0357HardHash TableStringBacktrackingBreadth-First SearchBidirectional Search

Call Sign Relay Routes

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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 relay station has to move from the call sign beginWord to the call sign endWord. All call signs are lowercase strings of the same length.

One hop rewrites exactly one letter of the current sign, and the sign it produces has to appear in the approved register wordList. A route is a list of signs that starts with beginWord, ends with endWord, and where each sign after the first is one hop from the sign before it. beginWord itself does not have to be in the register.

Return every route that uses the fewest hops possible, each route listed from beginWord to endWord. The routes themselves may come back in any order. Return an empty list when no route exists at all.

Examples

Example 1

Input: beginWord = "fin", endWord = "ban", wordList = ["fan","fun","ban","bin","bun"]
Output: [["fin","bin","ban"],["fin","fan","ban"]]

Two routes of two hops each are the shortest ones: `fin` to `bin` to `ban`, and `fin` to `fan` to `ban`. Every sign in them comes from the register and every hop rewrites exactly one letter.

Example 2

Input: beginWord = "tin", endWord = "ten", wordList = ["ten","tan"]
Output: [["tin","ten"]]

`ten` is in the register and differs from `tin` in one letter, so a single hop is enough and only one route is that short.

Example 3

Input: beginWord = "fin", endWord = "oak", wordList = ["fan","ban","oak"]
Output: []

`oak` is in the register but shares no letter position with any sign that can be reached from `fin`, so no route ever arrives and the answer is empty.

Constraints

1 <= beginWord.length <= 5
endWord.length == beginWord.length
1 <= wordList.length <= 500
wordList[i].length == beginWord.length
beginWord, endWord and every wordList[i] hold lowercase English letters only.
beginWord != endWord
The register holds no repeated call sign.
The shortest routes hold at most 10^5 signs in total.

The values you return may be in any order.

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 find_ladders(beginWord: str, endWord: str, wordList: list[str]) -> list[list[str]]:

Java

public List<List<String>> findLadders(String beginWord, String endWord, List<String> wordList)

The intended approach

Time O(len(wordList) * L * 26) for the sweep, plus the total size of the returned routesSpace O(len(wordList) * L) for the links, plus the returned routes

Grow rings of signs outward from `beginWord`, one hop per ring, and for each sign in the new ring record every sign of the current ring that hops to it. Signs are struck off the register only after their whole ring has been built, so two routes of the same length both keep their links. The sweep stops at the ring that first contains `endWord`, and the recorded links are then walked backwards with an explicit stack, which spells out each shortest route without recursing once per sign.

Pythonentry point: find_ladders

def find_ladders(beginWord, endWord, wordList):
    # Sweep outward one hop at a time, and for every sign reached in the new ring
    # remember all the signs of the previous ring that could produce it.  A ring's
    # signs leave the register only once the whole ring is built, so sibling routes of
    # equal length survive.  Once the target appears, walk the recorded links backwards
    # with an explicit stack to spell out every route.
    # Time O(len(wordList) * L * 26) for the sweep plus the size of the answer.
    # Space O(len(wordList) * L) for the links, plus the answer itself.
    register = set(wordList)
    if endWord not in register:
        return []
    register.discard(beginWord)
    letters = "abcdefghijklmnopqrstuvwxyz"
    came_from = {}
    ring = {beginWord}
    reached = False
    while ring and not reached:
        nxt = set()
        for sign in ring:
            for i in range(len(sign)):
                head = sign[:i]
                tail = sign[i + 1:]
                for ch in letters:
                    if ch == sign[i]:
                        continue
                    candidate = head + ch + tail
                    if candidate in register:
                        nxt.add(candidate)
                        if candidate in came_from:
                            came_from[candidate].append(sign)
                        else:
                            came_from[candidate] = [sign]
        register -= nxt
        if endWord in nxt:
            reached = True
        ring = nxt
    if not reached:
        return []
    routes = []
    stack = [(endWord, [endWord])]
    while stack:
        sign, trail = stack.pop()
        if sign == beginWord:
            routes.append(trail[::-1])
            continue
        for previous in came_from.get(sign, ()):
            stack.append((previous, trail + [previous]))
    routes.sort()
    return routes

The obvious version

Breadth-first over whole partial routes: hold every route of the current length, stop as soon as any of them ends at `endWord` and return all of those, otherwise extend each route by every legal hop into a sign no earlier layer has used.

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_all_shortest_routes(beginWord, endWord, wordList):
    register = set(wordList)
    if endWord not in register:
        return []
    letters = "abcdefghijklmnopqrstuvwxyz"
    layer = [[beginWord]]
    used = {beginWord}
    while layer:
        done = []
        for trail in layer:
            if trail[-1] == endWord:
                done.append(trail)
        if done:
            done.sort()
            return done
        fresh = set()
        nxt = []
        for trail in layer:
            sign = trail[-1]
            for i in range(len(sign)):
                for ch in letters:
                    candidate = sign[:i] + ch + sign[i + 1:]
                    if candidate != sign and candidate in register and candidate not in used:
                        fresh.add(candidate)
                        nxt.append(trail + [candidate])
        used |= fresh
        layer = nxt
    return []

How this problem was checked

Reference solution executed against 9 test cases — 9 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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