0603MediumArray

Matching Neighbours on a Tile Shelf

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Trains the technique from

LeetCode 2672Number of Adjacent Elements With the Same Color

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 display shelf carries n tiles in a row, numbered 0 through n - 1. Every tile starts out blank, meaning it carries no colour whatsoever.

You are given queries. Its i-th entry names a position index_i followed by a colour colour_i. Work through the queries in the order given: query i paints the tile at index_i in colour colour_i, wiping out whatever that tile carried before.

Straight after each query, count the neighbouring tile pairs that carry the same colour. Two tiles are neighbours when their positions are one apart. A pair only counts when both of its tiles have been painted, so a blank tile never takes part in a counted pair, not even alongside another blank tile.

Return answer, where answer[i] is that count taken right after query i is applied.

Examples

Example 1

Input: n = 4, queries = [[0, 2], [1, 2], [3, 2], [1, 7], [2, 7]]
Output: [0, 1, 1, 0, 1]

After the first query only tile 0 is painted, so no pair can count. The second query paints tile 1 in the same colour, giving one pair. Tile 3 is then painted but tile 2 is still blank, so the count stays at 1. Repainting tile 1 in colour 7 breaks that pair, and painting tile 2 in colour 7 forms a new one.

Example 2

Input: n = 1, queries = [[0, 5], [0, 9]]
Output: [0, 0]

A shelf of one tile has no neighbouring pairs at all, so both counts are 0.

Example 3

Input: n = 3, queries = [[0, 4], [2, 4], [1, 9]]
Output: [0, 0, 0]

Tiles 0 and 2 both carry colour 4, but they are two apart and so are not neighbours. Painting tile 1 in colour 9 matches neither of them, so every count is 0.

Constraints

1 <= n <= 10^5
1 <= queries.length <= 10^5
queries[i].length == 2
0 <= index_i <= n - 1
1 <= colour_i <= 10^5

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 color_the_array(n: int, queries: list[list[int]]) -> list[int]:

Java

public int[] colorTheArray(int n, int[][] queries)

The intended approach

Time O(n + q): O(n) to lay out the shelf and constant work per query.Space O(n) for the shelf, plus the O(q) answer that is returned.

A single query only touches the at most two pairs that include the painted tile, so the count from the previous query can be repaired instead of recomputed. If the tile was already painted, first drop any pairs its former colour was forming with its left and right neighbour. Then store the new colour and add a pair for each neighbour now carrying it. Blank is recorded as `0` while real colours start at `1`, so a blank neighbour can never match. Each query's repaired count is appended to the answer.

Pythonentry point: color_the_array

def color_the_array(n, queries):
    # O(n + q) time, O(n) space. Painting one tile can only make or break the
    # at most two pairs it belongs to, so keep a running count and adjust it by
    # inspecting the two neighbours: first undo what the old colour contributed,
    # then add what the new colour contributes. Blank is stored as 0 and real
    # colours are at least 1, so a blank tile never equals a neighbour's colour.
    shade = [0] * n
    matched = 0
    answer = []
    for index, colour in queries:
        old = shade[index]
        if old != 0:
            if index > 0 and shade[index - 1] == old:
                matched -= 1
            if index + 1 < n and shade[index + 1] == old:
                matched -= 1
        shade[index] = colour
        if index > 0 and shade[index - 1] == colour:
            matched += 1
        if index + 1 < n and shade[index + 1] == colour:
            matched += 1
        answer.append(matched)
    return answer

The obvious version

Apply each query to a plain array of colours, then walk the whole shelf from left to right and count the neighbouring pairs whose colours agree and are not blank.

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_color_the_array(n, queries):
    shade = [0] * n
    answer = []
    for index, colour in queries:
        shade[index] = colour
        tally = 0
        for i in range(n - 1):
            if shade[i] != 0 and shade[i] == shade[i + 1]:
                tally += 1
        answer.append(tally)
    return answer

How this problem was checked

Reference solution executed against 8 test cases — 8 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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