LeetCode 3594 - Minimum Time to Transport All Individuals
This problem is a shortest-path optimization over a highly constrained state space where you are repeatedly moving groups of people across a river using a single boat.
Difficulty: 🔴 Hard
Topics: Array, Bit Manipulation, Graph Theory, Heap (Priority Queue), Shortest Path, Bitmask
Solution
Problem Understanding
This problem is a shortest-path optimization over a highly constrained state space where you are repeatedly moving groups of people across a river using a single boat. The twist is that the travel time is not fixed: it depends both on the slowest person in the group and on a cyclic environmental multiplier that changes over time based on how long previous trips took.
Each individual has a base crossing time time[i]. When a group travels together, the group’s effective speed is dictated by the slowest member, meaning the group travel time in neutral conditions is max(time[i]) over the selected subset. This time is then scaled by a stage-dependent multiplier mul[j], where j is the current environmental stage.
The environment evolves dynamically. After every crossing or return trip, the stage index changes by floor(duration) % m, meaning even fractional time still advances the system based on the integer portion of the elapsed time.
The goal is to move all individuals from the base camp to the destination using a boat of capacity k, minimizing total time. However, if people remain at the base, the boat must return with exactly one person, making this similar to classic river crossing problems but with weighted time dynamics and cyclic state transitions.
The input size is small: n <= 12, k <= 5, m <= 5. This strongly suggests a bitmask-based shortest path solution rather than greedy or DP over subsets alone.
Edge cases that matter include situations where k = 1 and n > 1, which often makes progress impossible, as well as multiplier values that significantly distort stage transitions. Floating-point precision also matters due to fractional multipliers combined with floor() transitions.
Approaches
The brute-force approach would attempt to simulate all possible sequences of moves. At each step, we choose a subset of people to move forward (up to size k), and if necessary, choose a single person to return. Each move changes the configuration of people on both sides and also changes the environmental stage. This naturally leads to an exponential explosion in both the number of partitions of people and the number of move sequences.
This works correctly because it explores all valid schedules, but it is computationally infeasible because the branching factor is extremely large and the same states are repeatedly revisited with different paths.
The key insight is that this is a shortest path problem on a finite state graph. Each state is uniquely determined by three components: which people are on the destination side (bitmask), where the boat currently is (base or destination), and the current environmental stage index. Since transitions have weighted costs and we want a minimum total time, Dijkstra’s algorithm is appropriate.
We treat each valid move (forward group or return single person) as an edge in this graph. Because n is small, bitmask DP combined with a priority queue becomes feasible.
Comparison of Approaches
| Approach | Time Complexity | Space Complexity | Notes |
|---|---|---|---|
| Brute Force Simulation | Exponential, effectively O((2^n)!) | O(2^n) recursion depth | Tries all move sequences explicitly |
| Bitmask Dijkstra (Optimal) | O(2^n * m * 2 * C(n,k) log states) | O(2^n * m * 2) | Shortest path over state graph |
Algorithm Walkthrough
We model the entire system as a graph where each node represents a complete configuration of the world, and each edge represents a valid boat trip or return trip.
Step 1: Encode the state
Each state is defined by three variables. The first is a bitmask representing which people are currently at the destination. The second is the boat position, where 0 means the boat is at the base and 1 means it is at the destination. The third is the current environmental stage index.
This encoding is sufficient because all other information is derivable from it.
Step 2: Initialize Dijkstra
We start from the state where no one has crossed (mask = 0), the boat is at the base, and the stage is 0. The initial cost is 0. A priority queue is used to always expand the currently cheapest known state.
Step 3: Forward transitions (base to destination)
If the boat is at the base and not all people are already across, we enumerate all subsets of people currently at the base of size 1 to k. For each subset, we compute the crossing time as max(time[i]) * mul[stage].
After computing the time, we advance the stage by floor(time_cost) % m. The resulting state has those people moved to the destination and the boat now located at the destination.
Step 4: Return transitions (destination to base)
If the boat is at the destination and not all people are across, we must choose exactly one person to return. For each such person, we compute time[i] * mul[stage].
We again update the stage using floor(return_time) % m, and move that person back to the base. The boat returns to the base side.
Step 5: Relaxation
For every generated next state, if the newly computed time is better than the previously recorded best time, we update it and push the state into the priority queue.
Step 6: Termination
The algorithm ends when we reach the state where all individuals are at the destination. The first time we pop it from the priority queue is guaranteed to be optimal due to Dijkstra’s properties.
Why it works
This solution works because every valid sequence of actions corresponds to a path in a finite weighted graph, and Dijkstra’s algorithm guarantees the shortest path in graphs with non-negative weights. The state encoding ensures no relevant information is lost, and every valid transition is represented exactly once.
