1178HardArrayStringTreeDepth-First SearchGraph TheoryTopological Sort

The Longest Run of Unlike Posts

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Trains the technique from

LeetCode 2246Longest Path With Different Adjacent Characters

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 tree has posts numbered 0 through n - 1, and post 0 sits at the top. above[i] names the post directly above post i, and above[0] is -1. Each post carries one lowercase letter, given by marks.

A run is a path through the tree that never repeats a post and along which no two neighbouring posts carry the same letter.

Return the largest number of posts a run can hold.

Examples

Example 1

Input: above = [-1, 0, 1, 2, 3], marks = "abcde"
Output: 5

The posts form a single chain and no two neighbours share a letter, so the whole chain is one run.

Example 2

Input: above = [-1, 0], marks = "aa"
Output: 1

Both posts carry the same letter, so no run can hold the two of them.

Example 3

Input: above = [-1, 0, 0, 0, 0, 0], marks = "abcdef"
Output: 3

Five posts hang off the top one, all carrying different letters, so a run can come up one branch, through the top, and down another.

Constraints

1 <= above.length <= 10^5
above.length == marks.length
above[0] == -1
above describes a tree with post 0 at the top
marks holds 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 longest_path(above: list[int], marks: str) -> int:

Java

public int longestPath(int[] above, String marks)

The intended approach

Time O(n) over the posts.Space O(n) for the links and the reaches.

Every run has a highest post, and from there it goes down at most two ways: down one branch, through the highest post, and down another. So it is enough to know, for each post, how many posts the longest run reaching straight down from it can hold. Work the posts from the bottom upwards. For a post, look at each post directly below it and ignore any carrying the same letter, since the run cannot step there. Among the rest, the longest reach downwards is one more than the best of theirs, and the best run whose highest post is this one joins the two best of theirs through it, adding one for the post itself. Keeping the best and the runner-up among the posts below is what lets the run come up one branch and go down another. A post with nothing usable below it still counts as a run of one, which is why the reach starts at one.

Pythonentry point: longest_path

def longest_path(above, marks):
    n = len(above)
    below = [[] for _ in range(n)]
    for i in range(1, n):
        below[above[i]].append(i)
    order = []
    stack = [0]
    while stack:
        node = stack.pop()
        order.append(node)
        for kid in below[node]:
            stack.append(kid)
    reach = [1] * n
    best = 1
    for at in range(len(order) - 1, -1, -1):
        node = order[at]
        top = 0
        second = 0
        for kid in below[node]:
            if marks[kid] == marks[node]:
                continue
            span = reach[kid]
            if span > top:
                second = top
                top = span
            elif span > second:
                second = span
        reach[node] = top + 1
        through = top + second + 1
        if through > best:
            best = through
    return best

The obvious version

Walk out every run by hand. Treat the tree as plain links both ways, set off from each post in turn and follow every path that never repeats a post and never steps onto a matching letter, keeping the most posts any path holds. Costs a search per post with the whole path carried along.

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_longest_path(above, marks):
    n = len(above)
    links = [[] for _ in range(n)]
    for i in range(1, n):
        links[i].append(above[i])
        links[above[i]].append(i)
    best = 1
    for start in range(n):
        stack = [(start, (start,))]
        while stack:
            node, path = stack.pop()
            if len(path) > best:
                best = len(path)
            for nb in links[node]:
                if nb in path:
                    continue
                if marks[nb] == marks[node]:
                    continue
                stack.append((nb, path + (nb,)))
    return best

How this problem was checked

Reference solution executed against 15 test cases — 15 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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