LeetCode 3538 - Merge Operations for Minimum Travel Time

This problem presents a road of length l kilometers, segmented by n signs at strictly increasing positions, where the first sign is at the start of the road (position[0] = 0) and the last sign is at the end of the road (position[n-1] = l).

leetcodehardarraydynamic-programmingprefix-sum

LeetCode Problem 3538

Difficulty: 🔴 Hard
Topics: Array, Dynamic Programming, Prefix Sum

Solution

Problem Understanding

This problem presents a road of length l kilometers, segmented by n signs at strictly increasing positions, where the first sign is at the start of the road (position[0] = 0) and the last sign is at the end of the road (position[n-1] = l). Each segment between consecutive signs has a corresponding travel time per kilometer given in the time array. Your goal is to reduce the total travel time along the road by performing exactly k merge operations.

A merge operation allows combining two adjacent signs (not the first or last) into a single segment. When two segments are merged, the resulting segment’s travel time per km is the sum of the two merged segments' times. This affects the total travel time because the new segment will cover a longer distance but at a higher travel rate.

The input consists of integers l, n, and k, plus arrays position and time. The output is the minimum total travel time in minutes after exactly k merges.

Constraints indicate that n is relatively small (up to 50), and k is very small (up to 10). These limits suggest a dynamic programming solution is feasible. Edge cases to consider include k = 0 (no merges), k = n-2 (maximum merges possible without touching first or last signs), and segments with identical travel times.

The challenge is to choose which k merges minimize the total travel time across variable-length segments with variable travel times, balancing the trade-off between longer distances and higher travel rates.

Approaches

The brute-force approach is to try every possible combination of k merges among the n-2 eligible signs. For each combination, compute the total travel time and take the minimum. This works correctly but becomes infeasible quickly because the number of ways to pick k merges from n-2 signs grows combinatorially. For example, C(48, 10) is over 400 million possibilities, which is far too large for practical computation.

The optimal approach leverages dynamic programming with prefix sums. The key observation is that the total travel time can be expressed as a function of segment distances and travel times, and merges always combine consecutive segments. We can define dp[i][j] as the minimum travel time using the first i segments with exactly j merges. By iterating over possible merge positions for each segment and building up from smaller subproblems, we can compute the minimum total travel time efficiently. Precomputing segment distances with prefix sums allows constant-time access to any merged segment length, avoiding repeated summation.

Approach Time Complexity Space Complexity Notes
Brute Force O(C(n-2,k) * n) O(n) Enumerates all combinations of merges, too slow for n ~ 50
Dynamic Programming O(n^2 * k) O(n * k) Uses prefix sums and DP table to track minimal travel time for each merge count

Algorithm Walkthrough

  1. Compute segment distances: Compute the distances between consecutive positions using distance[i] = position[i+1] - position[i]. These distances remain fixed throughout the merges.
  2. Compute prefix sums of distances: Create a prefix sum array to allow O(1) calculation of total distance for any consecutive segment range.
  3. Initialize DP table: Define dp[i][j] as the minimal total travel time using the first i segments with exactly j merges. Initialize dp[0][0] = 0 and all other entries to infinity.
  4. Fill DP table: Iterate through all segments i and merge counts j. For each possible previous merge position prev, compute the total travel time if the last i-prev segments are merged into one segment, adding its contribution total_distance * merged_time to dp[prev][j-merged_count].
  5. Return result: The final answer is dp[n-1][k], representing the minimal total travel time after exactly k merges for all segments.

Why it works: This DP maintains the invariant that dp[i][j] always stores the minimum total travel time for the first i segments using j merges. By considering all possible merge ranges, the algorithm ensures that every legal combination of k merges is evaluated indirectly, without enumerating every combination explicitly. This problem is asking us to minimize the total travel time along a straight road after performing a fixed number of merge operations on road signs. The road has length l, and there are n signs positioned along it, starting at 0 and ending at l. Between consecutive signs, there is a "time per km" value that determines how long it takes to traverse that segment.

The input arrays position and time describe the placement and cost per km between the signs. A merge operation allows us to combine two adjacent segments into one by summing their travel times, but we must remove the earlier sign to maintain the integrity of the array. The goal is to select exactly k merges to minimize the total travel time across all segments.

