0277MediumStringDynamic Programming

Merged Two Feed Lanes

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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.

Two feed lanes discharge onto a single packing belt. Each lane holds a queue of parts written as a string of lowercase letters, one letter per part: left is the left lane and right the right lane, and the first letter of each is the part at the head of that lane.

An operator clears both lanes onto the belt one part at a time. At every step they pick a lane that still has parts and release the part at its head. Nothing is skipped, nothing is added, and neither lane may be reordered, so the parts of a single lane arrive on the belt in the order they were queued.

belt is the run that came off the belt, head part first. Return true when some sequence of lane picks produces exactly belt, and false otherwise.

Examples

Example 1

Input: left = "ab", right = "cd", belt = "acbd"
Output: true

Releasing a from the left lane, then c from the right, then b from the left, then d from the right lays down a, c, b, d. Within each lane the parts came off in their queued order.

Example 2

Input: left = "ab", right = "cd", belt = "abdc"
Output: false

The belt carries all four parts, but d comes off before c while the right lane queues c ahead of d, so no run of picks can produce it.

Example 3

Input: left = "", right = "xy", belt = "xy"
Output: true

The left lane is empty, so every pick comes from the right lane and the belt shows that lane's own order.

Constraints

0 <= left.length, right.length <= 100
0 <= belt.length <= 200
left, right, and belt consist of 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 is_merge(left: str, right: str, belt: str) -> bool:

Java

public boolean isMerge(String left, String right, String belt)

The intended approach

Time O(len(left) * len(right))Space O(len(right))

Reject at once unless the belt is exactly as long as both lanes together. Then decide, for each count of parts already released from each lane, whether the matching prefix of the belt is reachable: it is, when the previous state is reachable and the part at the head of the lane just used matches the belt at that position. Sweeping the left-lane count outward lets one row of flags be reused, because the entry to the left has already been rewritten for the current row while the entry in place still holds the previous row.

Pythonentry point: is_merge

def is_merge(left, right, belt):
    # dp[j] answers "with i parts already taken from the left lane and j from the
    # right, could the first i + j parts of the belt have come off that way?".
    # The rows are filled in increasing i and one row is reused: dp[j - 1] has
    # already been rewritten for the current i, which is the took-from-the-right
    # case, while dp[j] still holds row i - 1, the took-from-the-left case.
    # Time: O(len(left) * len(right)).  Space: O(len(right)).
    m = len(left)
    n = len(right)
    if m + n != len(belt):
        return False
    dp = [False] * (n + 1)
    dp[0] = True
    for j in range(1, n + 1):
        dp[j] = dp[j - 1] and right[j - 1] == belt[j - 1]
    for i in range(1, m + 1):
        dp[0] = dp[0] and left[i - 1] == belt[i - 1]
        for j in range(1, n + 1):
            c = belt[i + j - 1]
            took_left = dp[j] and left[i - 1] == c
            took_right = dp[j - 1] and right[j - 1] == c
            dp[j] = took_left or took_right
    return dp[n]

The obvious version

Walk every sequence of lane picks with an explicit stack, collecting each complete run the two lanes could produce, then check whether the belt is one of them.

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_is_merge(left, right, belt):
    if len(left) + len(right) != len(belt):
        return False
    m = len(left)
    n = len(right)
    produced = set()
    stack = [(0, 0, "")]
    while stack:
        i, j, so_far = stack.pop()
        if i == m and j == n:
            produced.add(so_far)
            continue
        if i < m:
            stack.append((i + 1, j, so_far + left[i]))
        if j < n:
            stack.append((i, j + 1, so_far + right[j]))
    return belt in produced

How this problem was checked

Reference solution executed against 11 test cases — 11 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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