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:
i + 1 so that its time becomes time[i] + time[i + 1].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.
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 |
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:
3 + 9 = 12.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 |
16 + 12 + 6 = 34, which is the minimum possible time after exactly 1 merge.
Constraints:
1 <= l <= 1052 <= n <= min(l + 1, 50)0 <= k <= min(n - 2, 10)position.length == nposition[0] = 0 and position[n - 1] = lposition is sorted in strictly increasing order.time.length == n1 <= time[i] <= 1001 <= sum(time) <= 100Problem Overview: You are given an array representing travel segments and can merge adjacent segments to reduce overall travel cost. The goal is to determine the minimum possible travel time after performing optimal merge operations. The challenge is deciding which segments to combine so that the cumulative cost stays minimal.
Approach 1: Brute Force Merge Simulation (Exponential Time, O(2^n) time, O(n) space)
The most direct idea tries every possible sequence of merge operations. For each pair of adjacent segments, merge them and recursively evaluate the remaining array. You compute the resulting travel time after every merge configuration and track the minimum. This method guarantees correctness because it explores all states, but the number of states grows exponentially as the array length increases. It mainly serves as a conceptual baseline showing that merge order affects the final result.
Approach 2: Dynamic Programming on Intervals (O(n^3) time, O(n^2) space)
Instead of recomputing results for the same subarrays repeatedly, store them using dynamic programming. Define dp[i][j] as the minimum travel time required to merge segments from index i to j. For each interval, try splitting it at every possible midpoint k. The total cost becomes dp[i][k] + dp[k+1][j] + cost(i,j), where cost(i,j) represents the travel cost of merging the entire range. This interval DP approach is similar to classic merge problems such as optimal matrix chain multiplication. While much faster than brute force, computing range costs repeatedly can still be expensive.
Approach 3: Dynamic Programming with Prefix Sum Optimization (O(n^2) time, O(n^2) space)
The optimal implementation precomputes cumulative sums using a prefix sum array so the merge cost of any interval (i, j) can be calculated in constant time. During the DP transition, you iterate through possible split points but avoid recalculating segment totals. This reduces redundant work and keeps the algorithm practical for larger inputs. The key insight is that every merge ultimately depends on the sum of the merged range, so prefix sums eliminate repeated iteration over the array.
Arrays are processed systematically from smaller intervals to larger ones, ensuring each state is solved only once. This approach combines array manipulation with interval dynamic programming to efficiently evaluate merge decisions.
Recommended for interviews: Start by describing the brute force idea to demonstrate understanding of the merge decision space. Then transition to interval dynamic programming with prefix sums. Interviewers expect the optimized DP solution because it shows familiarity with subproblem reuse, range cost calculation, and time complexity reduction from exponential to polynomial.
Solutions for this problem are being prepared.
Try solving it yourself| Approach | Time | Space | When to Use |
|---|---|---|---|
| Brute Force Merge Simulation | O(2^n) | O(n) | Useful only for conceptual understanding or very small arrays |
| Interval Dynamic Programming | O(n^3) | O(n^2) | General DP solution when computing range costs directly |
| DP with Prefix Sum Optimization | O(n^2) | O(n^2) | Best practical approach when merge cost depends on subarray sums |
Merge Operations for Minimum Travel Time | Thought Process | Leetcode 3538 | codestorywithMIK • codestorywithMIK • 3,789 views views
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