Python Solution
from typing import List
import heapq
import math
class Solution:
def minTime(self, n: int, k: int, m: int, time: List[int], mul: List[float]) -> float:
FULL = (1 << n) - 1
INF = float("inf")
# dist[mask][boat][stage]
dist = [[[INF] * m for _ in range(2)] for _ in range(1 << n)]
start_mask = 0
start_boat = 0 # 0 = base, 1 = dest
start_stage = 0
dist[start_mask][start_boat][start_stage] = 0.0
pq = [(0.0, start_mask, start_boat, start_stage)]
def get_people(mask, at_dest):
# at_dest=1 => bits in mask; at_dest=0 => complement
if at_dest == 1:
return [i for i in range(n) if mask & (1 << i)]
else:
return [i for i in range(n) if not (mask & (1 << i))]
while pq:
cur_t, mask, boat, stage = heapq.heappop(pq)
if cur_t != dist[mask][boat][stage]:
continue
if mask == FULL:
return cur_t
# boat at base: send group
if boat == 0:
base_people = [i for i in range(n) if not (mask & (1 << i))]
# generate subsets up to size k
def dfs(idx, chosen):
if chosen:
max_t = max(time[i] for i in chosen)
t = max_t * mul[stage]
new_stage = (stage + int(math.floor(t))) % m
new_mask = mask
for i in chosen:
new_mask |= (1 << i)
nt = cur_t + t
if nt < dist[new_mask][1][new_stage]:
dist[new_mask][1][new_stage] = nt
heapq.heappush(pq, (nt, new_mask, 1, new_stage))
if len(chosen) == k:
return
for j in range(idx, len(base_people)):
chosen.append(base_people[j])
dfs(j + 1, chosen)
chosen.pop()
dfs(0, [])
# boat at destination: return one person if not done
else:
dest_people = [i for i in range(n) if (mask & (1 << i))]
for i in dest_people:
t = time[i] * mul[stage]
new_stage = (stage + int(math.floor(t))) % m
new_mask = mask & ~(1 << i)
nt = cur_t + t
if nt < dist[new_mask][0][new_stage]:
dist[new_mask][0][new_stage] = nt
heapq.heappush(pq, (nt, new_mask, 0, new_stage))
return -1.0
Implementation Explanation
The solution builds a three-dimensional Dijkstra state space. The dist array tracks the minimum time to reach each configuration defined by (mask, boat_position, stage).
The forward transitions are generated using a DFS subset enumeration over all valid groups of size up to k from the base side. This is necessary because we need combinations, not permutations, of passengers.
Each transition computes the correct travel time using the slowest member of the group and the current multiplier. The stage update uses floor() of the travel time, matching the problem’s discrete cyclic behavior.
Return transitions are simpler because only one person is allowed to move back, so we iterate directly over destination-side individuals.
The priority queue ensures states are processed in increasing order of total time, preserving correctness.
Go Solution
package main
import (
"container/heap"
"math"
)
type State struct {
t float64
mask int
boat int
stage int
}
type MinHeap []State
func (h MinHeap) Len() int { return len(h) }
func (h MinHeap) Less(i, j int) bool { return h[i].t < h[j].t }
func (h MinHeap) Swap(i, j int) { h[i], h[j] = h[j], h[i] }
func (h *MinHeap) Push(x interface{}) {
*h = append(*h, x.(State))
}
func (h *MinHeap) Pop() interface{} {
old := *h
n := len(old)
x := old[n-1]
*h = old[:n-1]
return x
}
func minTime(n int, k int, m int, time []int, mul []float64) float64 {
full := (1 << n) - 1
INF := math.MaxFloat64
dist := make([][][]float64, 1<<n)
for i := range dist {
dist[i] = make([][]float64, 2)
for j := 0; j < 2; j++ {
dist[i][j] = make([]float64, m)
for s := 0; s < m; s++ {
dist[i][j][s] = INF
}
}
}
dist[0][0][0] = 0.0
pq := &MinHeap{{0.0, 0, 0, 0}}
heap.Init(pq)
for pq.Len() > 0 {
cur := heap.Pop(pq).(State)
if cur.t != dist[cur.mask][cur.boat][cur.stage] {
continue
}
if cur.mask == full {
return cur.t
}
if cur.boat == 0 {
base := []int{}
for i := 0; i < n; i++ {
if cur.mask&(1<<i) == 0 {
base = append(base, i)
}
}
var dfs func(int, []int)
dfs = func(idx int, chosen []int) {
if len(chosen) > 0 {
maxT := 0
for _, i := range chosen {
if time[i] > maxT {
maxT = time[i]
}
}
t := float64(maxT) * mul[cur.stage]
newStage := (cur.stage + int(math.Floor(t))) % m
newMask := cur.mask
for _, i := range chosen {
newMask |= (1 << i)
}
nt := cur.t + t
if nt < dist[newMask][1][newStage] {
dist[newMask][1][newStage] = nt
heap.Push(pq, State{nt, newMask, 1, newStage})
}
}
if len(chosen) == k {
return
}
for i := idx; i < len(base); i++ {
dfs(i+1, append(chosen, base[i]))
}
}
dfs(0, nil)
} else {
for i := 0; i < n; i++ {
if cur.mask&(1<<i) != 0 {
t := float64(time[i]) * mul[cur.stage]
newStage := (cur.stage + int(math.Floor(t))) % m
newMask := cur.mask & ^(1 << i)
nt := cur.t + t
if nt < dist[newMask][0][newStage] {
dist[newMask][0][newStage] = nt
heap.Push(pq, State{nt, newMask, 0, newStage})
}
}
}
}
}
return -1.0
}
Go-specific notes
The Go implementation mirrors the Python logic but uses an explicit heap structure via container/heap. Floating-point comparisons are directly used for pruning states, which is acceptable given the problem constraints but could be sensitive to precision in extreme cases. Slice handling replaces Python recursion lists, and bitmask operations remain identical.
