Combination Sum III

Medium

Apple

2 views

Topics:

RecursionArrays

Given two integers k and n, find all valid combinations of k numbers that sum up to n such that the following conditions are true:

Only numbers 1 through 9 are used.
Each number is used at most once.

Return a list of all possible valid combinations. The list must not contain the same combination twice, and the combinations may be returned in any order.

For example:

If k = 3 and n = 7, the output should be [[1, 2, 4]] because 1 + 2 + 4 = 7 and there are no other valid combinations using 3 distinct numbers from 1 to 9.
If k = 3 and n = 9, the output should be [[1, 2, 6], [1, 3, 5], [2, 3, 4]] because 1 + 2 + 6 = 9, 1 + 3 + 5 = 9, and 2 + 3 + 4 = 9 and there are no other valid combinations.
If k = 4 and n = 1, the output should be [] because it's impossible to find 4 distinct numbers between 1 and 9 that sum to 1 (the smallest possible sum is 1 + 2 + 3 + 4 = 10).

Explain your approach, analyze the time and space complexity, and handle potential edge cases. Provide code for your solution.

Combination Sum III

Medium

Apple

2 views

Topics:

RecursionArrays

Given two integers k and n, find all valid combinations of k numbers that sum up to n such that the following conditions are true:

Only numbers 1 through 9 are used.
Each number is used at most once.

Return a list of all possible valid combinations. The list must not contain the same combination twice, and the combinations may be returned in any order.

For example:

If k = 3 and n = 7, the output should be [[1, 2, 4]] because 1 + 2 + 4 = 7 and there are no other valid combinations using 3 distinct numbers from 1 to 9.
If k = 3 and n = 9, the output should be [[1, 2, 6], [1, 3, 5], [2, 3, 4]] because 1 + 2 + 6 = 9, 1 + 3 + 5 = 9, and 2 + 3 + 4 = 9 and there are no other valid combinations.
If k = 4 and n = 1, the output should be [] because it's impossible to find 4 distinct numbers between 1 and 9 that sum to 1 (the smallest possible sum is 1 + 2 + 3 + 4 = 10).

Explain your approach, analyze the time and space complexity, and handle potential edge cases. Provide code for your solution.

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What are the constraints on `k` (the number of elements in each combination) and `n` (the target sum)? Are they within reasonable bounds such as `1 <= k <= 9` and `1 <= n <= 45`?
Are the numbers 1 through 9 the only candidates for combinations, or can I assume a different range of numbers?
If no valid combinations exist that sum up to `n` with `k` distinct numbers from 1 to 9, what should the function return? Should I return an empty list?
Are the combinations I return expected to be in a specific order, either lexicographically sorted or sorted by the numbers within each combination? How are they ordered relative to each other and within themselves?
Should I be concerned about integer overflow, given the constraints on `k` and `n`?

Brute Force Solution

Approach

Imagine you have the numbers 1 through 9, and you need to find all the unique groups of a specific size that add up to a target number. The brute force approach is to simply try every possible combination of those numbers. We check if each combination is the right size and adds up to the target, and if so, we keep it.

Here's how the algorithm would work step-by-step:

Start by picking the smallest possible combination: the first numbers available.
Check if that combination contains the required number of numbers.
Also check if the sum of those numbers equals the target number.
If both checks pass, save that combination.
Now, methodically go through every other possible combination of numbers, one at a time.
For each new combination, perform the same two checks: count the numbers and calculate the sum.
If a combination passes both checks, save it.
Continue until you've tried every possible combination of numbers from 1 to 9.
Finally, you'll have a list of all valid combinations that meet both requirements.

Code Implementation

def combination_sum_iii_brute_force(k, target):
    all_combinations = []

    def backtrack(combination, next_number):
        if len(combination) == k:
            if sum(combination) == target:
                all_combinations.append(combination.copy())
            return

        # Only go up to 9, the maximum number possible
        for number in range(next_number, 10):
            combination.append(number)

            # Recursively explore combinations
            backtrack(combination, number + 1)

            # Remove last element to backtrack and explore other possibilities
            combination.pop()

    # Start with an empty combination and the smallest possible number, 1
    backtrack([], 1)

    return all_combinations

Big(O) Analysis

Time Complexity

O(9 choose k) – The algorithm explores all possible combinations of k numbers chosen from the set {1, 2, 3, 4, 5, 6, 7, 8, 9}. The number of such combinations is given by the binomial coefficient '9 choose k', which represents the number of ways to choose k elements from a set of 9 elements without regard to order. The algorithm effectively iterates through each of these combinations to check if their sum equals n. Therefore the time complexity is directly proportional to the number of combinations, or O(9 choose k). Since '9' is constant the time complexity could be expressed as O(C(9,k)).

