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Integer Break

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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.

Example 1:

Input: n = 2
Output: 1
Explanation: 2 = 1 + 1, 1 × 1 = 1.

Example 2:

Input: n = 10
Output: 36
Explanation: 10 = 3 + 3 + 4, 3 × 3 × 4 = 36.

Constraints:

2 <= n <= 58

Try coding it in LeetCode

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.

Example 1:

Input: n = 2
Output: 1
Explanation: 2 = 1 + 1, 1 × 1 = 1.

Example 2:

Input: n = 10
Output: 36
Explanation: 10 = 3 + 3 + 4, 3 × 3 × 4 = 36.

Constraints:

2 <= n <= 58

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 range of the input integer n? What are the upper and lower bounds?
The problem states that n must be broken into the sum of at least two positive integers. What should I return if n is less than or equal to 1?
Can you provide a small example, such as n=4, and the corresponding breakdown and product to ensure my understanding is correct?
Are we expected to return an integer or a long if the product becomes very large?
Is there a specific time or space complexity I should aim for beyond the general expectation of optimality?

Brute Force Solution

Approach

The brute force approach to this problem involves exploring every possible way to break the given number into smaller numbers. For each possible break, we calculate the product of those numbers. We then keep track of the largest product we find across all the possible breaks.

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

Consider all possible ways to split the number into at least two smaller positive integers.
For each split, calculate the product of those integers.
Compare the product to the current largest product found so far.
If the current product is larger, update the largest product.
Repeat this process, trying out all possible combinations of smaller integers until every combination is tried.
Once all combinations are tested, return the largest product that was found.

Code Implementation

def integer_break_brute_force(number):
    maximum_product = 0

    def explore_breaks(current_number, current_product, numbers_list):
        nonlocal maximum_product

        # Check if we've fully broken down the number.
        if current_number == 0:
            if len(numbers_list) >= 2:

                # Update max_product if necessary.
                maximum_product = max(maximum_product, current_product)
            return

        # Iterate through all possible numbers to subtract.
        for integer in range(1, number + 1):
            if integer <= current_number:

                # Explore this break by subtracting it.
                explore_breaks(current_number - integer, current_product * integer, numbers_list + [integer])

    explore_breaks(number, 1, [])
    return maximum_product

Big(O) Analysis

Time Complexity

O(2^n) – The brute force approach explores all possible ways to break the integer n into at least two positive integers. In the worst case, this means we consider all possible combinations of integers that sum up to n. The number of these combinations grows exponentially with n. Specifically, each number from 1 to n-1 can either be included in a split or not, leading to approximately 2^(n-1) possibilities. Therefore, the time complexity of exploring all splits is O(2^n).

Space Complexity

–

Optimal Solution

Approach

The problem asks how to break an integer into smaller integers to maximize their product. The key idea is that breaking the number into as many 3s as possible tends to yield the highest product. We leverage this property for an efficient solution.

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

If the given integer is 2, the best you can do is return 1, as you cannot break it into smaller positive integers.
If the given integer is 3, the best you can do is return 2, breaking it into 1 and 2.
For integers larger than 3, repeatedly break off as many 3s as possible.
If the remaining value is 0 after breaking off all the 3s, the product is just 3 raised to the power of how many 3s you broke off.
If the remaining value is 1, take one of the 3s back and combine it with the 1 to make 4, and use 2 and 2 instead, as 2*2 is greater than 3*1. Then multiply the result by 3 raised to the power of (number of 3s broken off -1).
If the remaining value is 2, multiply this 2 with all the 3s to get the maximum product.

Code Implementation

def integer_break(number):
    if number == 2:
        return 1

    if number == 3:
        return 2

    number_of_threes = 0
    while number > 4:
        number -= 3
        number_of_threes += 1

    # The remaining number is either 2, 3, or 4.
    if number == 2:
        return 3 ** number_of_threes * 2

    # The remaining number is 3.
    if number == 3:
        return 3 ** number_of_threes * 3

    # This case handles the scenario where the number is 4.
    return 3 ** number_of_threes * 4

Big(O) Analysis

Time Complexity

O(n) – The algorithm's runtime is dominated by the number of times we can subtract 3 from the input integer n. The while loop or equivalent logic reduces n by 3 in each iteration until n is less than 4. The number of iterations is therefore proportional to n/3. Since the loop's body performs constant-time operations (multiplication, subtraction, comparison), the overall time complexity is O(n/3), which simplifies to O(n).

Space Complexity

O(1) – The algorithm uses a constant amount of extra space. It only stores a few integer variables to keep track of the number of 3s and the remaining value after breaking down the input integer n. No auxiliary data structures like arrays, lists, or hash maps are created. Therefore, the space used does not depend on the input size n, resulting in O(1) space complexity.

