0549MediumHash TableMathString

Ratio Printed as a Decimal

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Trains the technique from

LeetCode 166Fraction to Recurring Decimal

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 reporting tool has to print the ratio numerator / denominator in ordinary decimal notation and return it as a string. Follow this format exactly.

A leading - appears when and only when the true value is below zero. A value of zero is printed as 0, never as -0.
When the ratio comes out whole, print only the digits of that whole number and no decimal point. So a ratio worth two is printed as 2.
Otherwise print the whole part, then a ., then the digits after the point. A value between -1 and 1 has a whole part of 0, as in 0.75 or -0.75.
The digits after the point may run on forever, in which case they finish with a block of digits that repeats over and over. Enclose that block in ( and ) and stop there. The ( goes at the earliest digit from which the repeat runs unbroken to the end, and the block inside the brackets must be as short as possible. So one ninth is 0.(1), one sixth is 0.1(6), and one forty-fourth is 0.02(27).
When the digits after the point stop, print them and nothing more, as in 0.875.

Return that string.

Examples

Example 1

Input: numerator = 7, denominator = 8
Output: "0.875"

Seven eighths is 0.875 and the digits after the point stop, so no brackets appear.

Example 2

Input: numerator = 5, denominator = 44
Output: "0.11(36)"

The digits after the point run 1, 1, 3, 6, 3, 6, 3, 6 and onward. The repeat is the pair 36 and the earliest place it starts running unbroken is the third digit, so the first two digits sit outside the brackets.

Example 3

Input: numerator = -9, denominator = 4
Output: "-2.25"

The value is below zero, so the string opens with a minus sign, and nine quarters is 2.25.

Example 4

Input: numerator = 8, denominator = 4
Output: "2"

Eight divided by four comes out whole, so only the digits of the whole number are printed and there is no decimal point.

Constraints

-2^31 <= numerator, denominator <= 2^31 - 1
denominator != 0
The inputs are chosen so that the returned string holds at most 10^4 characters.

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 fraction_to_decimal(numerator: int, denominator: int) -> str:

Java

public String fractionToDecimal(int numerator, int denominator)

The intended approach

Time O(d) where d is the number of printed digits after the point, which is below the magnitude of denominator.Space O(d) for the digit list and the remainder map.

Decide the sign from whether the two inputs disagree, then work purely with magnitudes so the digit arithmetic never has to think about signs. Take the whole part and the first remainder; a remainder of zero means the ratio is whole and the string ends there. Otherwise run long division: multiply the remainder by ten, take the next digit, keep the new remainder, and record in a map the position each remainder produced. The moment a remainder repeats, every digit after its first appearance is bound to repeat too, so the map gives the exact position where the bracket opens. A remainder reaching zero instead means the digits stop. A zero numerator is answered before anything else, which is what keeps a negative denominator from producing a signed zero.

Pythonentry point: fraction_to_decimal

def fraction_to_decimal(numerator, denominator):
    # Long division on magnitudes, with a map from remainder to the position of
    # the digit that remainder produced. A remainder that shows up twice means
    # every digit from that point on repeats, so the map hands back exactly
    # where the bracket opens. The sign is decided once, up front.
    if numerator == 0:
        return "0"
    sign = "-" if (numerator < 0) != (denominator < 0) else ""
    top = abs(numerator)
    bottom = abs(denominator)
    whole = top // bottom
    rem = top - whole * bottom
    if rem == 0:
        return sign + str(whole)
    seen = {}
    digits = []
    while rem != 0 and rem not in seen:
        seen[rem] = len(digits)
        rem *= 10
        digits.append(str(rem // bottom))
        rem = rem % bottom
    if rem == 0:
        tail = "".join(digits)
    else:
        cut = seen[rem]
        tail = "".join(digits[:cut]) + "(" + "".join(digits[cut:]) + ")"
    return sign + str(whole) + "." + tail

The obvious version

The same long division, but remainders are appended to a plain list and each new remainder is looked for by scanning that list from the front. When the scan finds it, the position it was found at is where the bracket opens.

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_fraction_to_decimal(numerator, denominator):
    if numerator == 0:
        return "0"
    neg = (numerator < 0) != (denominator < 0)
    top = abs(numerator)
    bottom = abs(denominator)
    whole = top // bottom
    rem = top % bottom
    head = ("-" if neg else "") + str(whole)
    if rem == 0:
        return head
    rems = []
    digits = []
    while rem != 0 and rem not in rems:
        rems.append(rem)
        rem *= 10
        digits.append(str(rem // bottom))
        rem = rem % bottom
    if rem == 0:
        return head + "." + "".join(digits)
    cut = rems.index(rem)
    return head + "." + "".join(digits[:cut]) + "(" + "".join(digits[cut:]) + ")"

How this problem was checked

Reference solution executed against 20 test cases — 20 passed
4 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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