1055MediumMathDynamic Programming

Fewest Presses to Reach n Copies

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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 sheet starts with a single mark on it and the clipboard empty. Two keys are available:

copy all: replace the clipboard with everything currently on the sheet;
paste: append the clipboard's contents to the sheet.

Return the fewest key presses that leave exactly n marks on the sheet.

Examples

Example 1

Input: n = 6
Output: 5

Copy and paste once reaches two marks in two presses, then copy and paste twice triples that to six in three more. Five presses in all, which is two and three added together.

Example 2

Input: n = 2
Output: 2

Copy the single mark, then paste it once.

Example 3

Input: n = 9
Output: 6

Three presses reach three marks, and three more triple that to nine.

Constraints

1 <= n <= 1000

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 min_steps(n: int) -> int:

Java

public int minSteps(int n)

The intended approach

Time O(sqrt(n)) to peel the factors off.Space O(1).

A run of presses is always one copy followed by some pastes, then another copy followed by some pastes, and so on, since a paste before any copy does nothing. A copy followed by `k - 1` pastes multiplies the marks by `k` and costs `k` presses. So reaching `n` means writing `n` as a product of whole numbers each at least two, and the cost is those numbers added together. Splitting a factor never costs more: replacing a factor `a` times `b` by the two factors `a` and `b` changes the cost from `a * b` to `a + b`, which is no larger whenever both are at least two. So the cheapest product is the one into primes, and the answer is the prime factors of `n` added together, counting repeats. Peel the factors off from two upwards, dividing them out as they are found. Whatever is left above one at the end is itself prime and joins the total. A sheet of one mark needs no presses at all.

Pythonentry point: min_steps

def min_steps(n):
    total = 0
    left = n
    factor = 2
    while factor * factor <= left:
        while left % factor == 0:
            total += factor
            left //= factor
        factor += 1
    if left > 1:
        total += left
    return total

The obvious version

Search the presses themselves. A position is how many marks are on the sheet and how many sit on the clipboard, and from it a copy sets the clipboard to the marks while a paste adds the clipboard to the marks. Walk outward from the starting position level by level until the wanted count of marks turns up, which gives the fewest presses. Costs the positions rather than a factor peel.

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_min_steps(n):
    from collections import deque
    if n == 1:
        return 0
    seen = set([(1, 0)])
    queue = deque([(1, 0, 0)])
    while queue:
        marks, board, presses = queue.popleft()
        for nxt in ((marks, marks), (marks + board, board)):
            if nxt[0] > n or nxt[0] < 1:
                continue
            if nxt[0] == n:
                return presses + 1
            if nxt in seen:
                continue
            seen.add(nxt)
            queue.append((nxt[0], nxt[1], presses + 1))
    return 0

How this problem was checked

Reference solution executed against 15 test cases — 15 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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