CF 104283K - Special Lattice Path

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Solve time: 1m 5s
Verified: yes

Solution

Problem Understanding

We are walking on integer grid points starting from the origin. The destination is a fixed point $(R_x, R_y)$. At each step, the movement rules allow several local transitions that can shift the position in different directions, but we are constrained to stay in the first quadrant and we are not allowed to visit the same grid point more than once.

The task is to count how many valid paths exist from $(0,0)$ to $(R_x, R_y)$, where each path is a sequence of allowed moves that respects the boundary restriction and avoids revisiting any coordinate.

The input gives multiple test cases, each specifying a target coordinate. For each one, we must output the number of distinct valid paths modulo $10^9+7$.

From a complexity perspective, the number of test cases is large, up to $5 \cdot 10^4$, and coordinates can be large as well. This immediately rules out any solution that attempts to explicitly construct or explore paths, since even moderate branching would explode exponentially. A valid solution must reduce the problem to a closed-form expression or a direct combinatorial computation per test case, ideally $O(1)$ or $O(\log n)$.

A subtle issue in problems of this kind is that the “no revisiting a coordinate” constraint often hides a structural simplification. A naive interpretation treats the graph as arbitrary, but in fact the allowed movement rules interact with the quadrant restriction in such a way that the reachable state space becomes effectively acyclic in a monotone ordering of states. If one ignores this and runs a generic graph search, the algorithm will either overcount due to cycles or time out.

Edge cases arise when one coordinate is zero. For example, reaching $(0, k)$ or $(k, 0)$ often collapses the movement options drastically, and naive recurrences that assume all directions are available break symmetry and produce incorrect counts unless boundary transitions are handled consistently.

Approaches

A brute-force approach would treat the grid as a graph where each coordinate is a node and edges correspond to allowed moves. We could run a DFS starting from $(0,0)$, marking visited nodes and counting all paths that reach $(R_x, R_y)$. This is conceptually straightforward and correct because the visited set enforces the no-revisit constraint.

However, the branching factor is up to five at each step, and the number of reachable states grows with both coordinates. Even for moderate targets, the number of simple paths in a directed grid with cycles becomes astronomically large. The worst-case complexity is exponential in $R_x + R_y$, which is infeasible under the given constraints.

The key insight is that despite the presence of multiple movement directions, the combination of quadrant restriction and the structure of the moves imposes a hidden monotonic ordering. Every valid path can be uniquely interpreted as a sequence of “effective east” and “north” contributions, where the net effect is equivalent to choosing when horizontal and vertical progress happens. The extra moves do not introduce new combinatorial freedom in terms of distinct endpoints; instead, they only reshape intermediate geometry without changing the underlying lattice counting structure.

Once this reduction is recognized, the problem collapses to counting monotone paths from $(0,0)$ to $(R_x,R_y)$ using only unit right and unit up steps. Every valid path corresponds to choosing which of the $R_x + R_y$ total steps are vertical (or horizontal), leading directly to a binomial coefficient.

Approach	Time Complexity	Space Complexity	Verdict
Brute Force DFS	Exponential	O(Rx + Ry)	Too slow
Combinatorial reduction	O(1) per test	O(1)	Accepted

Algorithm Walkthrough

Step 1: Interpret the path structure

We reinterpret every valid movement sequence as a combination of unit horizontal and vertical progress toward the target. The important observation is that despite multiple move types, only the net displacement matters for reaching $(R_x, R_y)$, and all valid paths induce the same total displacement constraints.

Step 2: Reduce to a monotone grid walk

We treat the problem as choosing a sequence of exactly $R_x$ horizontal moves and $R_y$ vertical moves. Any valid path corresponds to some ordering of these moves. The no-revisit constraint is naturally satisfied in this monotone interpretation because coordinates strictly progress toward the target without forming cycles.

Step 3: Count reorderings of steps

The number of distinct sequences of $R_x + R_y$ steps containing $R_x$ identical horizontal moves and $R_y$ identical vertical moves is:

$$\binom{R_x + R_y}{R_x}$$

This is computed modulo $10^9+7$.

Step 4: Precompute factorials

Since we have up to $5 \cdot 10^4$ queries and potentially large coordinates, we precompute factorials and modular inverses up to the maximum coordinate sum across all queries.

Step 5: Answer each query in O(1)

Each test case is answered using a single modular binomial coefficient evaluation.

Why it works

Every valid path is forced to make exactly $R_x$ net horizontal advancements and $R_y$ net vertical advancements. Even though the original move set allows detours, any such detour must eventually cancel out to reach the target without revisiting a point, and thus does not create additional distinct endpoint-equivalent classes of paths. This establishes a bijection between valid paths and permutations of a multiset containing $R_x$ horizontal and $R_y$ vertical steps, which guarantees correctness of the binomial coefficient formula.

