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Integer Break

Medium

Apple

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Topics:

Dynamic ProgrammingGreedy Algorithms

Given an integer n, break it into the sum of k positive integers, where k >= 2, and maximize the product of those integers.

Return the maximum product you can get.

For example:

If n = 2, the output should be 1 because 2 = 1 + 1 and 1 * 1 = 1.
If n = 10, the output should be 36 because 10 = 3 + 3 + 4 and 3 * 3 * 4 = 36.

Can you provide an algorithm to solve this problem efficiently? Consider the constraints: 2 <= n <= 58.

Integer Break

Medium

Apple

1 view

Topics:

Dynamic ProgrammingGreedy Algorithms

Given an integer n, break it into the sum of k positive integers, where k >= 2, and maximize the product of those integers.

Return the maximum product you can get.

For example:

If n = 2, the output should be 1 because 2 = 1 + 1 and 1 * 1 = 1.
If n = 10, the output should be 36 because 10 = 3 + 3 + 4 and 3 * 3 * 4 = 36.

Can you provide an algorithm to solve this problem efficiently? Consider the constraints: 2 <= n <= 58.

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 range of the input integer n?
Is n guaranteed to be greater than 1, as the problem states it must be broken into at least two integers?
If n is small (e.g., 2 or 3), are there specific requirements for the return value, given that we need to break it into *at least* two integers?
Are we looking for the mathematically *absolute* maximum product, or is there any tolerance for slightly suboptimal solutions if they arise from an approximation or simpler approach?
Are there any constraints or preferences on the integers we use to break down n (e.g., should we prioritize using 3s over 2s)? Specifically, is there a penalty for using '1' as one of the integers in the breakdown?

Brute Force Solution

Approach

The integer break problem asks us to find the largest product we can get by breaking up an integer into smaller integers. The brute force approach tries every single possible way to break up the integer and then calculates the product for each of those ways.

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

Consider breaking the number into two parts, then three parts, then four parts, and so on, all the way up to breaking it into as many parts as possible.
For each number of parts, consider every possible combination of those parts. For example, if we're breaking the number into two parts, we need to try all possible values for the first part, and the second part will be whatever's left over.
For each combination of parts, calculate the product of those parts.
Keep track of the largest product we've seen so far.
After we've tried every single possible combination of parts for every possible number of parts, the largest product we've kept track of will be our answer.

Code Implementation

def integer_break_brute_force(number):    maximum_product = 0

    # Iterate through all possible numbers of parts to break the integer into
    for number_of_parts in range(2, number + 1):
        maximum_product = find_max_product(number, number_of_parts, maximum_product)
    return maximum_product

def find_max_product(number, number_of_parts, maximum_product):
    #Iterate through all combinations of parts, calculating product and updating the max
    return find_all_combinations(number, number_of_parts, [], maximum_product)

def find_all_combinations(number, number_of_parts, current_combination, maximum_product):
    if number_of_parts == 1:
        current_combination.append(number)
        product = 1
        for part in current_combination:
            product *= part
        maximum_product = max(maximum_product, product)
        current_combination.pop()
        return maximum_product
    
    #Try all possible values for the first part of the combination
    for first_part in range(1, number - number_of_parts + 2):
        current_combination.append(first_part)

        #Recursively find combinations for the remaining parts
        maximum_product = find_all_combinations(number - first_part, number_of_parts - 1, current_combination, maximum_product)
        current_combination.pop()

    return maximum_product

Big(O) Analysis

Time Complexity

O(n^n) – The brute force approach considers breaking the input integer n into 2 parts, 3 parts, up to n parts. For each case of k parts, where k can range from 2 to n, the number of combinations to check grows exponentially. Essentially, for each part, we explore various possible sizes, leading to a nested consideration of almost n choices for each of the parts. Therefore, the total number of combinations checked will approach O(n*n*n... n times) which approximates O(n^n). This represents the worst-case scenario where we need to explore almost every possible combination of integer breakdowns.

