3594. Minimum Time to Transport All Individuals

Problem Link

Description

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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 g departing at stage j takes time equal to the maximum time[i] among its members, multiplied by mul[j] minutes to reach the destination.
• After the group crosses the river in time d, the stage advances by floor(d) % m steps.
• If individuals are left behind, one person must return with the boat. Let r be the index of the returning person, the return takes time[r] × mul[current_stage], defined as return_time, and the stage advances by floor(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 = 5.00 minutes.
• All team members are now at the destination. Thus, the total time taken is 5.00 minutes.

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 = 8.00 minutes. The stage advances by floor(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 = 1.50 minutes. The stage advances by floor(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 = 5.00 minutes. The stage advances by floor(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 = 14.50 minutes.

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 <= 12
• 1 <= k <= 5
• 1 <= m <= 5
• 1 <= time[i] <= 100
• m == mul.length
• 0.5 <= mul[i] <= 2.0

Solution

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Python3

class Solution:
    def minTime(self, n: int, k: int, m: int, time: List[int], mul: List[float]) -> float:
        seen = dict()
        pq = [(0.0, 0, 0, 0)] # (distance, where, mask, stage)
        full_mask = (1 << n) - 1
 
        while pq:
            t, p, mask, stage = heappop(pq)
            state = (p, mask, stage)
 
            if state in seen and t >= seen[state]:
                continue
            
            seen[state] = t
 
            if mask == full_mask and p == 1:
                return round(t, 5)
            
            # boat
            if p == 0:
                boat = [i for i in range(n) if mask & (1 << i) == 0]
                for take in range(1, min(k, len(boat)) + 1):
                    for group in combinations(boat, take):
                        cross = max(time[i] for i in group) * mul[stage]
                        newStage = (stage + floor(cross)) % m
                        newMask = mask
                        for i in group:
                            newMask |= (1 << i)
                        heappush(pq, (t + cross, 1, newMask, newStage))
 
            else:
                # destination
                dst = [i for i in range(n) if mask & (1 << i)]
                for i in dst:
                    returnTime = time[i] * mul[stage]
                    newStage = (stage + floor(returnTime)) % m
                    newMask = mask - (1 << i)
                    heappush(pq, (t + returnTime, 0, newMask, newStage))
 
        return -1.0