3538 - Merge Operations for Minimum Travel Time.

Hard

You are given a straight road of length l km, an integer n, an integer k ** , ** and two integer arrays, position and time, each of length n.

The array position lists the positions (in km) of signs in strictly increasing order (with position[0] = 0 and position[n - 1] = l).

Each time[i] represents the time (in minutes) required to travel 1 km between position[i] and position[i + 1].

You must perform exactly k merge operations. In one merge, you can choose any two adjacent signs at indices i and i + 1 (with i > 0 and i + 1 < n) and:

• Update the sign at index i + 1 so that its time becomes time[i] + time[i + 1].

• Remove the sign at index i.

Return the minimum total travel time (in minutes) to travel from 0 to l after exactly k merges.

Example 1:

Input: l = 10, n = 4, k = 1, position = 0,3,8,10, time = 5,8,3,6

Output: 62

Explanation:

• Merge the signs at indices 1 and 2. Remove the sign at index 1, and change the time at index 2 to 8 + 3 = 11.

• After the merge:

• Total Travel Time: 40 + 22 = 62, which is the minimum possible time after exactly 1 merge.

Example 2:

Input: l = 5, n = 5, k = 1, position = 0,1,2,3,5, time = 8,3,9,3,3

Output: 34

Explanation:

• Merge the signs at indices 1 and 2. Remove the sign at index 1, and change the time at index 2 to 3 + 9 = 12.

• After the merge:

• Total Travel Time: 16 + 12 + 6 = 34 ** , ** which is the minimum possible time after exactly 1 merge.

Constraints:

• <code>1 <= l <= 10⁵</code>

• 2 <= n <= min(l + 1, 50)

• 0 <= k <= min(n - 2, 10)

• position.length == n

• position[0] = 0 and position[n - 1] = l

• position is sorted in strictly increasing order.

• time.length == n

• 1 <= time[i] <= 100

• 1 <= sum(time) <= 100