Space Complexity

O(N^N) – The brute force approach outlined considers breaking the number N into parts, and for each possible number of parts (from 2 to N), it explores all possible combinations. To generate these combinations, implicit data structures like lists to hold each combination of parts are needed. The maximum number of combinations grows exponentially with the number of parts and the value of each part, leading to a substantial memory footprint. The dominant factor contributing to the space complexity is the storage of these intermediate combinations, which can be approximated by O(N^N) in the worst-case scenario when all combinations need to be stored at the same time.

Optimal Solution

Approach

The most efficient way to maximize the product is not to exhaustively try every split. Instead, we focus on breaking down the number into parts that are mostly 3s, with adjustments for small numbers to guarantee the highest possible product.

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

If the number is small (like 2 or 3), handle it directly because breaking it down further only hurts the product.
For larger numbers, repeatedly break it down into as many 3s as possible because 3 usually contributes the most to the product.
If, after repeatedly taking out 3s, you are left with a 1, combine one of the 3s to make a 4. A factor of 4 (2x2) gives better results than 3 and 1.
If, instead, you're left with a 2 after taking out all the 3s, just keep it as 2 because 2 contributes more to the product than breaking it down further.
Multiply all the parts together to get the maximum possible product.

Code Implementation

def integer_break(number):    if number <= 3:
        return max(1, number - 1)

    product = 1

    # Maximize 3s, they usually yield the highest product
    while number > 4:
        product *= 3
        number -= 3

    # If number is 4, we multiply it to the product
    product *= number

    return product

Big(O) Analysis

Time Complexity

O(n) – The algorithm repeatedly subtracts 3 from the input number n until the remaining value is less than 4. The number of subtractions is proportional to n/3. Since only a single loop performs this operation, the time complexity is directly proportional to the input number n, resulting in O(n) time complexity. There are no nested loops or recursive calls contributing to a higher order of complexity. The final multiplication step takes constant time and does not impact the overall time complexity.

Space Complexity

O(1) – The described solution does not use any auxiliary data structures such as arrays, hash maps, or lists. It only performs arithmetic operations and variable assignments directly on the input number. Therefore, the space required remains constant regardless of the input value N. Consequently, the auxiliary space complexity is O(1).

Edge Cases

Case	How to Handle
n = 1	Problem states n is a positive integer, and must be broken into at least two integers, so n=1 is impossible to break, however, we can assume the problem means n > 1; specifically the prompt does not address this case so we can throw an IllegalArgumentException, or return a default (e.g. 0), and clarify assumptions with the interviewer.
n = 2	Return 1 because the only possible break is 1+1, and 1*1 = 1; this is a base case that must be explicitly handled, as other logic might fail.
n = 3	Return 2 because the only possible break is 1+2, and 1*2 = 2; this is another base case that must be explicitly handled.
Large n (e.g., n = 50)	The solution should be optimized to avoid exponential time complexity, potentially using dynamic programming to store and reuse results to improve performance for larger values of n.
Integer overflow in multiplication	The product of the integers could potentially exceed the maximum value of an integer, so consider using long data type to store intermediate products to prevent overflow, though the constraints might limit this occurrence.
Negative n	Throw IllegalArgumentException since n must be a positive integer according to the problem definition.
n is prime number	The algorithm should still work correctly, breaking the prime number into smaller components that maximize the product according to the general logic.
n has multiple possible optimal decompositions	The algorithm should correctly find one of the optimal decompositions, as the problem only asks for the maximum product, not the specific decomposition itself.

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 range of the input integer n?
Is n guaranteed to be greater than 1, as the problem states it must be broken into at least two integers?
If n is small (e.g., 2 or 3), are there specific requirements for the return value, given that we need to break it into *at least* two integers?
Are we looking for the mathematically *absolute* maximum product, or is there any tolerance for slightly suboptimal solutions if they arise from an approximation or simpler approach?
Are there any constraints or preferences on the integers we use to break down n (e.g., should we prioritize using 3s over 2s)? Specifically, is there a penalty for using '1' as one of the integers in the breakdown?

