function with an empty current list, starting from index 0 in the candidates list, and the given target.

The result list stores all the valid combinations.

This approach ensures that we generate unique combinations without duplicates and avoids unnecessary calculations.

The time complexity of this solution is 2^n, where n is the number of candidates.

This is because, in the worst case, we explore all possible combinations.

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Conclusion

In this blog post, we’ve explored the Combination Sum II problem on LeetCode.

We’ve discussed the problem statement, constraints, and an efficient Python solution using backtracking.

This solution allows us to find all unique combinations of candidates that sum up to a given target while avoiding duplicate combinations.

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Additionally, if you have any questions, suggestions, or ideas, please feel free to share them in the comments section below.

Happy coding!

Question Link

That’s the solution to the Combination Sum II problem on LeetCode.

If you have any questions or need further clarification on any part of the solution, please feel free to ask.

Happy coding!