The core insight is that two vehicles i and j collide when x_i(t) = x_j(t), which reduces to a quadratic in t with coefficients from the differences in their kinematics.
Model each vehicle's position as a quadratic function of time and find the earliest positive time when any two positions coincide. Use a sweep line algorithm with a priority queue of potential collision events, updating as vehicles cross. Alternatively, solve pairwise equations and take the minimum valid time.
Pro tip: Emphasize that you would first clarify assumptions (e.g., whether vehicles are points, whether collisions are elastic, and if simultaneous collisions are possible) and discuss the trade-offs between O(n^2) pairwise checking and O(n log n) sweep line approaches.
Ask about the range of n, whether accelerations can be zero or negative, and if multiple collisions at the same time need special handling. Confirm that vehicles are points and collisions occur when positions are equal.
For each pair of vehicles i and j, set their position equations equal: (1/2)a_i t^2 + v_i t + x_i = (1/2)a_j t^2 + v_j t + x_j. This yields a quadratic equation in t; solve for t > 0.
For small n, compute all pairwise collision times and take the minimum. For large n, use a sweep line approach: sort vehicles by initial position, maintain a priority queue of adjacent collision events, and update when vehicles cross.
Consider cases with no real roots, negative times, or multiple collisions at the same time. Validate with simple examples (e.g., two vehicles with constant velocities).
Discuss time and space complexity: O(n^2) for brute force, O(n log n) for sweep line. Mention that sweep line is more complex but scales better for large n.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
Relative motion between two disks is also constant velocity, so the distance squared between centers is a quadratic in t.
Model each pair of disks as a quadratic inequality in time and solve for the earliest t ≥ 0 where the distance between centers equals the sum of radii. Handle already-overlapping pairs by returning t=0 immediately, and use an epsilon tolerance to avoid false positives from floating-point errors.
Pro tip: Mention that you would use squared distances to avoid square roots and that you'd sort or use a priority queue if you need the earliest collision among all pairs, but for a single earliest time, a linear scan with early termination is sufficient.
For two disks with centers p1, p2 and radii r1, r2, collision occurs when |p1 + v1*t - (p2 + v2*t)|^2 ≤ (r1 + r2)^2. Expand to a quadratic in t: a*t^2 + b*t + c ≤ 0, where a = |v1 - v2|^2, b = 2*(p1 - p2)·(v1 - v2), c = |p1 - p2|^2 - (r1 + r2)^2.
Find the roots of a*t^2 + b*t + c = 0. If a ≈ 0 (relative velocities parallel or zero), handle the linear case. Otherwise, compute discriminant D = b^2 - 4ac. If D < 0, no collision; else, the earliest time is the smallest root ≥ 0, or 0 if c ≤ 0 (already overlapping).
Check if c ≤ 0 at t=0 (i.e., initial distance ≤ sum of radii). If so, return t=0 immediately for that pair. This avoids unnecessary root computation and correctly identifies immediate collisions.
Use an epsilon (e.g., 1e-9) for comparisons: treat D < -epsilon as no collision, D > epsilon as two roots, and |D| ≤ epsilon as a double root. Also, when checking c ≤ 0, use c ≤ epsilon to account for rounding errors. For the earliest time, consider only roots ≥ -epsilon and clamp small negative roots to 0.
For each pair of disks, compute the earliest collision time (or infinity if none). Return the minimum over all pairs. If any pair yields t=0, return 0 immediately. For efficiency, consider spatial partitioning if the number of disks is large.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.