0024EasyTwo PointersString

Mirror Phrase Check

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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 word-game server accepts a submission phrase typed by a player and decides whether it counts as a mirror phrase.

The server judges it with two relaxations, applied before the comparison:

Only letters and digits are inspected. Every other character in phrase, including spaces, punctuation and symbols, is skipped over entirely.
Letter case carries no meaning. An uppercase letter and its lowercase twin are treated as one and the same character.

After those relaxations, phrase is a mirror phrase when the kept characters spell the same thing read left to right as they do read right to left. Return true if it does and false if it does not.

A submission whose characters are all skipped keeps nothing at all, and nothing trivially matches itself, so those submissions are mirror phrases too.

Examples

Example 1

Input: phrase = "Step on no pets!"
Output: true

Dropping the spaces and the exclamation mark keeps steponnopets, which spells itself in reverse. Compared as typed it would fail, so the skipping rule is what decides this one.

Example 2

Input: phrase = "Deed"
Output: true

Nothing is skipped here. The capital D at the front lines up with the small d at the back, and folding case away makes them equal.

Example 3

Input: phrase = "Bell hop"
Output: false

The kept characters spell bellhop, whose reverse begins with p, so the very first comparison already disagrees.

Example 4

Input: phrase = "@@ ##"
Output: true

Not one letter or digit survives the first rule, leaving an empty run of characters, which the server accepts.

Constraints

1 <= phrase.length <= 2 * 10^5
phrase is made up of printable ASCII 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 is_mirror_phrase(phrase: str) -> bool:

Java

public boolean isMirrorPhrase(String phrase)

The intended approach

Time O(n)Space O(1)

Run one cursor in from each end. Advance each cursor past anything that is not a letter or digit, then compare the two characters it stopped on after folding both to lowercase. A single mismatch settles the answer as false; if the cursors meet without disagreeing the phrase mirrors itself. No filtered copy is built, so the extra memory stays constant.

Pythonentry point: is_mirror_phrase

def is_mirror_phrase(phrase):
    left = 0
    right = len(phrase) - 1
    while left < right:
        while left < right and not phrase[left].isalnum():
            left += 1
        while left < right and not phrase[right].isalnum():
            right -= 1
        if phrase[left].lower() != phrase[right].lower():
            return False
        left += 1
        right -= 1
    return True

The obvious version

Materialise the kept characters in lowercase as a list and compare it against its own reverse. No cursors, no early exit.

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_mirror_phrase(phrase):
    kept = []
    for ch in phrase:
        if ch.isalnum():
            kept.append(ch.lower())
    return kept == kept[::-1]

How this problem was checked

Reference solution executed against 14 test cases — 14 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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