Implement `cheapest_round_trip(outbound, return_fares, min_gap)`. `outbound[i]` is the fare for departing on day `i`, and `return_fares[j]` is the fare for returning on day `j`. Choose one departure and one return such that `j - i >= min_gap`, and return the minimum total fare. The two fare arrays have the same length, `min_gap` is positive, and at least one valid pair exists. Target `O(n)` time and `O(n)` auxiliary space, or explain how to reduce the extra space while preserving linear time. ```hint Precompute future choices For each day, record the cheapest return fare available from that day through the end of the array. ``` ```hint Shift by the required stay When considering departure day `i`, combine its fare with the suffix minimum beginning at `i + min_gap`, not at `i`. ``` ### Discussion Extensions - How does the solution simplify when only the minimum total cost is required rather than the chosen dates? - How would you return deterministic dates when several pairs have the same minimum fare?