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3594. Minimum Time to Transport All Individuals
Description
You are given n individuals at a base camp who need to cross a river to reach a destination using a single boat. The boat can carry at most k people at a time. The trip is affected by environmental conditions that vary cyclically over m stages.
Each stage j has a speed multiplier mul[j]:
- If
mul[j] > 1, the trip slows down. - If
mul[j] < 1, the trip speeds up.
Each individual i has a rowing strength represented by time[i], the time (in minutes) it takes them to cross alone in neutral conditions.
Rules:
- A group
gdeparting at stagejtakes time equal to the maximumtime[i]among its members, multiplied bymul[j]minutes to reach the destination. - After the group crosses the river in time
d, the stage advances byfloor(d) % msteps. - If individuals are left behind, one person must return with the boat. Let
rbe the index of the returning person, the return takestime[r] × mul[current_stage], defined asreturn_time, and the stage advances byfloor(return_time) % m.
Return the minimum total time required to transport all individuals. If it is not possible to transport all individuals to the destination, return -1.
Example 1:
Input: n = 1, k = 1, m = 2, time = [5], mul = [1.0,1.3]
Output: 5.00000
Explanation:
- Individual 0 departs from stage 0, so crossing time =
5 × 1.00 = 5.00minutes. - All team members are now at the destination. Thus, the total time taken is
5.00minutes.
Example 2:
Input: n = 3, k = 2, m = 3, time = [2,5,8], mul = [1.0,1.5,0.75]
Output: 14.50000
Explanation:
The optimal strategy is:
- Send individuals 0 and 2 from the base camp to the destination from stage 0. The crossing time is
max(2, 8) × mul[0] = 8 × 1.00 = 8.00minutes. The stage advances byfloor(8.00) % 3 = 2, so the next stage is(0 + 2) % 3 = 2. - Individual 0 returns alone from the destination to the base camp from stage 2. The return time is
2 × mul[2] = 2 × 0.75 = 1.50minutes. The stage advances byfloor(1.50) % 3 = 1, so the next stage is(2 + 1) % 3 = 0. - Send individuals 0 and 1 from the base camp to the destination from stage 0. The crossing time is
max(2, 5) × mul[0] = 5 × 1.00 = 5.00minutes. The stage advances byfloor(5.00) % 3 = 2, so the final stage is(0 + 2) % 3 = 2. - All team members are now at the destination. The total time taken is
8.00 + 1.50 + 5.00 = 14.50minutes.
Example 3:
Input: n = 2, k = 1, m = 2, time = [10,10], mul = [2.0,2.0]
Output: -1.00000
Explanation:
- Since the boat can only carry one person at a time, it is impossible to transport both individuals as one must always return. Thus, the answer is
-1.00.
Constraints:
1 <= n == time.length <= 121 <= k <= 51 <= m <= 51 <= time[i] <= 100m == mul.length0.5 <= mul[i] <= 2.0