The constraints suggest that n is relatively small (≤50) and k is even smaller (≤10). This indicates that a solution involving dynamic programming or careful combinatorial exploration is feasible. Edge cases that could trip up naive implementations include k = 0 (no merges allowed), k = n-2 (almost all inner signs must be merged), and segments with uniform or maximum time values, which can affect optimal merge selection.

The key challenge is that the cost of a merge is not just local - merging earlier segments affects distances and times downstream - so a simple greedy approach may fail. The problem is essentially an optimization over sequences, which hints at dynamic programming.

Approaches

The brute-force approach would involve trying all combinations of k merges among the n-2 possible inner signs. For each combination, we would compute the total travel time. While correct, this is computationally infeasible because the number of combinations grows as the binomial coefficient C(n-2, k), which could be as large as 1,225 for n=50 and k=10. Each evaluation requires computing the total travel time in O(n) time, making the total complexity O(n * C(n-2, k)), which is too slow for repeated evaluation.

The key insight for an optimal approach is to model the problem as a dynamic programming problem over subarrays. Define dp[i][j] as the minimum total travel time considering the first i signs after performing j merges. For each segment, we can decide whether to merge with the previous segment or not, and use prefix sums of distances to compute the contribution of each segment efficiently. Since k is small, we only need to track up to k merges, which keeps the DP table manageable. This reduces the problem to a tractable O(n^2 * k) complexity.

Approach Time Complexity Space Complexity Notes
Brute Force O(n * C(n-2, k)) O(n) Enumerate all combinations of merges and compute travel time
Optimal O(n^2 * k) O(n * k) DP over subarrays with merge states and prefix sums

Algorithm Walkthrough

  1. Precompute segment distances using the position array. The distance between position[i] and position[i+1] is position[i+1] - position[i].
  2. Initialize a DP table dp[i][j] where i is the index of the current segment considered and j is the number of merges used. dp[i][j] stores the minimum total travel time up to segment i after j merges.
  3. Set base cases: dp[0][0] = 0, as no segments processed yet means zero travel time.
  4. Iterate through segments i from 1 to n-1. For each i, iterate through possible merge counts j from 0 to k.
  5. For each segment, consider all possible previous segments p to merge with, where merging i-p segments uses p merges. Update dp[i][j] by considering the new segment travel time after merging. The travel time is computed as the sum of time values multiplied by the corresponding distances for the merged segments.
  6. After filling the DP table, the answer is dp[n-1][k], the minimum travel time to reach the last segment using exactly k merges.

Why it works: The DP invariant ensures that for each prefix of segments and a given number of merges, we track the minimum achievable travel time. By considering all valid merge options for each segment and propagating the minimum cost, we guarantee the global optimum is found.

Python Solution

from typing import List

class Solution:
    def minTravelTime(self, l: int, n: int, k: int, position: List[int], time: List[int]) -> int:
        import math

        # Step 1: Compute distances between positions
        distances = [position[i+1] - position[i] for i in range(n-1)]
        
        # Step 2: Initialize dp table
        dp = [[math.inf] * (k+1) for _ in range(n)]
        dp[0][0] = 0

        # Step 3: Dynamic programming
        for i in range(1, n):
            for j in range(min(i, k)+1):
                total_distance = 0
                merged_time = 0
                for p in range(i-1, -1, -1):
                    total_distance += distances[p]
                    merged_time += time[p+1]
                    if j - (i - p - 1) >= 0:
                        dp[i][j] = min(dp[i][j], dp[p][j - (i - p - 1)] + total_distance * merged_time)

        return dp[n-1][k]

The Python code first calculates distances, then sets up a DP table. For each segment, it considers merging with previous segments in all possible ways that do not exceed k merges. The innermost loop computes the total distance and merged travel time efficiently using cumulative addition. The final DP entry contains the minimum travel time. dist = [position[i+1] - position[i] for i in range(n-1)] INF = float('inf') dp = [[INF] * (k+1) for _ in range(n)] dp[0][0] = 0

    for i in range(1, n):
        for merges in range(0, min(i, k)+1):
            total_time = 0
            for take in range(i, 0, -1):
                total_time += time[take] * dist[take-1]
                used_merges = i - take
                if used_merges <= merges:
                    dp[i][merges] = min(dp[i][merges], dp[take-1][merges - used_merges] + total_time)
    return dp[n-1][k]


The Python implementation first calculates segment distances, then sets up a DP table with dimensions `[n][k+1]`. It iterates over all segments and considers all ways to merge previous segments into the current segment, updating the minimum travel time. The nested loop over `take` allows us to sum segment travel times after merging, and the inner condition ensures we do not exceed the allowed number of merges. Finally, the result is found at `dp[n-1][k]`.

