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Factor Combinations

#120 Most AskedMedium

Factor Combinations #120 Most Asked

Medium

Uber

28 views

Topics:

Recursion

Numbers can be regarded as the product of their factors.

For example, 8 = 2 x 2 x 2 = 2 x 4.

Given an integer n, return all possible combinations of its factors. You may return the answer in any order.

Note: The factors should be in the range [2, n - 1].

Example 1:

Input: n = 1
Output: []

Example 2:

Input: n = 12
Output: [[2,6],[2,2,3],[3,4]]
Explanation: 12 = 2 x 6 = 2 x 2 x 3 = 3 x 4.

Example 3:

Input: n = 37
Output: []

Constraints:

1 <= n <= 10⁷

Try coding it in LeetCode

Factor Combinations

#120 Most AskedMedium

Factor Combinations #120 Most Asked

Medium

Uber

28 views

Topics:

Recursion

Numbers can be regarded as the product of their factors.

For example, 8 = 2 x 2 x 2 = 2 x 4.

Given an integer n, return all possible combinations of its factors. You may return the answer in any order.

Note: The factors should be in the range [2, n - 1].

Example 1:

Input: n = 1
Output: []

Example 2:

Input: n = 12
Output: [[2,6],[2,2,3],[3,4]]
Explanation: 12 = 2 x 6 = 2 x 2 x 3 = 3 x 4.

Example 3:

Input: n = 37
Output: []

Constraints:

1 <= n <= 10⁷

Try coding it in LeetCode

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 is the maximum possible value of 'n'? I want to consider potential integer overflow issues or performance implications with very large numbers.
If 'n' is a prime number, or if its only factors are 1 and itself, should I return an empty list?
Is the order of factors within each combination important? For example, is [2, 3] the same as [3, 2]?
Is the order of the combinations within the list of lists important?
Are we looking for all possible factor combinations, or is there a specific requirement regarding the size or nature of the factors included (e.g., only prime factors)?

Brute Force Solution

Approach

The brute force approach to factor combinations involves exploring every possible way to break down a number into its factors. It's like trying every combination of ingredients to see what dishes you can make. We'll generate sets of factors and check if their product equals the original number.

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

Start by finding all the numbers that can divide evenly into the target number, excluding 1 and the number itself. These are potential factors.
Try using only one factor from the list and see if it multiplies to the target number (it won't, since we excluded the original number as a factor).
Next, try using two factors from the list. Multiply them together and see if the result equals the original number. If it does, you've found a valid combination.
Then, try using three factors, then four factors, and so on, checking each combination's product against the original number.
For each number of factors, consider all possible combinations of factors from the list you made in the first step. Make sure to avoid duplicates (e.g., 2 * 3 * 5 is the same as 5 * 2 * 3).
Keep a record of all the unique combinations of factors whose product equals the original number. These are your factor combinations.

Code Implementation

def get_factor_combinations_brute_force(number):
    factor_combinations = []
    factors = []

    # Find potential factors (excluding 1 and the number itself)
    for possible_factor in range(2, number):
        if number % possible_factor == 0:
            factors.append(possible_factor)

    def backtrack(current_combination, remaining_product, start_index):
        if remaining_product == 1:
            factor_combinations.append(current_combination[:])
            return

        if remaining_product < 1:
            return

        # Avoid duplicate combinations by only considering factors
        # from the current index onwards
        for i in range(start_index, len(factors)):
            factor = factors[i]
            if remaining_product % factor == 0:

                current_combination.append(factor)
                backtrack(current_combination, remaining_product % factor == 0 and remaining_product // factor, i)

                # Backtrack: Remove the last added factor to explore other combinations
                current_combination.pop()

    # Start the backtracking process
    backtrack([], number, 0)

    # Filter combinations to remove single element list that is the number itself
    filtered_combinations = [combo for combo in factor_combinations if len(combo) > 0]

    return filtered_combinations

Big(O) Analysis

Time Complexity

O(2^n) – The algorithm first finds potential factors, which can be up to n/2 in the worst case, where n is the input number. It then attempts to generate all possible combinations of these factors. Generating all combinations of a set of size k requires 2^k operations. Since k can be up to n/2, the number of combinations can be as high as 2^(n/2), which is still exponential in terms of n. Therefore, the dominant operation is generating and checking these combinations, making the overall time complexity O(2^n).

Space Complexity

O(N) – The algorithm uses a list to store combinations of factors. In the worst-case scenario, the number N is a prime number raised to a power, like 2^k. This leads to a maximum recursion depth and temporary list size related to the number of factors possible. Therefore, the space used to store the factor combinations can grow linearly with N, where N is the input number, leading to an auxiliary space complexity of O(N).

Optimal Solution

Approach

The core idea is to break down the given number into its factors and then recursively find all combinations of these factors that multiply back to the original number. We want to avoid redundant calculations and keep our solutions organized. Think of building factor combinations like a tree, where each branch represents a factor.

