Solution Explanation for Integer Break

The problem asks to find the maximum product obtained by breaking a given integer n into a sum of at least two positive integers.

Approach 1: Dynamic Programming

This approach uses dynamic programming to build up a solution iteratively.

1. State Definition:

Let f[i] represent the maximum product achievable by breaking the integer i. f[1] = 1 (base case: no break possible).

2. State Transition:

For each integer i, we consider all possible ways to split it into two parts j and i-j where 1 <= j < i. The maximum product for i is the maximum of:

• f[i-j] * j: The product obtained by breaking i-j optimally and multiplying by j.
• (i-j) * j: The product obtained without further breaking i-j.

Thus, the recurrence relation is:

f[i] = max(f[i], max(f[i-j] * j, (i-j) * j)) for all j in [1, i-1]

3. Base Case:

f[1] = 1

4. Algorithm:

Iterate from i = 2 to n, applying the recurrence relation to compute f[i]. The final result is f[n].

Time Complexity: O(n^2) due to the nested loops. Space Complexity: O(n) to store the f array.

Approach 2: Mathematical Optimization

This approach leverages mathematical insights to derive a more efficient solution.

1. Observation:

For n >= 4, repeatedly dividing by 3 gives the best result because 3 * 3 > 2 * 2 * 2 and 2 * 3 > 1 * 1 * 1 * 1.

2. Algorithm:

• If n < 4, return n - 1.
• Otherwise, repeatedly divide n by 3 until the remainder is 0, 1, or 2.
• If the remainder is 0, the maximum product is 3^(n/3).
• If the remainder is 1, the maximum product is 4 * 3^((n-4)/3) (because 4 > 1*1).
• If the remainder is 2, the maximum product is 2 * 3^(n/3).

Time Complexity: O(log n) due to repeated division by 3 (or O(1) if considering the number of divisions as constant). Space Complexity: O(1).

Code Examples (Python):

Approach 1 (Dynamic Programming):

def integerBreakDP(n):
    f = [1] * (n + 1)
    for i in range(2, n + 1):
        for j in range(1, i):
            f[i] = max(f[i], f[i - j] * j, (i - j) * j)
    return f[n]
 

Approach 2 (Mathematical Optimization):

def integerBreakMath(n):
    if n < 4:
        return n - 1
    if n % 3 == 0:
        return 3**(n // 3)
    if n % 3 == 1:
        return 4 * 3**((n - 4) // 3)
    return 2 * 3**(n // 3)

Both approaches are included in the detailed markdown response above, in multiple programming languages. The mathematical approach is significantly faster for larger values of n.