## Go Solution

```go
import "math"

func minTravelTime(l int, n int, k int, position []int, time []int) int {
    distances := make([]int, n-1)
    for i := 0; i < n-1; i++ {
        distances[i] = position[i+1] - position[i]
    }

func minTravelTime(l int, n int, k int, position []int, time []int) int {
    dist := make([]int, n-1)
    for i := 0; i < n-1; i++ {
        dist[i] = position[i+1] - position[i]
    }

    INF := 1 << 60
    dp := make([][]int, n)
    for i := range dp {
        dp[i] = make([]int, k+1)
        for j := range dp[i] {
            dp[i][j] = math.MaxInt64
            dp[i][j] = INF
        }
    }
    dp[0][0] = 0

    for i := 1; i < n; i++ {
        for j := 0; j <= min(i, k); j++ {
            totalDistance := 0
            mergedTime := 0
            for p := i - 1; p >= 0; p-- {
                totalDistance += distances[p]
                mergedTime += time[p+1]
                if j-(i-p-1) >= 0 {
                    if dp[p][j-(i-p-1)]+totalDistance*mergedTime < dp[i][j] {
                        dp[i][j] = dp[p][j-(i-p-1)] + totalDistance*mergedTime
        for merges := 0; merges <= min(i, k); merges++ {
            totalTime := 0
            for take := i; take > 0; take-- {
                totalTime += time[take] * dist[take-1]
                usedMerges := i - take
                if usedMerges <= merges {
                    if dp[take-1][merges - usedMerges] + totalTime < dp[i][merges] {
                        dp[i][merges] = dp[take-1][merges - usedMerges] + totalTime
                    }
                }
            }
        }
    }
    return dp[n-1][k]
}

func min(a, b int) int {
    if a < b {
        return a
    }
    if a < b { return a }
    return b
}

The Go implementation mirrors the Python version. The main difference is explicit initialization of slices and the use of math.MaxInt64 instead of math.inf. The cumulative distance and merged time calculations are identical.

Worked Examples

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

Compute distances: [3,5,2]

DP iterations find that merging segments 1 and 2 gives the minimal total travel time of 62:

Segment Distances Times Merge Travel Time
0→1 3 5 no 15
1→2 5 8+3=11 merged 55
2→3 2 6 no 12

Total = 62

Example 2: l = 5, n = 5, k = 1, position = [0,1,2,3,5], time = [8,3,9,3,3]

Distances: [1,1,1,2]

Optimal merge is segments 1 and 2: 1 km * 3+9=12, total travel time = 16+12+6=34 The Go solution mirrors the Python DP structure, using slices for distances and the DP table. Go requires explicit initialization of the DP table with a large INF value. The nested loops are identical in logic, iterating over segments and possible merges. Helper min function simplifies comparisons.

Worked Examples

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

Step 1: Compute distances: [3,5,2].

Step 2: Initialize dp[0][0] = 0.

Step 3: Iterate i=1, only 0 merges possible. dp[1][0] = dp[0][0] + 8*5 = 40.

Step 4: i=2, consider 0 and 1 merge. Merging segments 1 and 2 (8 + 3) with distance 5 gives travel time 11*2 = 22, total = 40 + 22 = 62. dp[3][1] = 62.

Step 5: Return dp[3][1] = 62.

Example 2: l = 5, n = 5, k = 1, position = [0,1,2,3,5], time = [8,3,9,3,3]

Step 1: Compute distances: [1,1,1,2].

Step 2: Initialize dp[0][0] = 0.

Step 3: Iterate i=1..4, consider merging segments 1 and 2, travel time = 12*1 = 12, cumulative total = 16 + 12 + 6 = 34.

Step 4: Return dp[4][1] = 34.

Complexity Analysis

Measure Complexity Explanation
Time O(n^2 * k) Outer loop over n, inner loop over previous segments, and merge count up to k
Space O(n * k) DP table stores minimal travel times for each segment and merge count

Because n ≤ 50 and k ≤ 10, this complexity is acceptable for the given constraints.

Test Cases

sol = Solution()

# Provided examples
assert sol.minTravelTime(10, 4, 1, [0,3,8,10], [5,

| Time | O(n^2 * k) | |