343. Integer Break
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Given a positive integer
n, our task is to divide it into the sum of
k positive integers, where $k \geq 2$, in such a way that the product of these integers is maximized. Our goal is to determine the maximum possible product.
A straightforward or naive way to solve this would be to consider all potential combinations to divide the number
n and calculate the product for each division. This method, while comprehensive, would be inefficient and impractical for larger values of
Hints & Tips
- Try to break down
ninto smaller parts and observe the pattern of the results.
- The number 3 plays a significant role, so try to understand its impact on the problem.
A pattern emerges when observing how the number can be broken down to maximize the product: the number 3 becomes significant. This realization stems from the fact that 3 multiplied by any number $x \geq 3$ is always greater than $x \times 2$ and $x \times 1$.
The only exception is 4, where $2 \times 2$ is preferable to 3 and 1.
Therefore, the optimized approach becomes:
When $n = 2$, the answer is 1 (because $1 \times 1 = 1$).
For $n = 3$, the answer becomes 2 (as $2 \times 1 = 2$).
If $n = 4$, the result is 4 (as $2 \times 2 = 4$).
For any $n > 4$, we can repeatedly subtract 3 from
nand multiply the resulting product by 3.
After all the 3s are extracted, the remaining
n(which will be less than 4) will contribute its optimal value to the product (either 1, 2, or 4).
def integerBreak(n: int) -> int: if n == 2: # base cases return 1 if n == 3: return 2 if n == 4: return 4 product = 1 while n > 4: # As long as n is greater than 4, product *= 3 # increase the product by a factor of 3 n -= 3 # and keep reducing n by 3 product *= n # Multiply the remaining value of n to the product return product