Minimum Time to Transport All Individuals - Solution & Explanation

Problem 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.