0050EasyMath

Mirrored Meter Reading

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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 two-way flow meter shows one signed whole number, reading. Flow drawn off the main drives the counter up and flow pushed back into the main drives it down, so the display can sit below zero.

The control room marks a reading as mirrored when its digits run the same way in both directions: sweeping the display from the left gives the same digit sequence as sweeping it from the right. The minus sign is printed only at the leading edge of the display and never at the trailing edge, so anything below zero cannot be mirrored. A reading of one digit, zero included, is mirrored.

Return true for a mirrored reading, false for anything else.

The meter firmware ships without any text handling, so your routine has to work on the number arithmetically. Converting the reading into characters and inspecting them is not available to you.

Examples

Example 1

Input: reading = 4884
Output: true

Sweeping left to right gives 4, 8, 8, 4 and sweeping the other way gives the same sequence.

Example 2

Input: reading = -717
Output: false

The digits alone would match, but the minus sign sits only at the leading edge, so the display cannot read alike in both directions.

Example 3

Input: reading = 250
Output: false

From the right the digits run 0, 5, 2, which is not the sequence 2, 5, 0.

Constraints

-2^31 <= reading <= 2^31 - 1
No conversion of the reading to a string is permitted

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 is_mirrored(reading: int) -> bool:

Java

public boolean isMirrored(int reading)

The intended approach

Time O(d) where d is the number of digitsSpace O(1)

Reject anything below zero straight away, and reject any positive reading whose last digit is zero since its leading digit cannot be zero. Then peel digits off the back with remainder by ten and build up a reversed tail, stopping once the tail has caught up with what is left of the front. The reading is mirrored when the two halves match, and for an odd digit count the middle digit is dropped from the tail before comparing.

Pythonentry point: is_mirrored

def is_mirrored(reading):
    if reading < 0:
        return False
    if reading != 0 and reading % 10 == 0:
        return False
    tail = 0
    while reading > tail:
        tail = tail * 10 + reading % 10
        reading //= 10
    return reading == tail or reading == tail // 10

The obvious version

Free of the firmware restriction, render the reading as characters and compare it against its own reversal. Negative readings carry the sign into the text and fail the comparison on their own.

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_is_mirrored(reading):
    if reading < 0:
        return False
    digits = str(reading)
    return digits == digits[::-1]

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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