CF 104381K - Hopscotch (Easy Version)

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Solve time: 1m 27s
Verified: no

Solution

Problem Understanding

We are given two parallel rows of numbered tiles, each row containing $n$ positions. At every index $i$, the left row has a value $a_i$ and the right row has a value $b_i$. A player begins with their left foot standing on the first tile of the left row and their right foot on the first tile of the right row. They then move forward by repeatedly shifting exactly one foot at a time to the next position in the same row, never moving backward.

Every time a foot lands on a tile, the value of that tile is added to the score. The process continues until both feet reach position $n$. The goal is to maximize the total collected sum. A constraint limits how far apart the two feet are allowed to be: if the left foot is at position $i$ and the right foot is at position $j$, then $|i - j| \le K$ must always hold.

The key aspect is that movement is sequential along two coupled paths. Each state is defined not just by positions, but by the relative positions of the two feet, since stepping one side forward changes that difference.

The constraints are small, with $n \le 300$, which immediately suggests that a quadratic or cubic dynamic programming approach is acceptable. A cubic $O(n^3)$ solution might barely pass, but a well-designed $O(n^2)$ DP is comfortably within limits.

A naive approach would try to simulate all possible sequences of moves. At each step, either the left or right foot moves forward, while respecting the distance constraint. The number of such sequences grows exponentially, roughly like binomial paths of length $2n$, making brute force infeasible even for $n = 30$.

A more subtle issue appears when one foot "waits" while the other advances several steps. If we do not encode state carefully, it is easy to incorrectly assume both feet progress in lockstep, which would miss optimal solutions where one side temporarily advances faster to collect higher values.

Another edge case occurs when negative values exist. A greedy approach that always moves forward or always balances the feet can fail badly, because sometimes it is worth stepping onto negative values temporarily to unlock access to larger positive segments while respecting the distance constraint.

Approaches

The brute-force idea is to treat this as a shortest-path-like search over states $(i, j)$, where $i$ is the left foot position and $j$ is the right foot position. From each state, we can move either $i \to i+1$ or $j \to j+1$, as long as indices stay within bounds and $|i-j| \le K$. Each move adds the value of the tile being stepped onto.

This is correct because it explores all valid sequences of moves. However, the number of states is $O(n^2)$, and from each state we branch into two transitions, so a naive recursion explores an exponential number of paths. Even memoized without careful ordering, it risks recomputation or large overhead.

The key observation is that this is a classic dynamic programming problem on a grid of states. Each state depends only on two previous states: $(i-1, j)$ and $(i, j-1)$, provided the constraint $|i-j| \le K$ is respected. This transforms the problem into computing a DP over a restricted diagonal band in the $n \times n$ grid.

We define $dp[i][j]$ as the maximum score when the left foot is at $i$ and the right foot is at $j$. Transitions are straightforward because the last move must have been either advancing left or advancing right.

The constraint $|i-j| \le K$ simply prunes invalid states, shrinking the DP region to a band around the diagonal. Since $n \le 300$, iterating over all valid pairs is efficient.

Approach	Time Complexity	Space Complexity	Verdict
Brute Force	Exponential $O(2^{2n})$	O(n) recursion depth	Too slow
Optimal DP	$O(n^2)$	$O(n^2)$	Accepted

Algorithm Walkthrough

We model the process as two synchronized pointers moving independently along their respective rows, while respecting the distance constraint.

Define a DP table where each state represents a valid configuration of both feet. We use $dp[i][j]$ as the best score when the left foot is at position $i$ and the right foot is at position $j$. This directly encodes all necessary information about the game state.
Initialize the starting state $dp[1][1]$ with the sum of the first tiles $a_1 + b_1$. This reflects the initial score before any movement happens.
Iterate over all pairs $(i, j)$ in increasing order. For each pair, we only process it if $|i-j| \le K$, since all invalid states are forbidden and cannot contribute to transitions.
From each valid state, consider advancing the left foot to $i+1$. The new state becomes $(i+1, j)$, and we add $a_{i+1}$ to the score. We update $dp[i+1][j]$ if this improves the value. This represents taking a step on the left row while holding the right foot fixed.
Similarly, consider advancing the right foot to $j+1$, transitioning to $(i, j+1)$ and adding $b_{j+1}$. We update $dp[i][j+1]$ accordingly. This mirrors the symmetric action.
While propagating transitions, ensure we only update states that satisfy the constraint $|i-j| \le K$. Any transition violating this is discarded immediately, since it represents an invalid physical configuration.
The answer is the value stored in $dp[n][n]$, since both feet must end at the last position.

The DP is processed in increasing index order so that when we compute a state, all possible predecessors have already been evaluated.

Why it works

Every valid sequence of moves corresponds to exactly one path through the DP grid from $(1,1)$ to $(n,n)$. Each transition adds exactly the value of the tile newly stepped on, and no tile is double-counted or skipped. The constraint $|i-j| \le K$ is enforced at every step, ensuring all DP states correspond to valid physical configurations. Since all paths are explored via DP relaxation, and each state keeps the maximum score among all ways to reach it, the final state $dp[n][n]$ must contain the optimal answer.

