The Minimum Skips to Arrive at Meeting On Time problem gives you a list of road distances and a travel speed. Normally, after each road you must rest (which adds a fixed rest time). You can "skip" a rest on any road, but you want to minimize the number of skips while arriving at the destination within a given time limit. This Minimum Skips to Arrive at Meeting On Time coding problem is a hard dynamic programming problem involving floating point precision traps.
Google asks this problem because it tests sophisticated DP reasoning combined with precision awareness. The naive approach uses floating point, which causes incorrect comparisons due to rounding. The key insight — using integer arithmetic by scaling all values — is what separates strong candidates. The array and dynamic programming interview pattern is central, but the precision subtlety elevates this to a truly challenging question.
Define dp[i][j] = minimum time to travel i roads with exactly j skips. Recurrence:
dp[i][j] = dp[i-1][j-1] + dist[i-1]/speed (no rest added).dp[i][j] = ceil(dp[i-1][j] + dist[i-1]/speed) (rest added by rounding up to next integer multiple).To avoid floating point, multiply all times by speed and work entirely in integers. Final answer: smallest j where dp[n][j] ≤ hoursBefore * speed.
Distances: [1, 3, 2], speed = 4, hoursBefore = 2. With 0 skips: times after each road: ceil(1/4)=1hr, ceil(1+3/4)=2hr → 2+2/4=ceil=3hr. Total > 2hrs. With 1 skip: skip rest after first road. Time = 1/4 + 3/4 = 1hr, then ceil(1+2/4)=2hr. Exactly 2hrs. Minimum skips = 1.
speed to convert to integer arithmetic.When a DP problem involves division and ceiling, always consider scaling up to integers to avoid floating point entirely. Multiply all relevant values by the divisor and work in integer space. This eliminates rounding errors that cause WA on edge cases. Practice identifying "precision trap" problems — they're relatively rare but come up in Google-level interviews and require awareness that going pure-integer is sometimes the only reliable approach.