0193EasyTwo PointersString

Braided Bead Strands

Tracked in this browser only

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 braids two bead strands onto a single cord. Each strand arrives as text with one letter per bead, listed from the clasp outward: strandA and strandB.

The braid grows one bead at a time. First the clasp-most bead of strandA goes on, then the clasp-most bead of strandB, then the second bead of strandA, then the second bead of strandB, and the turns keep alternating like that. The moment one strand has nothing left to give, every bead still waiting on the other strand goes on in its own order and finishes the braid.

Return the finished braid as text.

Examples

Example 1

Input: strandA = "ruby", strandB = "jade"
Output: "rjuabdye"

Both strands hold four beads, so the turns alternate the whole way and neither strand has a leftover.

Example 2

Input: strandA = "opal", strandB = "tin"
Output: "otpianl"

Six beads alternate, then `strandB` is empty and the last bead of `strandA` closes the braid.

Example 3

Input: strandA = "gem", strandB = "amber"
Output: "gaemmber"

After six alternating beads `strandA` runs dry, so the two beads still on `strandB` follow in order.

Constraints

1 <= strandA.length, strandB.length <= 100
strandA and strandB hold lowercase 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 braid_strands(strandA: str, strandB: str) -> str:

Java

public String braidStrands(String strandA, String strandB)

The intended approach

Time O(n + m)Space O(n + m)

Keep one cursor per strand. While both cursors are still inside their strand, push a bead from each and step both. Once the loop ends at most one strand still has beads, so append its remaining slice.

Pythonentry point: braid_strands

def braid_strands(strandA, strandB):
    # Time O(n + m): every bead is pushed exactly once.
    # Space O(n + m): only the braid being returned.
    braid = []
    i = j = 0
    while i < len(strandA) and j < len(strandB):
        braid.append(strandA[i])
        braid.append(strandB[j])
        i += 1
        j += 1
    while i < len(strandA):
        braid.append(strandA[i])
        i += 1
    while j < len(strandB):
        braid.append(strandB[j])
        j += 1
    return "".join(braid)

The obvious version

Recursion that swaps whose turn it is: emit the head of the strand on turn, then braid the other strand against the shortened one. An exhausted strand ends the recursion by handing back whatever the other still holds.

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_braid_strands(strandA, strandB):
    if not strandA:
        return strandB
    return strandA[0] + bf_braid_strands(strandB, strandA[1:])

How this problem was checked

Reference solution executed against 10 test cases — 10 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

Create a free account to unlock the solution

See the step-by-step animation, the intuition, and clean code in every language — free, no credit card.

Create a free account Already have an account? Sign in