1144MediumArrayDynamic ProgrammingGreedy

Sideways Jumps Down the Three-Track Chute

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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.

Three tracks numbered 1, 2 and 3 run side by side from point 0 to the last point of a chute. blocks[i] names the one track shut at point i, or is 0 when no track is shut there.

A trolley starts at point 0 on track 2 and rolls forward to the last point. It cannot roll onto a point where its own track is shut.

At any point the trolley may jump sideways to either of the other two tracks, so long as that track is not shut at that point. A jump from track 1 straight to track 3 counts as one jump. Rolling forward costs nothing.

Return the fewest jumps needed to reach the last point on any track. A way through always exists.

Examples

Example 1

Input: blocks = [0, 2, 0]
Output: 1

Track 2 is shut at the middle point, so a single jump onto either other track carries the trolley through.

Example 2

Input: blocks = [0, 2, 2, 2, 2, 2, 0]
Output: 1

Track 2 is shut at every point along the way, so one jump at the start onto another track is enough and nothing further is needed.

Example 3

Input: blocks = [0, 3, 2, 1, 0]
Output: 2

Track 3 is shut at the first point so the trolley jumps to track 1. Track 1 is shut at the third point and track 2 is shut where the trolley stands, so the second jump has to go straight across from track 1 to track 3.

Constraints

2 <= blocks.length <= 500001
0 <= blocks[i] <= 3
blocks[0] == 0
the last entry of blocks is 0

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 min_side_jumps(blocks: list[int]) -> int:

Java

public int minSideJumps(int[] blocks)

The intended approach

Time O(n) over the points.Space O(1) for the three numbers.

Rolling forward is free and jumping costs one, so what matters at each point is the cheapest way of standing on each of the three tracks there. Three running numbers carry that along, starting at nothing for track `2` and one apiece for the other two, since reaching them at the very start takes a single jump. Moving to the next point happens in two steps. First, whichever track is shut there becomes unreachable, since the trolley cannot roll onto a shut point. Then jumps at that point are allowed: any open track can be reached for one more than the cheapest open track at that point, and because a jump across the middle counts as one, the cheapest open track is the only source worth looking at. Barring the shut track before allowing jumps is what stops the trolley slipping onto it, and the answer is the cheapest of the three numbers once the last point is done.

Pythonentry point: min_side_jumps

def min_side_jumps(blocks):
    big = 1 << 30
    cost = [big, 1, 0, 1]
    for i in range(1, len(blocks)):
        shut = blocks[i]
        if shut:
            cost[shut] = big
        low = big
        for lane in (1, 2, 3):
            if cost[lane] < low:
                low = cost[lane]
        for lane in (1, 2, 3):
            if lane != shut and cost[lane] > low + 1:
                cost[lane] = low + 1
    answer = cost[1]
    if cost[2] < answer:
        answer = cost[2]
    if cost[3] < answer:
        answer = cost[3]
    return answer

The obvious version

Treat every pairing of point and track as a place to be and walk outwards from the start, always taking the place reached with the fewest jumps so far. Rolling forward to an open point keeps the count, and jumping to either open track at the current point adds one. The first time the last point is drawn out, its count is settled.

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_min_side_jumps(blocks):
    import heapq
    last = len(blocks) - 1
    settled = set()
    heap = [(0, 0, 2)]
    while heap:
        jumps, at, lane = heapq.heappop(heap)
        if (at, lane) in settled:
            continue
        settled.add((at, lane))
        if at == last:
            return jumps
        if blocks[at + 1] != lane:
            heapq.heappush(heap, (jumps, at + 1, lane))
        for other in (1, 2, 3):
            if other != lane and blocks[at] != other:
                heapq.heappush(heap, (jumps + 1, at, other))
    return -1

How this problem was checked

Reference solution executed against 15 test cases — 15 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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