0572MediumMathDynamic Programming

Cut the Rod for the Largest Product

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Trains the technique from

This is an original problem, written from a brief that listed the technique, the difficulty, the topics, the function shape and the input bounds — none of that problem's wording, examples, hints or editorials. The link is there so you can map your practice onto the standard set.

Same function shape, different story and different numbers.

A supplier holds a single rod of length n centimetres and has to saw it into two or more pieces. Every piece must be a whole number of centimetres long, and the piece lengths have to add up to exactly n. Pieces may repeat a length, and their order does not matter.

The pieces are then sold as a set priced at the product of their lengths. Return the largest product any allowed set of pieces can reach.

Examples

Example 1

Input: n = 7
Output: 12

Cutting the rod into pieces of 3 and 4 gives lengths that add to 7 and a product of 12.

Example 2

Input: n = 4
Output: 4

Cutting into 2 and 2 gives a product of 4, and the two pieces add up to 4.

Example 3

Input: n = 43
Output: 6377292

One allowed set is thirteen pieces of 3 and one piece of 4, which add up to 43 and multiply to 6377292.

Constraints

2 <= n <= 58
The answer never exceeds 1549681956.

The signature

The editor is preloaded with this. It matches the parent problem's shape, so a solution that works here transfers to a judge unchanged.

Python

def integer_break(n: int) -> int:

Java

public int integerBreak(int n)

The intended approach

Time O(n^2): each length tries every split point.Space O(n) for the table.

Fill a table where entry `i` is the best product for a rod of length `i` sawn into at least two pieces. For a length, try every possible size for one piece: what remains is either sold as a single piece of that leftover length, or sawn again, in which case its own table entry applies, so take the larger of the two and multiply. Reading the table entry for `n` gives the answer.

Pythonentry point: integer_break

def integer_break(n):
    # O(n^2) time, O(n) space: best[i] is the largest product for a rod of length i
    # cut into at least two pieces; for each first piece j the rest is either left
    # whole or cut again, whichever is larger.
    best = [0] * (n + 1)
    for length in range(2, n + 1):
        top = 0
        for j in range(1, length):
            rest = length - j
            other = rest if rest > best[rest] else best[rest]
            value = j * other
            if value > top:
                top = value
        best[length] = top
    return best[n]

The obvious version

Enumerate every way of writing `n` as a sum of whole numbers in non-decreasing order, multiply the parts of each such set, and keep the largest product among the sets with at least two parts.

This is not filler. It was written independently of the fast solution and the two were run against thousands of random inputs to confirm they agree — which is how the answer above is known to be right. Read it when the intended approach feels like a rabbit pulled out of a hat.

Python

def bf_integer_break(n):
    best = [0]

    def walk(remaining, smallest, product, pieces):
        if remaining == 0:
            if pieces >= 2 and product > best[0]:
                best[0] = product
            return
        for piece in range(smallest, remaining + 1):
            walk(remaining - piece, piece, product * piece, pieces + 1)

    walk(n, 1, 1, 0)
    return best[0]

How this problem was checked

Reference solution executed against 11 test cases — 11 passed
3 worked examples confirmed by running the code
0 eight-word sequences shared with the parent problem's text
Agrees with an independent brute-force implementation on random inputs
Test set proven to catch deliberately broken solutions
Reviewed by a separate quality pass
Function arity matches the parent problem's

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