0525MediumHash TableStringSorting

Matching Tile Runs With Parity Swaps

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Trains the technique from

LeetCode 2840Check if Strings Can be Made Equal With Operations II

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.

Two display rails, left and right, each carry the same number of tiles. A tile is written as a lowercase letter, its colour. Seats on a rail are numbered from 0.

A fitter may repeat this move as often as they like, on either rail: pick two seats on the same rail whose numbers are both even or both odd, and swap the tiles sitting in them. Seats do not have to be next to each other.

Return true if the two rails can be made to show the same colours in the same seats, and false otherwise.

Examples

Example 1

Input: left = "abcd", right = "cbad"
Output: true

Swapping the tiles in seats 0 and 2 of `left` turns it into "cbad", which is what `right` already shows.

Example 2

Input: left = "axaxbx", right = "axbxbx"
Output: false

The even seats of `left` carry a, a and b, while the even seats of `right` carry a, b and b, and no permitted move changes either of those groups. So the rails cannot be made to agree.

Example 3

Input: left = "ab", right = "ba"
Output: false

Seats 0 and 1 have different parity, so no permitted move touches this pair of tiles at all and the rails stay different.

Constraints

n == left.length == right.length
1 <= n <= 10^5
left and right hold only lowercase English letters.

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 can_match_runs(left: str, right: str) -> bool:

Java

public boolean canMatchRuns(String left, String right)

The intended approach

Time O(n): one pass over each rail.Space O(1), at most 26 counters per pool.

A permitted move always keeps a tile on a seat of the same parity, so the tiles on even seats and the tiles on odd seats form two pools that never exchange members, and within a pool the fitter can reach any arrangement. So walk both rails once, adding each of `left`'s tiles into the tally for its pool and subtracting each of `right`'s, and answer yes exactly when every tally lands back on zero.

Pythonentry point: can_match_runs

def can_match_runs(left, right):
    # O(n) time: a tally of the letters on even seats and another on odd seats,
    # because a swap never moves a tile between the two groups.
    even = {}
    odd = {}
    for i in range(len(left)):
        box = even if i % 2 == 0 else odd
        box[left[i]] = box.get(left[i], 0) + 1
    for i in range(len(right)):
        box = even if i % 2 == 0 else odd
        box[right[i]] = box.get(right[i], 0) - 1
    for count in even.values():
        if count != 0:
            return False
    for count in odd.values():
        if count != 0:
            return False
    return True

The obvious version

Grow the set of arrangements each rail can reach by applying every permitted swap over and over until nothing new appears, then answer yes if the two sets share an arrangement.

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_can_match_runs(left, right):
    def reach(s):
        seen = {s}
        stack = [s]
        n = len(s)
        while stack:
            cur = stack.pop()
            for i in range(n):
                for j in range(i + 2, n, 2):
                    lst = list(cur)
                    lst[i], lst[j] = lst[j], lst[i]
                    nxt = "".join(lst)
                    if nxt not in seen:
                        seen.add(nxt)
                        stack.append(nxt)
        return seen
    here = reach(left)
    there = reach(right)
    for s in here:
        if s in there:
            return True
    return False

How this problem was checked

Reference solution executed against 14 test cases — 14 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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