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Leetcode Problem 216, “Combination Sum III”, requires you to find all possible combinations of k numbers that add up to a number n, given that only numbers from 1 to 9 can be used and each combination should be a unique set of numbers.

A combination is considered unique if it contains numbers in a different sequence or different sets of numbers. Given these constraints, ensure that your solution does not include duplicate combinations and utilizes the numbers 1 through 9 only once in each combination.

Return the list of all possible combinations.

Clarifying Questions

  1. Are the numbers in a combination required to be in any particular order?
    • No, the unique combinations should be reported regardless of the order.
  2. Can a number be used more than once in the same combination?
    • No, each number from 1 to 9 can be used only once in each combination.
  3. Is the order of the combinations in the output important?
    • No, the order of the combinations in the output does not matter.
  4. What should be returned if no combination is found?
    • An empty list should be returned if no valid combination is found.

Let’s proceed with the strategy for solving this problem.

Strategy

  1. Backtracking Approach:
    • We will use backtracking to generate all possible combinations of k numbers from 1 to 9.
    • We need to keep track of the current combination and the sum of its elements.
    • If the length of the current combination is k and the sum is n, we add the combination to the results.
    • If the sum exceeds n or the length exceeds k, we backtrack.
  2. Optimize Early Stopping:
    • If at any point the sum exceeds n, there is no point in exploring further as adding more numbers will only increase the sum.

Code

Here’s the Python implementation of the above strategy:

def combinationSum3(k, n):
    def backtrack(start, comb, remaining_sum):
        if len(comb) == k:
            if remaining_sum == 0:
                result.append(list(comb))
            return
        for i in range(start, 10):
            if i > remaining_sum:
                break  # No need to continue if i exceeds remaining sum
            comb.append(i)
            backtrack(i + 1, comb, remaining_sum - i)
            comb.pop()  # Backtrack to explore the next number
    
    result = []
    backtrack(1, [], n)
    return result

# Example usage:
print(combinationSum3(3, 7))  # Output: [[1, 2, 4]]
print(combinationSum3(3, 9))  # Output: [[1, 2, 6], [1, 3, 5], [2, 3, 4]]

Time Complexity

The time complexity of this backtracking approach is difficult to express exactly due to the branching factor. In the worst case, it explores a large part of the power set of the numbers [1-9], leading towards O(2^N) where N is 9. However, since we prune the search space by limiting the sum and the combination length, the practical performance is significantly better than the worst case in most scenarios.

Complexity Breakdown:

Feel free to ask further questions or run through specific test cases if needed.

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