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Merge Operations for Minimum Travel Time - Solution & Explanation

HardArrayDynamic ProgrammingPrefix Sum5 min readAsked at: Google
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Problem Statement

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​​​​​​

Approach Overview

Problem 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.

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Detailed Complexity Analysis

ApproachTimeSpaceWhen to Use
Brute Force Merge SimulationO(2^n)O(n)Useful only for conceptual understanding or very small arrays
Interval Dynamic ProgrammingO(n^3)O(n^2)General DP solution when computing range costs directly
DP with Prefix Sum OptimizationO(n^2)O(n^2)Best practical approach when merge cost depends on subarray sums

Video Solution

Merge Operations for Minimum Travel Time | Thought Process | Leetcode 3538 | codestorywithMIK • codestorywithMIK • 4,055 views views

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Frequently Asked Questions

Is Merge Operations for Minimum Travel Time easy or hard?
Merge Operations for Minimum Travel Time is classified as a Hard problem because it requires interval dynamic programming and careful optimization with prefix sums. Understanding how to structure DP states and compute range costs efficiently is the main challenge.
Merge Operations for Minimum Travel Time Python/Java solution
The typical implementation builds a prefix sum array and a 2D DP table. The algorithm iterates over interval lengths, computes dp transitions for each split point, and updates the minimum cost. This approach can be implemented efficiently in Python, Java, C++, and Go.
How to solve Merge Operations for Minimum Travel Time in O(n^2)?
Use interval dynamic programming and precompute a prefix sum array for the input. Define dp[i][j] as the minimum cost to merge segments from i to j. Iterate over increasing interval lengths and try all split points while computing merge costs using prefix sums in constant time.
What is the best approach for Merge Operations for Minimum Travel Time?
The most efficient approach uses interval dynamic programming combined with prefix sums. DP stores the minimum cost for merging every subarray, while prefix sums compute the merge cost of a range in O(1). This reduces redundant work and leads to an overall time complexity around O(n^2) with O(n^2) space.
Is Merge Operations for Minimum Travel Time asked at Google/Amazon/Meta?
Problems involving interval dynamic programming and merge cost optimization commonly appear in interviews at companies like Google, Amazon, and Meta. Variants of this problem are related to classic interview topics such as stone merging, matrix chain multiplication, and optimal partitioning.
What data structure is used in Merge Operations for Minimum Travel Time?
The solution primarily relies on arrays for storing prefix sums and a 2D dynamic programming table. Prefix sums allow constant-time subarray cost queries, while the DP table tracks the optimal cost for merging every interval.
What is the time complexity of Merge Operations for Minimum Travel Time?
The optimal solution using dynamic programming with prefix sums runs in O(n^2) time and uses O(n^2) space. A simpler interval DP without prefix sum optimization can take O(n^3) time because each state must recompute the sum of elements in the range.

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