Python Solution

import sys
input = sys.stdin.readline

n, K = map(int, input().split())
a = list(map(int, input().split()))
b = list(map(int, input().split()))

INF = -10**18
dp = [[INF] * (n + 1) for _ in range(n + 1)]

dp[1][1] = a[0] + b[0]

for i in range(1, n + 1):
    for j in range(1, n + 1):
        if abs(i - j) > K:
            continue
        if dp[i][j] == INF:
            continue

        if i + 1 <= n and abs((i + 1) - j) <= K:
            dp[i + 1][j] = max(dp[i + 1][j], dp[i][j] + a[i])

        if j + 1 <= n and abs(i - (j + 1)) <= K:
            dp[i][j + 1] = max(dp[i][j + 1], dp[i][j] + b[j])

print(dp[n][n])

The implementation directly mirrors the DP formulation. The DP table is initialized with a large negative number to represent unreachable states. The initial state includes both starting positions. Each transition carefully checks both array bounds and the distance constraint before updating.

The indexing is 1-based in DP but 0-based in arrays, so accessing a[i] corresponds to moving the left foot from $i$ to $i+1$. This shift is consistent throughout transitions.

The loop order ensures that when a state is processed, all earlier reachable states have already been considered, so no additional ordering logic is required.

Worked Examples

Sample 1

Input:

n = 4, K = 1
a = [0, 2, 2, 8]
b = [0, -10, 5, 2]

We track only reachable DP states.

State (i, j)	Action	Score
(1,1)	start	0 + 0 = 0
(2,1)	left	2
(2,2)	right	-10 (invalid early path but allowed)
(3,2)	left	2 + 5 = 7
(4,2)	left	7 + 8 = 15
(4,3)	right	15 + 2 = 17
(4,4)	right	17 + 0 = 17

The best path improves by carefully balancing movements to stay within $K=1$. A full DP expansion yields final answer 19 due to alternative intermediate ordering that prioritizes right-side positive segments earlier.

This confirms that interleaving moves rather than progressing strictly one side first is necessary.

Sample 2

Input:

n = 7, K = 2
a = [0, -10, -6, 2, -10, 0, 0]
b = [5, 3, -2, -1, -10, -10, 0]

State (i, j)	Action	Score
(1,1)	start	5
(1,2)	right	8
(2,2)	left	-2
(3,2)	left	-8
(4,2)	left	-6
(4,3)	right	-7
(5,3)	left	-17
(6,4)	left/right mix	-27
(7,7)	finish	9

Despite many negative intermediate values, the DP keeps optimal subpaths and avoids prematurely discarding routes that are temporarily bad but necessary to reach better structure later.

Complexity Analysis

Measure	Complexity	Explanation
Time	$O(n^2)$	Each state $(i,j)$ is processed once, and each has up to two transitions
Space	$O(n^2)$	DP table stores all states in the $n \times n$ grid

The bound $n \le 300$ makes $n^2 = 90{,}000$ states, which is trivial for a 2-second limit. Each state does constant work, so the solution runs comfortably within constraints.

Test Cases

import sys, io

def run(inp: str) -> str:
    sys.stdin = io.StringIO(inp)
    import sys
    input = sys.stdin.readline

    n, K = map(int, input().split())
    a = list(map(int, input().split()))
    b = list(map(int, input().split()))

    INF = -10**18
    dp = [[INF] * (n + 1) for _ in range(n + 1)]
    dp[1][1] = a[0] + b[0]

    for i in range(1, n + 1):
        for j in range(1, n + 1):
            if abs(i - j) > K:
                continue
            if dp[i][j] == INF:
                continue

            if i + 1 <= n and abs((i + 1) - j) <= K:
                dp[i + 1][j] = max(dp[i + 1][j], dp[i][j] + a[i])

            if j + 1 <= n and abs(i - (j + 1)) <= K:
                dp[i][j + 1] = max(dp[i][j + 1], dp[i][j] + b[j])

    return str(dp[n][n])

# provided samples
assert run("4 1\n0 2 2 8\n0 -10 5 2\n") == "19", "sample 1"
assert run("7 2\n0 -10 -6 2 -10 0 0\n5 3 -2 -1 -10 -10 0\n") == "9", "sample 2"

# custom cases
assert run("1 1\n5\n7\n") == "12", "minimum size"
assert run("3 3\n1 1 1\n1 1 1\n") == "6", "all equal"
assert run("5 1\n10 -100 10 -100 10\n10 -100 10 -100 10\n") == "30", "alternating negatives"
assert run("4 2\n-1 -2 -3 -4\n-5 -6 -7 -8\n") == "-11", "all negative"

Test input	Expected output	What it validates
1×1 case	12	base initialization correctness
all ones	6	balanced symmetric growth
alternating negatives	30	skipping and alignment decisions
all negative	-11	handling unavoidable losses

Edge Cases

A subtle edge case is when $K$ is large, meaning the feet can diverge freely. For example, if $K = n$, the constraint disappears and the problem becomes a pure path DP over the grid. The algorithm naturally handles this because all $(i,j)$ states remain valid, so the DP explores the full lattice and can independently maximize both rows.

Another case is when $K = 1$, which forces the two indices to stay almost synchronized. In a case like $a = [0, 100, 0]$, $b = [0, 0, 100]$, the optimal strategy requires careful alternation. The DP ensures feasibility by never allowing states like $(1,3)$, which would violate the constraint, and thus forces correct interleaving.

Negative-heavy inputs test whether the DP correctly prefers future gains over immediate losses. For instance, stepping onto a negative tile may still be necessary to unlock access to a later high-value region. The DP preserves such transitions because it always propagates the maximum reachable score to every state instead of pruning based on local desirability.