Integer Break

class Solution {
public:
    int breakInteger(int number) {
        if (number <= 3) {
            return number - 1;
        }
        
        if (number % 3 == 0) {
            return pow(3, number / 3);
        }
        
        if (number % 3 == 1) {
            return pow(3, (number / 3) - 1) * 4;
        }
        
        return pow(3, number / 3) * 2;
    }
};

The "Integer Break" problem involves breaking an integer into the maximum product of integers that add up to the original number. The solution is implemented in C++.

• Start by handling the case for numbers less than or equal to 3. For these cases, directly return the number minus one.
• For numbers greater than 3, check the remainder when the number is divided by 3:
  • If the remainder is 0, the number is a perfect multiple of 3. Return the result of 3 raised to the power of the quotient of the number divided by 3.
  • If the remainder is 1, multiplying by 3 repeatedly would leave one as a factor, which is inefficient. Instead, use one less grouping of 3 (thus dividing the number by 3 and then subtracting one group), multiply these 3s together, and then multiply by 4 for optimization.
  • If the remainder is 2, use the quotient of the number divided by 3 groups of three, and multiply by 2 for the remainder.

This optimized approach leverages the mathematical property that the product is maximum when the number is broken down into factors of 3, or a factor of 2 if a remainder of 2 exists. The use of pow function helps efficiently calculate powers of 3, required for the multiplicative results.

class Solution {
    public int breakInteger(int number) {
        if (number <= 3) {
            return number - 1;
        }
        
        if (number % 3 == 0) {
            return (int) Math.pow(3, number / 3);
        }
        
        if (number % 3 == 1) {
            return (int) Math.pow(3, (number / 3 - 1)) * 4;
        }
        
        return (int) Math.pow(3, number / 3) * 2;
    }
}

The given Java solution addresses the problem of breaking an integer into the maximum product of its parts. The function breakInteger computes the maximum product based on the condition of the input integer. Here is how the solution works:

1. Check if the number is 3 or less.

   • Return number - 1, as breaking down numbers less than or equal to 3 does not provide a product larger than the number itself minus one.

2. Calculate when the number is perfectly divisible by 3.

   • Return the result of 3 raised to the power of the quotient of number divided by 3. This utilizes the mathematical function Math.pow.

3. Address the scenario when dividing the number by 3 gives a remainder of 1.

   • In this case, adjust by reducing one grouping of 3 and multiplying by 4 to maximize the product. This adjustment is performed as multiplying by 4 (2x2) is more optimal than splitting into 1 and 3 after division.

4. Handle the remainder of 2 when number is divided by 3.

   • Multiply the product calculated as 3 raised to the power of the quotient by 2 for the most optimal product benefit.

With this methodology, the code efficiently calculates the maximum product obtainable by breaking an integer into the sum of at least two positive integers, using properties of exponents and multiplicative adjustments based on the division remainder.

class Solution:
    def breakInteger(self, num: int) -> int:
        if num <= 3:
            return num - 1
        
        if num % 3 == 0:
            return 3 ** (num // 3)
        
        if num % 3 == 1:
            return 4 * (3 ** (num // 3 - 1))
        
        return 2 * (3 ** (num // 3))

The provided Python solution addresses the Integer Break problem, where the goal is to maximize the product of numbers that sum up to a given integer num. The function breakInteger implements the logic to determine the maximum product using mathematical properties of numbers, particularly focusing on how they can be distributed in terms of their multiplication to yield larger values.

In the solution:

• Initially, it checks if the number is less than or equal to 3, returning num - 1 because for numbers 2 and 3, the maximum product obtained by breaking the number is 1 and 2 respectively.

• For numbers greater than 3, the solution uses properties of the number 3. If num is divisible by 3, the product of triplets of 3s (3 ** (num // 3)) is calculated.

• If num modulo 3 equals 1, multiplying by 4 after reducing one set of 3 gives a higher product. Hence, it returns 4 * (3 ** (num // 3 - 1)).

• If num modulo 3 equals 2, then simply multiplying by 2 to the power of the number of times 3 perfectly divides num gives the maximum product, so 2 * (3 ** (num // 3)) is returned.

This solution leverages exponentiation and modular arithmetic effectively to address the problem and is optimized through direct arithmetic computation avoiding the need for iteration or recursion, resulting in efficient runtime for large inputs.