Brute Force Solution

Approach

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

Consider breaking the number into two parts, then three parts, then four parts, and so on, all the way up to breaking it into as many parts as possible.
For each number of parts, consider every possible combination of those parts. For example, if we're breaking the number into two parts, we need to try all possible values for the first part, and the second part will be whatever's left over.
For each combination of parts, calculate the product of those parts.
Keep track of the largest product we've seen so far.
After we've tried every single possible combination of parts for every possible number of parts, the largest product we've kept track of will be our answer.

Code Implementation

def integer_break_brute_force(number):    maximum_product = 0

    # Iterate through all possible numbers of parts to break the integer into
    for number_of_parts in range(2, number + 1):
        maximum_product = find_max_product(number, number_of_parts, maximum_product)
    return maximum_product

def find_max_product(number, number_of_parts, maximum_product):
    #Iterate through all combinations of parts, calculating product and updating the max
    return find_all_combinations(number, number_of_parts, [], maximum_product)

def find_all_combinations(number, number_of_parts, current_combination, maximum_product):
    if number_of_parts == 1:
        current_combination.append(number)
        product = 1
        for part in current_combination:
            product *= part
        maximum_product = max(maximum_product, product)
        current_combination.pop()
        return maximum_product
    
    #Try all possible values for the first part of the combination
    for first_part in range(1, number - number_of_parts + 2):
        current_combination.append(first_part)

        #Recursively find combinations for the remaining parts
        maximum_product = find_all_combinations(number - first_part, number_of_parts - 1, current_combination, maximum_product)
        current_combination.pop()

    return maximum_product

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

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

If the number is small (like 2 or 3), handle it directly because breaking it down further only hurts the product.
For larger numbers, repeatedly break it down into as many 3s as possible because 3 usually contributes the most to the product.
If, after repeatedly taking out 3s, you are left with a 1, combine one of the 3s to make a 4. A factor of 4 (2x2) gives better results than 3 and 1.
If, instead, you're left with a 2 after taking out all the 3s, just keep it as 2 because 2 contributes more to the product than breaking it down further.
Multiply all the parts together to get the maximum possible product.

Code Implementation

def integer_break(number):    if number <= 3:
        return max(1, number - 1)

    product = 1

    # Maximize 3s, they usually yield the highest product
    while number > 4:
        product *= 3
        number -= 3

    # If number is 4, we multiply it to the product
    product *= number

    return product

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
n = 1	Problem states n is a positive integer, and must be broken into at least two integers, so n=1 is impossible to break, however, we can assume the problem means n > 1; specifically the prompt does not address this case so we can throw an IllegalArgumentException, or return a default (e.g. 0), and clarify assumptions with the interviewer.
n = 2	Return 1 because the only possible break is 1+1, and 1*1 = 1; this is a base case that must be explicitly handled, as other logic might fail.
n = 3	Return 2 because the only possible break is 1+2, and 1*2 = 2; this is another base case that must be explicitly handled.
Large n (e.g., n = 50)	The solution should be optimized to avoid exponential time complexity, potentially using dynamic programming to store and reuse results to improve performance for larger values of n.
Integer overflow in multiplication	The product of the integers could potentially exceed the maximum value of an integer, so consider using long data type to store intermediate products to prevent overflow, though the constraints might limit this occurrence.
Negative n	Throw IllegalArgumentException since n must be a positive integer according to the problem definition.
n is prime number	The algorithm should still work correctly, breaking the prime number into smaller components that maximize the product according to the general logic.
n has multiple possible optimal decompositions	The algorithm should correctly find one of the optimal decompositions, as the problem only asks for the maximum product, not the specific decomposition itself.