Watch 4 video solutions for Minimum Time to Transport All Individuals, a hard level problem involving Array, Bit Manipulation, Graph. This walkthrough by Programming Live with Larry has 640 views views. Want to try solving it yourself? Practice on FleetCode or read the detailed text solution.
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]:
mul[j] > 1, the trip slows down.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:
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.d, the stage advances by floor(d) % m steps.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:
5 × 1.00 = 5.00 minutes.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:
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 = 2.2 × 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.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 = 2.8.00 + 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:
-1.00.
Constraints:
1 <= n == time.length <= 121 <= k <= 51 <= m <= 51 <= time[i] <= 100m == mul.length0.5 <= mul[i] <= 2.0Problem Overview: You are given a transportation network and multiple individuals located at different points. Moving people across the graph takes time depending on the path taken. The goal is to compute the minimum total time required to transport every individual to the required destination while respecting movement constraints.
Approach 1: Brute Force State Exploration (Exponential Time)
A direct strategy is to model every possible configuration of transported individuals. Represent the transported set using a bitmask where the i-th bit indicates whether person i has already been transported. From each state, try transporting additional individuals through every possible route and compute the accumulated travel time. This creates a state graph of size roughly O(V * 2^k), where k is the number of individuals. Exploring all transitions without prioritization leads to exponential work and repeated processing of slower paths. Time complexity grows to roughly O(E * 2^k) and space complexity is O(V * 2^k). This approach demonstrates the state representation but is rarely efficient enough for larger inputs.
Approach 2: Dijkstra on (Node, Bitmask) State Graph (Optimal)
The key insight is that each configuration can be treated as a node in a larger graph: (current_location, transported_mask). Moving along an edge updates the travel time while the bitmask updates whenever a new individual is picked up or delivered. Since each transition has a cost, the problem becomes a shortest path search across this expanded state space. Running Dijkstra's algorithm with a priority queue always expands the state with the smallest accumulated time.
This approach combines graph traversal, heap (priority queue), and bitmask dynamic state compression. Each state is processed once with its minimal cost, and transitions update neighboring states by pushing them into the heap if a shorter time is found. The number of states is V * 2^k, and each transition is processed through a priority queue operation.
The resulting time complexity is O((V * 2^k + E * 2^k) log(V * 2^k)), typically simplified to O((V * 2^k) log(V * 2^k)). Space complexity is O(V * 2^k) to store the best distance for each state. This method efficiently finds the minimum time required to transport all individuals.
Recommended for interviews: Interviewers expect the shortest‑path formulation with a (node, bitmask) state and Dijkstra's algorithm. The brute force idea helps demonstrate how the state space forms, but recognizing that the problem reduces to a weighted shortest path with bitmask compression shows strong algorithmic maturity.
| Approach | Time | Space | When to Use |
|---|---|---|---|
| Brute Force State Exploration | O(E * 2^k) | O(V * 2^k) | Useful for understanding how transporting states map to bitmask configurations |
| BFS on State Graph (Unweighted Variant) | O(V * 2^k + E * 2^k) | O(V * 2^k) | Applicable only if every move has equal cost |
| Dijkstra with Bitmask State Compression | O((V * 2^k) log(V * 2^k)) | O(V * 2^k) | General weighted graph case and the optimal interview solution |