Minimum Skips to Arrive at Meeting On Time - Solution & Explanation

Problem Overview: You travel through multiple roads with known distances at a constant speed. After each road, you normally wait until the next integer hour before continuing. You can skip this waiting a limited number of times. The task is to determine the minimum number of skips required to reach the meeting within hoursBefore.

The challenge comes from the rounding rule: if you do not skip, the travel time after each road is rounded up to the next integer hour. This rounding accumulates delays, so choosing the right segments to skip becomes a classic optimization problem.

Approach 1: Dynamic Programming (O(n²) time, O(n²) space)

The standard solution models the problem using dynamic programming. Let dp[i][k] represent the minimum time needed to travel the first i roads using exactly k skips. For each road, you have two choices: do not skip and round the cumulative time up using ceil(), or skip and keep the fractional time. The transition compares these two options and stores the smaller result.

Iterate through all roads and possible skip counts. After filling the table, find the smallest k such that dp[n][k] ≤ hoursBefore. This approach directly simulates the waiting rule and guarantees the optimal solution. It works well because the number of roads is small enough to allow an O(n²) DP state space.

Approach 2: Optimized Dynamic Programming (O(n²) time, O(n) space)

The same DP idea can be implemented with rolling arrays to reduce memory. Instead of storing the full matrix, keep only the previous row while iterating through the roads. Each iteration updates skip states from high to low to avoid overwriting values. The logic stays the same, but memory drops from O(n²) to O(n).

This version is typically preferred in production or interviews when space efficiency matters. The algorithm still iterates through roads and skip counts, performing rounding calculations after each segment.

Approach 3: Greedy with Priority Structure (O(n log n) time, O(n) space)

An alternative perspective focuses on the waiting time introduced by rounding. Every time you round up to the next integer hour, you effectively add a delay equal to the fractional remainder. Skipping a wait removes that delay. The strategy becomes selecting the largest rounding penalties to skip.

While iterating through the roads, track fractional delays and push them into a structure such as a max-heap. When total travel time exceeds the allowed limit, remove the largest delay (simulate skipping that wait). This greedy strategy minimizes the number of skips needed to stay within the deadline.

This method relies on insights about rounding penalties rather than explicitly enumerating DP states. It can be faster in practice but is trickier to derive during interviews.

Recommended for interviews: The dynamic programming solution is what most interviewers expect. It clearly models the decision of skipping versus waiting and demonstrates strong state design. Understanding how rounding affects cumulative travel time is the key insight. The greedy optimization is impressive but less commonly required unless the interviewer pushes for alternative strategies involving arrays and scheduling trade‑offs.