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Leetcode problem 343 reads as follows:

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.

Clarifying Questions

  1. Is the input always a positive integer greater than 1?
    • Yes, based on the problem constraints.
  2. Should we handle any special cases, such as non-integer inputs or negative numbers?
    • No, we can assume the input is valid as per the problem statement.

Strategy

To solve this problem, notice:

The key observation is that products of smaller integers like 2 and particularly 3 (due to Euler’s and other mathematical proofs) tend to yield higher products. More formally:

  1. If n <= 3, then the result is simply n - 1 because:
    • For n = 2, we can only break it into 1 + 1.
    • For n = 3, we can only break it into 1 + 2.
  2. For n >= 4, breaking n into factors of 3 as much as possible maximizes the product:
    • If n modulo 3 equals 1, then use one 4 (e.g. n = 10 -> 3 + 3 + 4).
    • If n modulo 3 equals 2, use one 2 (e.g. n = 8 -> 3 + 3 + 2).
    • Otherwise, keep breaking into 3’s (e.g. n = 9 -> 3 + 3 + 3).

Code

def integerBreak(n: int) -> int:
    # Handle base cases
    if n <= 3:
        return n - 1
    
    # Initialize product
    product = 1
    
    # Use as many 3's as possible
    while n > 4:
        product *= 3
        n -= 3
    
    # Multiplying the remainder n directly to product
    return product * n

# Example usage:
print(integerBreak(2))  # Output: 1
print(integerBreak(10)) # Output: 36

Time Complexity

The time complexity of the above algorithm is O(1) (constant time), because the while loop iterates at most n / 3 times, but in practical sense, the operations are minimal and do not depend on input size proportionally. Therefore, it is very efficient. The space complexity is also O(1) as no extra space besides the variables to hold the product and the remainder is used.

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