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

Start by finding all potential factors of the number. We only need to check up to the square root of the number.
Begin building combinations of factors by picking one factor at a time.
For each factor you pick, recursively find factor combinations of the remaining number (the original number divided by the current factor).
Make sure that the factors in each combination are in non-decreasing order to avoid duplicate combinations. For example, include 2 x 3 but not 3 x 2.
If the product of your chosen factors equals the original number, save that combination.
Continue exploring different branches of factor combinations until you've exhausted all possibilities.
Return all the valid factor combinations you've found.

Code Implementation

def get_factors(number): 
    result = []
    
    def find_factor_combinations(remaining_number, current_combination, start_factor): 
        if remaining_number == 1: 
            if len(current_combination) > 1:
                result.append(current_combination[:])
            return

        # Iterate from the start_factor to the square root.
        for factor in range(start_factor, int(remaining_number**0.5) + 1): 
            if remaining_number % factor == 0: 
                current_combination.append(factor)

                # Recursively find more factors.
                find_factor_combinations(remaining_number // factor, current_combination, factor)
                current_combination.pop()

        # Handle the case where the remaining number is prime or has no smaller factors.
        if remaining_number >= start_factor:
            current_combination.append(remaining_number)
            if len(current_combination) > 1:
                result.append(current_combination[:])
            current_combination.pop()

    find_factor_combinations(number, [], 2)
    return result

def factor_combinations(number):
    # Handle edge cases.
    if number <= 1: 
        return []

    # The outer function to initiate the recursive process.
    return get_factors(number)

Big(O) Analysis

Time Complexity

O(2^n) – Finding factor combinations involves exploring a decision tree where each node represents a choice of including or excluding a factor. In the worst case, where the number is a product of distinct prime numbers close to the square root of n (where n is the input number), the number of factors can be proportional to the square root of n. Recursively exploring all combinations of factors leads to an exponential number of possibilities, roughly 2 raised to the power of the number of factors. Therefore, the time complexity is O(2^sqrt(n)), which can be simplified to O(2^n) in the worst-case scenario where the number of factors grows linearly with n (although this is highly improbable).

Space Complexity

O(sqrt(N) + logN) – The space complexity is determined by the recursion depth and the space used to store the intermediate factor combinations. The maximum recursion depth can be proportional to the number of distinct prime factors, which is bounded by logN in the worst case (e.g., when N is a power of 2). Additionally, we need to store a temporary list of factors whose maximum size is related to the number of factors checked up to the square root of N. These factors are stored at each recursive call to keep track of intermediate combinations. Therefore, the overall space complexity is O(sqrt(N) + logN), where sqrt(N) represents the space for factors and logN represents the maximum recursion depth.

Edge Cases

Case	How to Handle
Input n is less than or equal to 1	Return an empty list since factors must be greater than 1 and less than n.
Input n is a prime number	Return an empty list since prime numbers have no factors other than 1 and themselves.
Input n is 2 or 3	Return an empty list since no factors exist between 1 and n.
Large input n leading to deep recursion	Optimize the solution to potentially use an iterative approach or memoization to avoid stack overflow errors.
Input n is a perfect square	Ensure the square root is considered as a potential factor and avoid duplicate combinations.
Input n has repeated prime factors (e.g., n = 8)	Ensure all combinations of repeated factors are generated correctly and without duplicates.
Integer overflow when calculating factors or products	Use appropriate data types (e.g., long) or check for potential overflows before calculations.
The order of factors within each combination should not matter, but the combinations themselves must be unique.	Sort each combination to guarantee only unique combinations get generated and stored.

Input n is less than or equal to 1

How to Handle:

Return an empty list since factors must be greater than 1 and less than n.

Input n is a prime number

How to Handle:

Return an empty list since prime numbers have no factors other than 1 and themselves.

Input n is 2 or 3

How to Handle:

Return an empty list since no factors exist between 1 and n.

Large input n leading to deep recursion

How to Handle:

Optimize the solution to potentially use an iterative approach or memoization to avoid stack overflow errors.

Input n is a perfect square

How to Handle:

Ensure the square root is considered as a potential factor and avoid duplicate combinations.

Input n has repeated prime factors (e.g., n = 8)

How to Handle:

Ensure all combinations of repeated factors are generated correctly and without duplicates.

Integer overflow when calculating factors or products

How to Handle:

Use appropriate data types (e.g., long) or check for potential overflows before calculations.

The order of factors within each combination should not matter, but the combinations themselves must be unique.

How to Handle:

Sort each combination to guarantee only unique combinations get generated and stored.

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 is the maximum possible value of 'n'? I want to consider potential integer overflow issues or performance implications with very large numbers.
If 'n' is a prime number, or if its only factors are 1 and itself, should I return an empty list?
Is the order of factors within each combination important? For example, is [2, 3] the same as [3, 2]?
Is the order of the combinations within the list of lists important?
Are we looking for all possible factor combinations, or is there a specific requirement regarding the size or nature of the factors included (e.g., only prime factors)?

