0074MediumStringDynamic ProgrammingBacktracking

Bead Strand Cuts

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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 jeweller has one long strand of beads. strand is a string of lowercase letters giving the bead colour codes from the clasp outwards, one letter per bead.

She wants to snip the strand into consecutive pieces so that every piece is symmetric: reading a piece from either end must give the same sequence of codes. A piece of one bead counts as symmetric, and every bead must end up in exactly one piece.

Return every way of snipping the strand. Each way is the list of its pieces from the clasp outwards; the ways themselves may come back in any order.

Examples

Example 1

Input: strand = "abb"
Output: [["a", "b", "b"], ["a", "bb"]]

Cutting after every bead always works. The only other option is keeping the two b beads together, since "ab" and "abb" are not symmetric.

Example 2

Input: strand = "cbbc"
Output: [["c", "b", "b", "c"], ["c", "bb", "c"], ["cbbc"]]

Single beads always work, the middle pair can be kept whole, and the entire strand already reads the same from either end.

Example 3

Input: strand = "z"
Output: [["z"]]

A single bead admits no cut at all, so there is exactly one way and it is the whole strand.

Constraints

1 <= strand.length <= 16
strand contains lowercase English letters only

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 strand_cuts(strand: str) -> list[list[str]]:

Java

public List<List<String>> strandCuts(String strand)

The intended approach

Time O(n * 2^n)Space O(n^2)

First fill a table saying whether the piece between two bead positions is symmetric, working from the right so the shorter inner piece is already decided when the outer one is examined. Then walk the strand recursively: from the current bead, try every symmetric piece that starts there, recurse past its end, and undo the choice on the way back out.

Pythonentry point: strand_cuts

def strand_cuts(strand):
    n = len(strand)
    mirrored = [[False] * n for _ in range(n)]
    for left in range(n - 1, -1, -1):
        for right in range(left, n):
            if strand[left] == strand[right] and (right - left < 2 or mirrored[left + 1][right - 1]):
                mirrored[left][right] = True
    ways = []
    pieces = []

    def cut_from(start):
        if start == n:
            ways.append(pieces[:])
            return
        for end in range(start, n):
            if mirrored[start][end]:
                pieces.append(strand[start:end + 1])
                cut_from(end + 1)
                pieces.pop()

    cut_from(0)
    return ways

The obvious version

Treat the n-1 gaps between beads as a bitmask, build the pieces for all 2^(n-1) cut choices, and keep the choices whose every piece equals its own reverse.

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_strand_cuts(strand):
    n = len(strand)
    found = []
    for mask in range(1 << (n - 1)):
        pieces = []
        current = strand[0]
        for i in range(1, n):
            if (mask >> (i - 1)) & 1:
                pieces.append(current)
                current = strand[i]
            else:
                current += strand[i]
        pieces.append(current)
        good = True
        for piece in pieces:
            if piece != piece[::-1]:
                good = False
                break
        if good:
            found.append(pieces)
    return found

How this problem was checked

Reference solution executed against 12 test cases — 12 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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