0971MediumArrayHash TableTwo PointersStringDynamic ProgrammingSorting

Longest Chain of Grown Labels

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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.

Labels are given as words, each a string of lowercase letters.

One label grows into another when inserting a single letter somewhere into the first, without reordering anything else, gives the second exactly.

A chain is a run of labels from the list where each grows into the next. Return the greatest number of labels a chain can hold. A single label is a chain of one.

Examples

Example 1

Input: words = ["bd", "abd", "abcd", "cd", "abdz", "x"]
Output: 3

The chain "bd", "abd", "abcd", "abdz" does not work, since "abcd" does not grow into "abdz". The longest chain is "bd", "abd", "abcd", holding three labels.

Example 2

Input: words = ["ba", "ab"]
Output: 1

The two labels are the same length, so neither grows into the other and a chain holds one.

Example 3

Input: words = ["a", "ab", "abc", "abcd", "abcde"]
Output: 5

Each label grows into the next by adding a letter at the end, so all five form one chain.

Constraints

1 <= words.length <= 1000
1 <= words[i].length <= 16
Every entry of words consists of lowercase English letters only

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_str_chain(words: list[str]) -> int:

Java

public int longestStrChain(String[] words)

The intended approach

Time O(n log n + n * L^2) for n labels of length at most L: the sort, then each label builds L shorter strings of length L.Space O(n * L) for the table of best chains.

Growing a label always makes it one letter longer, so a chain runs strictly upward in length and every label in it is settled once the shorter ones are. So work through the labels shortest first, holding for each the longest chain ending on it. For a label, the labels it could follow are exactly those made by deleting one of its letters, and there are only as many of those as the label is long. So build each of them, look it up among the labels seen, and take the best chain found plus one. Looking up rather than comparing against every other label is what keeps this affordable, since the alternative is testing every pair.

Pythonentry point: longest_str_chain

def longest_str_chain(words):
    order = sorted(words, key=len)
    best = {}
    answer = 0
    for word in order:
        here = 1
        for at in range(len(word)):
            shorter = word[:at] + word[at + 1:]
            if shorter in best and best[shorter] + 1 > here:
                here = best[shorter] + 1
        if word not in best or here > best[word]:
            best[word] = here
        if here > answer:
            answer = here
    return answer

The obvious version

Test every ordered pair of labels to see whether one grows into the other by walking both with two pointers, build a table of which labels may follow which, then search out the longest run through it. Quadratic in the label count times their length, and it never looks anything up.

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_str_chain(words):
    n = len(words)

    def grows(a, b):
        if len(a) + 1 != len(b):
            return False
        i = 0
        j = 0
        skipped = False
        while i < len(a) and j < len(b):
            if a[i] == b[j]:
                i += 1
                j += 1
            elif not skipped:
                skipped = True
                j += 1
            else:
                return False
        return True

    follows = [[] for _ in range(n)]
    for i in range(n):
        for j in range(n):
            if i != j and grows(words[i], words[j]):
                follows[i].append(j)
    memo = {}

    def longest(at):
        if at in memo:
            return memo[at]
        best = 1
        for nxt in follows[at]:
            here = 1 + longest(nxt)
            if here > best:
                best = here
        memo[at] = best
        return best

    answer = 0
    for i in range(n):
        here = longest(i)
        if here > answer:
            answer = here
    return answer

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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