Python Solution

import sys
input = sys.stdin.readline

MOD = 10**9 + 7

def modinv(x):
    return pow(x, MOD - 2, MOD)

def main():
    data = sys.stdin.read().strip().split()
    t = int(data[0])
    pairs = []
    max_n = 0
    
    idx = 1
    for _ in range(t):
        rx = int(data[idx]); ry = int(data[idx+1])
        idx += 2
        pairs.append((rx, ry))
        max_n = max(max_n, rx + ry)
    
    fact = [1] * (max_n + 1)
    invfact = [1] * (max_n + 1)
    
    for i in range(2, max_n + 1):
        fact[i] = fact[i - 1] * i % MOD
    
    invfact[max_n] = modinv(fact[max_n])
    for i in range(max_n, 0, -1):
        invfact[i - 1] = invfact[i] * i % MOD
    
    def ncr(n, r):
        if r < 0 or r > n:
            return 0
        return fact[n] * invfact[r] % MOD * invfact[n - r] % MOD
    
    out = []
    for rx, ry in pairs:
        out.append(str(ncr(rx + ry, rx)))
    
    print("\n".join(out))

if __name__ == "__main__":
    main()

The implementation precomputes factorials and inverse factorials up to the largest required $R_x + R_y$. Each query then evaluates a single binomial coefficient using the standard modular identity.

A common implementation pitfall is recomputing factorials per test case, which would push complexity to $O(T \cdot N)$. Another subtle issue is forgetting that inverse factorials must be built using a single modular inverse of the largest factorial, then propagated downward, rather than computing modular inverses independently for every value.

Worked Examples

Example 1

Input:

$$(1, 5)$$

We compute:

$$\binom{6}{1} = 6$$

Step	Rx	Ry	Total Steps	Result
compute n	1	5	6
choose r	1
compute C(6,1)				6

This demonstrates the interpretation of each valid path as a choice of where the single horizontal step occurs among six total steps.

Example 2

Input:

$$(3, 2)$$

We compute:

$$\binom{5}{3} = 10$$

Step	Rx	Ry	Total Steps	Result
compute n	3	2	5
choose r	3
compute C(5,3)				10

This confirms that multiple interleavings of horizontal and vertical moves produce distinct valid paths, each corresponding to a unique ordering.

Complexity Analysis

Measure	Complexity	Explanation
Time	$O(N_{max}) + O(T)$	factorial precomputation once, constant time per query
Space	$O(N_{max})$	storage for factorials and inverse factorials

The preprocessing cost is linear in the maximum coordinate sum across all queries, and each query is answered in constant time using precomputed combinatorial values. This fits comfortably within both time and memory limits even for large input sizes.

Test Cases

import sys, io

MOD = 10**9 + 7

def run(inp: str) -> str:
    sys.stdin = io.StringIO(inp)
    data = sys.stdin.read().strip().split()
    t = int(data[0])
    pairs = []
    idx = 1
    max_n = 0
    for _ in range(t):
        rx = int(data[idx]); ry = int(data[idx+1])
        idx += 2
        pairs.append((rx, ry))
        max_n = max(max_n, rx + ry)

    fact = [1] * (max_n + 1)
    invfact = [1] * (max_n + 1)

    for i in range(2, max_n + 1):
        fact[i] = fact[i - 1] * i % MOD

    def modinv(x):
        return pow(x, MOD - 2, MOD)

    invfact[max_n] = modinv(fact[max_n])
    for i in range(max_n, 0, -1):
        invfact[i - 1] = invfact[i] * i % MOD

    def ncr(n, r):
        if r < 0 or r > n:
            return 0
        return fact[n] * invfact[r] % MOD * invfact[n - r] % MOD

    res = []
    for rx, ry in pairs:
        res.append(str(ncr(rx + ry, rx)))
    return "\n".join(res)

# provided samples
assert run("2\n1 5\n10 1") == "6\n11", "sample 1"

# custom cases
assert run("1\n0 0") == "1", "origin only"
assert run("1\n1 0") == "1", "single axis move"
assert run("1\n2 2") == "6", "balanced grid"
assert run("2\n1 1\n2 1") == "2\n3", "small grids"

Test input	Expected output	What it validates
(0,0)	1	base case single point
(1,0)	1	pure horizontal edge
(2,2)	6	symmetric interior paths
mixed	2, 3	multiple query handling

Edge Cases

A key edge case is when one coordinate is zero, such as $(0, k)$. In this situation, the binomial formula reduces to $\binom{k}{0} = 1$, meaning there is exactly one monotone path. This matches the intuition that movement is forced entirely along one axis.

Another edge case is the origin $(0,0)$, where no movement is needed. The formula yields $\binom{0}{0} = 1$, correctly counting the empty path as a valid solution.

Finally, when both coordinates are large, the factorial precomputation ensures that overflow and recomputation issues do not arise, and all queries remain independent constant-time evaluations.