Minimum Time to Transport All Individuals

Minimum Time to Transport All Individuals - Problem

Array Dynamic Programming Bit Manipulation Graph Heap (Priority Queue) Shortest Path Bitmask Hard

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] minutes, 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.

Input & Output

Example 1 — Basic Transport

$ Input: time = [10, 15, 8], k = 2, mul = [1.5, 1.2]

› Output: 51

💡 Note: Person 2 crosses alone (8×1.5=12), returns (8×1.2=9.6≈10), persons 0&1 cross together (15×1.2=18), person 2 returns and crosses with remaining person for total time of 51.

Example 2 — Single Person

$ Input: time = [5], k = 1, mul = [2.0]

› Output: 10

💡 Note: Only one person needs to cross: 5 × 2.0 = 10 minutes.

Example 3 — Optimal Grouping

$ Input: time = [1, 2, 5, 10], k = 2, mul = [1.0, 0.8]

› Output: 19

💡 Note: Fast people cross first, then optimal return strategy minimizes total time.

Constraints

1 ≤ n ≤ 20
1 ≤ k ≤ n
1 ≤ m ≤ 10
1 ≤ time[i] ≤ 1000
0.1 ≤ mul[j] ≤ 10.0

Visualization

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Understanding the Visualization

Initial Setup

People with different rowing times, boat capacity k, environmental stages

Transport Process

Groups cross, stages change, someone returns if needed

Optimization

Find minimum total time considering all factors

Key Takeaway

🎯 Key Insight: Model as shortest path with state (remaining people, current stage) and use BFS to find optimal solution

Asked in

G Google 15 a Amazon 12 f Meta 8 M Microsoft 6

The key insight is to model this as a shortest path problem where each state represents the set of people remaining and the current environmental stage. The best approach uses BFS with state compression to systematically explore all possible transportation strategies. Time: O(2^n × m × k!), Space: O(2^n × m)

Common Approaches

Approach	Time	Space	Notes
✓ BFS with State Tracking	O(2^n × m × k!)	O(2^n × m)	Use BFS to explore states efficiently with pruning
Brute Force - Try All Combinations	O(k^n × m)	O(n)	Generate all possible ways to group and transport individuals

BFS with State Tracking — Algorithm Steps

Create initial state (all people remaining, stage 0)
Use BFS queue to explore states level by level
For each state, generate all possible group combinations
Prune states with higher time than current best
Return minimum time when no people remain

Visualization

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Step-by-Step Walkthrough

Initial State

All people on start side, stage 0

Generate Transitions

Create all possible group movements

Prune & Continue

Eliminate suboptimal paths, continue BFS

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <math.h>

typedef struct {
    int mask;
    int stage;
    double time;
} State;

typedef struct {
    State* data;
    int front, rear, size, capacity;
} Queue;

Queue* createQueue(int capacity) {
    Queue* q = malloc(sizeof(Queue));
    q->data = malloc(capacity * sizeof(State));
    q->front = q->rear = q->size = 0;
    q->capacity = capacity;
    return q;
}

void enqueue(Queue* q, State s) {
    q->data[q->rear] = s;
    q->rear = (q->rear + 1) % q->capacity;
    q->size++;
}

State dequeue(Queue* q) {
    State s = q->data[q->front];
    q->front = (q->front + 1) % q->capacity;
    q->size--;
    return s;
}

int isEmpty(Queue* q) {
    return q->size == 0;
}

int solution(int* time, int n, int k, double* mul, int m) {
    Queue* queue = createQueue(100000);
    enqueue(queue, (State){(1 << n) - 1, 0, 0});
    
    while (!isEmpty(queue)) {
        State current = dequeue(queue);
        
        if (current.mask == 0) {
            free(queue->data);
            free(queue);
            return (int)current.time;
        }
        
        // Get remaining people
        int remaining[20], remainingCount = 0;
        for (int i = 0; i < n; i++) {
            if (current.mask & (1 << i)) {
                remaining[remainingCount++] = i;
            }
        }
        
        // Try different group sizes (simplified)
        int maxGroupSize = (k < remainingCount) ? k : remainingCount;
        for (int groupSize = 1; groupSize <= maxGroupSize; groupSize++) {
            // Simple approach: take first groupSize people
            int maxTime = 0;
            for (int i = 0; i < groupSize; i++) {
                if (time[remaining[i]] > maxTime) {
                    maxTime = time[remaining[i]];
                }
            }
            
            double crossingTime = maxTime * mul[current.stage];
            int newStage = (current.stage + (int)crossingTime) % m;
            double newTotal = current.time + crossingTime;
            
            int newMask = current.mask;
            for (int i = 0; i < groupSize; i++) {
                newMask &= ~(1 << remaining[i]);
            }
            
            if (newMask == 0) {
                enqueue(queue, (State){0, newStage, newTotal});
            } else {
                // First person returns
                int returner = remaining[0];
                double returnTime = time[returner] * mul[newStage];
                int finalStage = (newStage + (int)returnTime) % m;
                double finalTotal = newTotal + returnTime;
                
                int returnMask = newMask | (1 << returner);
                enqueue(queue, (State){returnMask, finalStage, finalTotal});
            }
        }
    }
    
    free(queue->data);
    free(queue);
    return -1;
}

int main() {
    char line[1000];
    
    fgets(line, sizeof(line), stdin);
    line[strcspn(line, "\n")] = 0;
    
    int time[20], n = 0;
    char* token = strtok(line + 1, ",]");
    while (token != NULL) {
        time[n++] = atoi(token);
        token = strtok(NULL, ",]");
    }
    
    int k;
    scanf("%d", &k);
    getchar();
    
    fgets(line, sizeof(line), stdin);
    line[strcspn(line, "\n")] = 0;
    
    double mul[20];
    int m = 0;
    token = strtok(line + 1, ",]");
    while (token != NULL) {
        mul[m++] = atof(token);
        token = strtok(NULL, ",]");
    }
    
    int result = solution(time, n, k, mul, m);
    printf("%d\n", result);
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(2^n × m × k!)

Each person can be in different states, with stage transitions and group permutations

⚠ Quadratic Growth

Space Complexity

O(2^n × m)

Queue stores states representing subsets of people and stages

⚠ Quadratic Space

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

BFS with State Tracking — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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