Brute Force Solution

Approach

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

Start by finding all the numbers that can divide evenly into the target number, excluding 1 and the number itself. These are potential factors.
Try using only one factor from the list and see if it multiplies to the target number (it won't, since we excluded the original number as a factor).
Next, try using two factors from the list. Multiply them together and see if the result equals the original number. If it does, you've found a valid combination.
Then, try using three factors, then four factors, and so on, checking each combination's product against the original number.
For each number of factors, consider all possible combinations of factors from the list you made in the first step. Make sure to avoid duplicates (e.g., 2 * 3 * 5 is the same as 5 * 2 * 3).
Keep a record of all the unique combinations of factors whose product equals the original number. These are your factor combinations.

Code Implementation

def get_factor_combinations_brute_force(number):
    factor_combinations = []
    factors = []

    # Find potential factors (excluding 1 and the number itself)
    for possible_factor in range(2, number):
        if number % possible_factor == 0:
            factors.append(possible_factor)

    def backtrack(current_combination, remaining_product, start_index):
        if remaining_product == 1:
            factor_combinations.append(current_combination[:])
            return

        if remaining_product < 1:
            return

        # Avoid duplicate combinations by only considering factors
        # from the current index onwards
        for i in range(start_index, len(factors)):
            factor = factors[i]
            if remaining_product % factor == 0:

                current_combination.append(factor)
                backtrack(current_combination, remaining_product % factor == 0 and remaining_product // factor, i)

                # Backtrack: Remove the last added factor to explore other combinations
                current_combination.pop()

    # Start the backtracking process
    backtrack([], number, 0)

    # Filter combinations to remove single element list that is the number itself
    filtered_combinations = [combo for combo in factor_combinations if len(combo) > 0]

    return filtered_combinations

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

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

Start by finding all potential factors of the number. We only need to check up to the square root of the number.
Begin building combinations of factors by picking one factor at a time.
For each factor you pick, recursively find factor combinations of the remaining number (the original number divided by the current factor).
Make sure that the factors in each combination are in non-decreasing order to avoid duplicate combinations. For example, include 2 x 3 but not 3 x 2.
If the product of your chosen factors equals the original number, save that combination.
Continue exploring different branches of factor combinations until you've exhausted all possibilities.
Return all the valid factor combinations you've found.

Code Implementation

def get_factors(number): 
    result = []
    
    def find_factor_combinations(remaining_number, current_combination, start_factor): 
        if remaining_number == 1: 
            if len(current_combination) > 1:
                result.append(current_combination[:])
            return

        # Iterate from the start_factor to the square root.
        for factor in range(start_factor, int(remaining_number**0.5) + 1): 
            if remaining_number % factor == 0: 
                current_combination.append(factor)

                # Recursively find more factors.
                find_factor_combinations(remaining_number // factor, current_combination, factor)
                current_combination.pop()

        # Handle the case where the remaining number is prime or has no smaller factors.
        if remaining_number >= start_factor:
            current_combination.append(remaining_number)
            if len(current_combination) > 1:
                result.append(current_combination[:])
            current_combination.pop()

    find_factor_combinations(number, [], 2)
    return result

def factor_combinations(number):
    # Handle edge cases.
    if number <= 1: 
        return []

    # The outer function to initiate the recursive process.
    return get_factors(number)

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
Input n is less than or equal to 1	Return an empty list since factors must be greater than 1 and less than n.
Input n is a prime number	Return an empty list since prime numbers have no factors other than 1 and themselves.
Input n is 2 or 3	Return an empty list since no factors exist between 1 and n.
Large input n leading to deep recursion	Optimize the solution to potentially use an iterative approach or memoization to avoid stack overflow errors.
Input n is a perfect square	Ensure the square root is considered as a potential factor and avoid duplicate combinations.
Input n has repeated prime factors (e.g., n = 8)	Ensure all combinations of repeated factors are generated correctly and without duplicates.
Integer overflow when calculating factors or products	Use appropriate data types (e.g., long) or check for potential overflows before calculations.
The order of factors within each combination should not matter, but the combinations themselves must be unique.	Sort each combination to guarantee only unique combinations get generated and stored.

Input n is less than or equal to 1

How to Handle:

Return an empty list since factors must be greater than 1 and less than n.

Input n is a prime number

How to Handle:

Return an empty list since prime numbers have no factors other than 1 and themselves.

Input n is 2 or 3

How to Handle:

Return an empty list since no factors exist between 1 and n.

Large input n leading to deep recursion

How to Handle:

Optimize the solution to potentially use an iterative approach or memoization to avoid stack overflow errors.

Input n is a perfect square

How to Handle:

Ensure the square root is considered as a potential factor and avoid duplicate combinations.

Input n has repeated prime factors (e.g., n = 8)

How to Handle:

Ensure all combinations of repeated factors are generated correctly and without duplicates.

Integer overflow when calculating factors or products

How to Handle:

Use appropriate data types (e.g., long) or check for potential overflows before calculations.

The order of factors within each combination should not matter, but the combinations themselves must be unique.

How to Handle:

Sort each combination to guarantee only unique combinations get generated and stored.

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