Edge Cases

Case	How to Handle
Input n = 1	Since we need to break n into at least two positive integers, n=1 is invalid, but per description, n will be > 1; return 0 if still encountered to indicate an error state.
Input n = 2	The only possible breakdown is 1 + 1, so return 1 as 1*1.
Input n = 3	The only possible breakdown is 1 + 2, so return 2 as 1*2.
Large input n (e.g., n = 50)	Ensure the solution (e.g., dynamic programming) is efficient and avoids excessive recursion depth to prevent stack overflow or timeouts.
Integer overflow when calculating the product	If using languages like Java/C++, carefully choose the data type (e.g., long) to store the intermediate products to avoid overflow.
Negative n	The problem description states n is a positive integer, so return an error value (e.g. -1 or throw an exception) or simply return 0 to denote an error state.
n is close to INT_MAX	Be wary of integer overflows, especially when multiplying the decomposed numbers; use long long if necessary to prevent integer overflow.
Optimal decomposition involves many 3s	The core logic should correctly handle scenarios where the optimal decomposition heavily favors breaking down into as many 3s as possible; verify this is handled efficiently in the implemented approach.

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? What are the upper and lower bounds?
The problem states that n must be broken into the sum of at least two positive integers. What should I return if n is less than or equal to 1?
Can you provide a small example, such as n=4, and the corresponding breakdown and product to ensure my understanding is correct?
Are we expected to return an integer or a long if the product becomes very large?
Is there a specific time or space complexity I should aim for beyond the general expectation of optimality?

Brute Force Solution

Approach

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

Consider all possible ways to split the number into at least two smaller positive integers.
For each split, calculate the product of those integers.
Compare the product to the current largest product found so far.
If the current product is larger, update the largest product.
Repeat this process, trying out all possible combinations of smaller integers until every combination is tried.
Once all combinations are tested, return the largest product that was found.

Code Implementation

def integer_break_brute_force(number):
    maximum_product = 0

    def explore_breaks(current_number, current_product, numbers_list):
        nonlocal maximum_product

        # Check if we've fully broken down the number.
        if current_number == 0:
            if len(numbers_list) >= 2:

                # Update max_product if necessary.
                maximum_product = max(maximum_product, current_product)
            return

        # Iterate through all possible numbers to subtract.
        for integer in range(1, number + 1):
            if integer <= current_number:

                # Explore this break by subtracting it.
                explore_breaks(current_number - integer, current_product * integer, numbers_list + [integer])

    explore_breaks(number, 1, [])
    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 given integer is 2, the best you can do is return 1, as you cannot break it into smaller positive integers.
If the given integer is 3, the best you can do is return 2, breaking it into 1 and 2.
For integers larger than 3, repeatedly break off as many 3s as possible.
If the remaining value is 0 after breaking off all the 3s, the product is just 3 raised to the power of how many 3s you broke off.
If the remaining value is 1, take one of the 3s back and combine it with the 1 to make 4, and use 2 and 2 instead, as 2*2 is greater than 3*1. Then multiply the result by 3 raised to the power of (number of 3s broken off -1).
If the remaining value is 2, multiply this 2 with all the 3s to get the maximum product.

Code Implementation

def integer_break(number):
    if number == 2:
        return 1

    if number == 3:
        return 2

    number_of_threes = 0
    while number > 4:
        number -= 3
        number_of_threes += 1

    # The remaining number is either 2, 3, or 4.
    if number == 2:
        return 3 ** number_of_threes * 2

    # The remaining number is 3.
    if number == 3:
        return 3 ** number_of_threes * 3

    # This case handles the scenario where the number is 4.
    return 3 ** number_of_threes * 4

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
Input n = 1	Since we need to break n into at least two positive integers, n=1 is invalid, but per description, n will be > 1; return 0 if still encountered to indicate an error state.
Input n = 2	The only possible breakdown is 1 + 1, so return 1 as 1*1.
Input n = 3	The only possible breakdown is 1 + 2, so return 2 as 1*2.
Large input n (e.g., n = 50)	Ensure the solution (e.g., dynamic programming) is efficient and avoids excessive recursion depth to prevent stack overflow or timeouts.
Integer overflow when calculating the product	If using languages like Java/C++, carefully choose the data type (e.g., long) to store the intermediate products to avoid overflow.
Negative n	The problem description states n is a positive integer, so return an error value (e.g. -1 or throw an exception) or simply return 0 to denote an error state.
n is close to INT_MAX	Be wary of integer overflows, especially when multiplying the decomposed numbers; use long long if necessary to prevent integer overflow.
Optimal decomposition involves many 3s	The core logic should correctly handle scenarios where the optimal decomposition heavily favors breaking down into as many 3s as possible; verify this is handled efficiently in the implemented approach.