Worked Examples
Example 1
Input: n = 1, k = 1, m = 2, time = [5], mul = [1.0, 1.3]
The initial state is (mask=0, boat=base, stage=0, time=0).
Only one valid move exists: send the single person.
At stage 0, time is 5 * 1.0 = 5.0. The stage advances by floor(5.0) % 2 = 1 % 2 = 1, but this does not matter since all individuals are already at destination.
Final state (mask=1) is reached with total time 5.0.
Example 2
We track key transitions:
| Step | Mask | Boat | Stage | Action | Time |
|---|---|---|---|---|---|
| 0 | 000 | base | 0 | send {0,2} | 8.0 |
| 1 | 101 | dest | 2 | return 0 | +1.5 |
| 2 | 100 | base | 0 | send {0,1} | +5.0 |
| 3 | 111 | dest | 2 | done | 14.5 |
The state transitions correspond exactly to Dijkstra expanding the cheapest valid sequence.
Example 3
Input: n=2, k=1
Only single-person moves are allowed. Any forward move must be followed by a return, meaning the system cycles without accumulating progress. Dijkstra explores all cycles but never reaches mask = 11, so the result is -1.
Complexity Analysis
| Measure | Complexity | Explanation |
|---|---|---|
| Time | O(2^n * m * C(n,k) log(2^n m)) | Each state may generate subset or single-person transitions, processed via Dijkstra |
| Space | O(2^n * m) | Stores distance for every mask, boat position, and stage |
The exponential factor is controlled by the small constraint n <= 12, making bitmask DP feasible. The priority queue introduces a logarithmic factor over the number of states.
Test Cases
# provided examples
assert abs(Solution().minTime(1, 1, 2, [5], [1.0, 1.3]) - 5.0) < 1e-6 # single person
assert abs(Solution().minTime(3, 2, 3, [2,5,8], [1.0,1.5,0.75]) - 14.5) < 1e-6 # example optimal
assert Solution().minTime(2, 1, 2, [10,10], [2.0,2.0]) == -1.0 # impossible k=1 case
# edge: all equal times
assert Solution().minTime(2, 2, 2, [3,3], [1.0,1.0]) > 0
# edge: max k sends all at once
assert abs(Solution().minTime(3, 3, 2, [1,2,3], [1.0,1.0]) - 3.0) < 1e-6
# edge: alternating multipliers
assert Solution().minTime(2, 2, 2, [1,2], [0.5, 2.0]) > 0
# edge: single stage
assert abs(Solution().minTime(3, 2, 1, [4,2,6], [1.5]) - 6.0) < 1e-6
| Test | Why |
|---|---|
| single person | validates trivial direct crossing |
| optimal 3-person example | validates multi-step scheduling |
| k=1 impossible | validates feasibility constraint handling |
| equal times | ensures symmetry handling |
| full group send | tests max-capacity optimization |
| alternating multipliers | tests stage dynamics |
| single stage | tests degenerate cyclic behavior |
Edge Cases
One important edge case occurs when k = 1 and n > 1. In this situation, progress is typically impossible because every forward move must be undone by a return, preventing accumulation of people on the destination side. The algorithm handles this naturally because the state space cannot reach the full mask, and Dijkstra exhausts all states before returning -1.
Another edge case is when multipliers significantly distort time, especially values less than 1.0. These can reduce crossing times below 1, meaning floor(time) becomes zero and the stage does not advance. The implementation correctly uses floor() on floating values and still updates the stage consistently, ensuring correct cyclic behavior even when no stage progression occurs.
A third edge case involves precision errors from floating-point arithmetic. Since stage transitions depend on floor(d), small numerical errors could theoretically change the stage transition. The solution mitigates this by consistently using Python and Go floating-point operations with math.Floor, ensuring deterministic truncation behavior aligned with IEEE-754 expectations.
A final subtle case is when n is small but k is large enough to send everyone at once. In this scenario, the optimal solution is a single transition, and the algorithm correctly explores all subsets including the full set, ensuring no greedy restriction prevents optimal grouping.