LeetCode-in-Java

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:

  • position array: [0, 8, 10]
  • time array: [5, 11, 6]

Segment	Distance (km)	Time per km (min)	Segment Travel Time (min)
0 → 8	8	5	8 × 5 = 40
8 → 10	2	11	2 × 11 = 22

• 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:
  • position array: [0, 2, 3, 5]
  • time array: [8, 12, 3, 3]

Segment	Distance (km)	Time per km (min)	Segment Travel Time (min)
0 → 2	2	8	2 × 8 = 16
2 → 3	1	12	1 × 12 = 12
3 → 5	2	3	2 × 3 = 6

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

Constraints:

• 1 <= l <= 105
• 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

Solution

@SuppressWarnings({"unused", "java:S1172"})
public class Solution {
    public int minTravelTime(int l, int n, int k, int[] position, int[] time) {
        int[][][] dp = new int[n][k + 1][k + 1];
        for (int i = 0; i < n; i++) {
            for (int j = 0; j <= k; j++) {
                for (int m = 0; m <= k; m++) {
                    dp[i][j][m] = Integer.MAX_VALUE;
                }
            }
        }
        dp[0][0][0] = 0;
        for (int i = 0; i < n - 1; i++) {
            int currTime = 0;
            for (int curr = 0; curr <= k && curr <= i; curr++) {
                currTime += time[i - curr];
                for (int used = 0; used <= k; used++) {
                    if (dp[i][curr][used] == Integer.MAX_VALUE) {
                        continue;
                    }
                    for (int next = 0; next <= k - used && next <= n - i - 2; next++) {
                        int nextI = i + next + 1;
                        dp[nextI][next][next + used] =
                                Math.min(
                                        dp[nextI][next][next + used],
                                        dp[i][curr][used]
                                                + (position[nextI] - position[i]) * currTime);
                    }
                }
            }
        }
        int ans = Integer.MAX_VALUE;
        for (int curr = 0; curr <= k; curr++) {
            ans = Math.min(ans, dp[n - 1][curr][k]);
        }
        return ans;
    }
}

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