Space Complexity

O(k) – The algorithm stores valid combinations in a result list. In the worst-case scenario, many combinations of size k (where k is an input) might sum up to the target. Therefore, the space used for storing these combinations grows proportionally to the number of valid combinations found. Each combination is of size k, and the number of combinations is difficult to precisely determine without further constraints, but we can say that the storage of the intermediate 'combination' list during the recursive calls contributes O(k) space. While there may be multiple valid combinations, the space used for the combination list within the recursive calls dominates the auxiliary space complexity. The auxiliary space is primarily determined by the size of the combination list during the recursive calls, where the size of the list is at most k.

Optimal Solution

Approach

The goal is to find all combinations of 'k' distinct numbers between 1 and 9 that add up to 'n'. The optimal approach uses a method that explores possibilities in a structured way, avoiding redundant calculations by making choices and then undoing them (backtracking).

Here's how the algorithm would work step-by-step:

Start with an empty combination.
Consider the numbers from 1 to 9 one at a time.
For each number, decide whether to include it in the current combination.
If you include the number, add it to the combination and reduce the target sum accordingly.
If the combination now has 'k' numbers and the target sum is zero, you've found a valid combination.
Whether you included the number or not, backtrack: remove the number from the combination (if you added it) and explore the next number.
To avoid duplicate combinations, only consider numbers greater than the last number included in the combination.
Stop exploring a branch if the combination already has 'k' numbers or the target sum becomes negative.
Continue this process until all possible combinations have been explored.

Code Implementation

def combinationSum3(k, n):
    results = []
    def backtrack(combination, remaining_number, start_number):
        # If the combination is valid, add it to the result.
        if len(combination) == k and remaining_number == 0:
            results.append(combination.copy())
            return

        # If the combination cannot be valid, stop exploring this branch.
        if len(combination) == k or remaining_number < 0:
            return

        for number in range(start_number, 10):
            # Explore including the current number in the combination.
            combination.append(number)
            backtrack(combination, remaining_number - number, number + 1)

            # Backtrack by removing the current number.
            combination.pop()

    backtrack([], n, 1)
    return results

Big(O) Analysis

Time Complexity

O(C(9,k)) – The backtracking algorithm explores the combinations of k numbers chosen from the set {1, 2, ..., 9}. In the worst-case scenario, it may need to generate and check all possible combinations. The number of such combinations is given by the binomial coefficient C(9, k), which represents "9 choose k". While the recursion depth is limited by 'k' and each call involves constant-time operations, the number of calls is dictated by the number of possible valid combinations. Thus, the time complexity is dominated by the number of combinations considered, resulting in O(C(9, k)).

Space Complexity

O(k) – The primary space usage comes from the recursion stack and the temporary list to store the current combination. In the worst-case scenario, the depth of the recursion can reach 'k' (the number of elements in a combination), leading to O(k) space for the call stack. Additionally, a list is used to store the combination of up to 'k' numbers, also contributing O(k) space. Therefore, the overall auxiliary space complexity is dominated by O(k).

Edge Cases

Case	How to Handle
k is zero or negative	Return an empty list since no combination of zero or negative size is possible.
n is zero or negative	Return an empty list since no combination can sum to a non-positive target.
k is greater than 9	Return an empty list because the numbers can only range from 1 to 9.
n is less than the sum of the smallest k distinct numbers (1+2+...+k)	Return an empty list as the sum cannot be reached.
n is greater than the sum of the largest k distinct numbers (9+8+...+(10-k))	Return an empty list as the sum cannot be reached.
k is 1 and n is greater than 9	Return an empty list, since only numbers from 1 to 9 can be used.
k is 1 and n is between 1 and 9	Return a list containing a single list with the number n if n is between 1 and 9.
Large values of k and n that lead to deep recursion stacks	Ensure efficient recursion by using tail call optimization if the language supports it, or consider an iterative approach if stack overflow becomes a concern.