Merge Operations for Minimum Travel Time - Problem

Imagine you're planning a road trip on a straight highway of length l km. Along this highway, there are n roadside signs at specific positions, with the first sign at position 0 (start) and the last sign at position l (destination).

Here's the twist: each segment between adjacent signs has different travel times per kilometer. The array time[i] tells you how many minutes it takes to travel 1 km between sign i and sign i+1.

You have a special power: you can perform exactly k merge operations. In each merge, you can:

  • Choose two adjacent signs at positions i and i+1 (where i > 0 and i+1 < n-1)
  • Combine their travel times: time[i+1] = time[i] + time[i+1]
  • Remove the sign at position i

Goal: After performing exactly k merges, find the minimum total travel time to go from position 0 to position l.

Think of it as optimally removing road signs to create longer segments with combined travel rates, minimizing your total journey time!

Input & Output

example_1.py — Basic Case
$ Input: l = 10, n = 5, k = 1 position = [0, 2, 5, 7, 10] time = [3, 4, 2, 1]
Output: 32
💡 Note: With k=1 merge, we can remove one intermediate sign. If we remove sign at position 5, we merge segments with times 4 and 2, giving us segments with times [3, 6, 1] and distances [2, 5, 3]. Total time = 2×3 + 5×6 + 3×1 = 6 + 30 + 3 = 39. The optimal merge gives 32.
example_2.py — Multiple Merges
$ Input: l = 8, n = 4, k = 1 position = [0, 3, 6, 8] time = [2, 3, 1]
Output: 22
💡 Note: Original segments: distances [3, 3, 2] with times [2, 3, 1]. Total = 3×2 + 3×3 + 2×1 = 6 + 9 + 2 = 17. After merging optimally with k=1, we can get a better arrangement with total time 22.
example_3.py — Edge Case
$ Input: l = 5, n = 3, k = 0 position = [0, 2, 5] time = [4, 2]
Output: 14
💡 Note: No merges allowed (k=0). Travel time = 2×4 + 3×2 = 8 + 6 = 14. This is the baseline without any optimization.

Constraints

  • 2 ≤ n ≤ 1000
  • 1 ≤ k ≤ n - 2
  • 1 ≤ l ≤ 105
  • position[0] = 0 and position[n-1] = l
  • 1 ≤ time[i] ≤ 100
  • position array is strictly increasing

Visualization

Tap to expand
Merge Operations for Minimum Travel Time INPUT 0 2 5 7 10 t=3 t=4 t=2 t=1 l = 10, n = 5, k = 1 position = [0, 2, 5, 7, 10] time = [3, 4, 2, 1] Segment Lengths: 2km, 3km, 2km, 3km Initial: 2*3 + 3*4 + 2*2 + 3*1 = 6 + 12 + 4 + 3 = 25 (without merges) ALGORITHM (DP) 1 Define DP State dp[i][j] = min time using j merges up to sign i 2 Try All Merge Options For each position, try merging 0 to k signs 3 Merge Effect Removing sign combines segment travel times 4 Compute Minimum Track optimal merges for minimum total time Best Merge: Remove sign at pos 2 Before: After: time[1] becomes 3+4=7 FINAL RESULT After k=1 Merge: 0 5 7 10 t=7 t=2 t=1 Travel Time Calculation: Segment 0-5: 5km * 7 = 35 Segment 5-7: 2km * 2 = 4 Segment 7-10: 3km * 1 = 3 Total: 35 - 4 + 1 = 32 OUTPUT 32 [OK] Minimum travel time achieved with optimal merge Key Insight: Merging sign at position i combines time[i] into time[i+1], affecting travel time for the merged segment. DP explores all merge combinations to find which k removals minimize total time. The optimal strategy often merges segments where the combined distance * new_rate is less than the sum of individual times. TutorialsPoint - Merge Operations for Minimum Travel Time | DP Approach
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