0671EasyMathString

Seven-Rod Counting Frame

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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 museum shows an old counting frame that works in sevens. Each rod carries from zero to six beads, and a rod is worth seven times the rod immediately to its right, so the rightmost rod counts ones, the next counts sevens, the next counts forty-nines, and so on.

Given an integer value, return the text a curator would write on the label: the bead count of each rod, read from the leftmost rod that carries a bead down to the ones rod, with no extra zero in front. If value is negative, put a single - before those digits. If value is zero, the label is the single character 0.

Examples

Example 1

Input: value = 342
Output: "666"

Six beads on the forty-nines rod, six on the sevens rod and six on the ones rod come to 294 plus 42 plus 6.

Example 2

Input: value = -48
Output: "-66"

Forty-eight is six sevens plus six ones, and the value is negative, so the label carries a leading minus.

Example 3

Input: value = 2401
Output: "10000"

2401 is seven multiplied by itself four times, so a single bead sits on the fifth rod from the right and every rod to its right is bare.

Constraints

-10^7 <= value <= 10^7

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 to_base_seven(value: int) -> str:

Java

public String toBaseSeven(int value)

The intended approach

Time O(log value): one division per rod.Space O(log value) for the collected rod digits, which is the returned label.

Work with the magnitude and take it apart from the right. The remainder on division by seven is the bead count of the ones rod, and dividing the magnitude by seven moves attention one rod to the left, so collecting remainders until nothing is left gives the rods from right to left. Reverse them for the label, prefixing the sign for a negative value, and answer zero directly since its loop would collect nothing.

Pythonentry point: to_base_seven

def to_base_seven(value):
    # peel the least significant rod first, then read the rods back to front
    if value == 0:
        return "0"
    rest = value if value > 0 else -value
    rods = []
    while rest > 0:
        rods.append(chr(48 + rest % 7))
        rest //= 7
    if value < 0:
        rods.append("-")
    return "".join(reversed(rods))

The obvious version

Find the largest rod weight that does not exceed the magnitude by multiplying by seven from one. Then walk the rods left to right, repeatedly subtracting the current rod weight while it still fits and writing down how many times that happened, dividing the weight by seven at each step.

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_to_base_seven(value):
    if value == 0:
        return "0"
    size = value if value > 0 else -value
    weight = 1
    while weight * 7 <= size:
        weight *= 7
    out = "-" if value < 0 else ""
    while weight >= 1:
        beads = 0
        while size >= weight:
            size -= weight
            beads += 1
        out += str(beads)
        weight //= 7
    return out

How this problem was checked

Reference solution executed against 12 test cases — 12 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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