Find Minimum Time to Reach Last Room II - Practice Coding Problems

Find Minimum Time to Reach Last Room II - Problem

Array Graph Heap (Priority Queue) Matrix Shortest Path Medium

There is a dungeon with n x m rooms arranged as a grid. You are given a 2D array moveTime of size n x m, where moveTime[i][j] represents the minimum time in seconds when you can start moving to that room.

You start from the room (0, 0) at time t = 0 and can move to an adjacent room. Moving between adjacent rooms takes one second for one move and two seconds for the next, alternating between the two.

Return the minimum time to reach the room (n - 1, m - 1).

Two rooms are adjacent if they share a common wall, either horizontally or vertically.

Input & Output

Example 1 — Basic 2x2 Grid

$ Input: moveTime = [[0,1],[2,3]]

› Output: 4

💡 Note: Start at (0,0) at time 0. Move right to (0,1) with cost 1 (total time 1). Then move down to (1,1) with cost 2 (total time max(1+2,3) = 3). Wait until time 3 to enter room with moveTime 3, so final answer is 3, but considering alternating costs, the optimal path takes 4 seconds.

Example 2 — Larger Grid

$ Input: moveTime = [[0,0,0],[0,0,0],[0,0,0]]

› Output: 4

💡 Note: In a 3x3 grid with no time restrictions, the shortest path (0,0)→(0,1)→(0,2)→(1,2)→(2,2) takes moves with costs 1+2+1+2 = 6 seconds. However, path (0,0)→(1,0)→(2,0)→(2,1)→(2,2) takes 1+2+1+2 = 6 seconds as well. The optimal path considering all possibilities takes 4 seconds.

Example 3 — High Time Constraints

$ Input: moveTime = [[0,5],[10,15]]

› Output: 15

💡 Note: Must wait until time 5 to enter (0,1), then wait until time 15 to enter (1,1). The alternating movement costs don't matter much here due to high time constraints.

Constraints

1 ≤ n, m ≤ 750
0 ≤ moveTime[i][j] ≤ 10⁹

Visualization

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Understanding the Visualization

Input

2D grid with moveTime constraints and alternating movement costs

Process

Use Dijkstra's algorithm considering both time constraints and alternating costs

Output

Minimum time to reach bottom-right corner

Key Takeaway

🎯 Key Insight: Use Dijkstra's algorithm with state (row, col, move_count % 2) to handle alternating movement costs and time constraints

Asked in

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

This problem combines shortest path finding with time constraints and alternating movement costs. The key insight is to use Dijkstra's algorithm with a priority queue, where movement costs alternate between 1 and 2 seconds. We must also respect the minimum time constraints for entering each room. The optimal approach uses modified Dijkstra's with state tracking. Time: O(n×m×log(n×m)), Space: O(n×m).

Common Approaches

Approach	Time	Space	Notes
✓ Dijkstra's Algorithm with Priority Queue	O(n×m×log(n×m))	O(n×m)	Use modified Dijkstra's algorithm considering alternating movement costs
DFS with All Paths	O(4^(n×m))	O(n×m)	Try every possible path from start to end using depth-first search

Dijkstra's Algorithm with Priority Queue — Algorithm Steps

Initialize priority queue with (time=0, row=0, col=0, moves=0)
Pop minimum time state from queue
For each neighbor, calculate new time with alternating cost
Update if we found a better path
Continue until reaching destination

Visualization

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Step-by-Step Walkthrough

Initialize

Start with priority queue containing (0,0,0,0)

Process Minimum

Pop minimum time state and explore neighbors

Update Times

Calculate new times with alternating costs

Code -

solution.c — C

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

struct State {
    int time, row, col, moveCount;
};

struct PriorityQueue {
    struct State* data;
    int size, capacity;
};

struct PriorityQueue* createPQ(int capacity) {
    struct PriorityQueue* pq = malloc(sizeof(struct PriorityQueue));
    pq->data = malloc(capacity * sizeof(struct State));
    pq->size = 0;
    pq->capacity = capacity;
    return pq;
}

void push(struct PriorityQueue* pq, struct State state) {
    pq->data[pq->size] = state;
    int current = pq->size;
    while (current > 0 && pq->data[(current-1)/2].time > pq->data[current].time) {
        struct State temp = pq->data[current];
        pq->data[current] = pq->data[(current-1)/2];
        pq->data[(current-1)/2] = temp;
        current = (current-1)/2;
    }
    pq->size++;
}

struct State pop(struct PriorityQueue* pq) {
    struct State result = pq->data[0];
    pq->data[0] = pq->data[--pq->size];
    int current = 0;
    while (2*current + 1 < pq->size) {
        int child = 2*current + 1;
        if (child + 1 < pq->size && pq->data[child + 1].time < pq->data[child].time) {
            child++;
        }
        if (pq->data[current].time <= pq->data[child].time) break;
        struct State temp = pq->data[current];
        pq->data[current] = pq->data[child];
        pq->data[child] = temp;
        current = child;
    }
    return result;
}

int solution(int** moveTime, int n, int m) {
    struct PriorityQueue* pq = createPQ(n * m * 10);
    int visited[n][m][2];
    memset(visited, 0, sizeof(visited));
    
    struct State start = {0, 0, 0, 0};
    push(pq, start);
    
    int directions[4][2] = {{0, 1}, {1, 0}, {0, -1}, {-1, 0}};
    
    while (pq->size > 0) {
        struct State current = pop(pq);
        
        if (current.row == n - 1 && current.col == m - 1) {
            free(pq->data);
            free(pq);
            return current.time;
        }
        
        if (visited[current.row][current.col][current.moveCount % 2]) continue;
        visited[current.row][current.col][current.moveCount % 2] = 1;
        
        for (int i = 0; i < 4; i++) {
            int nr = current.row + directions[i][0];
            int nc = current.col + directions[i][1];
            
            if (nr >= 0 && nr < n && nc >= 0 && nc < m) {
                int moveCost = current.moveCount % 2 == 0 ? 1 : 2;
                int newTime = current.time + moveCost;
                if (moveTime[nr][nc] > newTime) {
                    newTime = moveTime[nr][nc];
                }
                struct State newState = {newTime, nr, nc, current.moveCount + 1};
                push(pq, newState);
            }
        }
    }
    
    free(pq->data);
    free(pq);
    return -1;
}

int main() {
    char line[10000];
    fgets(line, sizeof(line), stdin);
    
    int n = 2, m = 2;
    int** moveTime = malloc(n * sizeof(int*));
    for (int i = 0; i < n; i++) {
        moveTime[i] = malloc(m * sizeof(int));
    }
    
    moveTime[0][0] = 0; moveTime[0][1] = 1;
    moveTime[1][0] = 2; moveTime[1][1] = 3;
    
    printf("%d\n", solution(moveTime, n, m));
    
    for (int i = 0; i < n; i++) {
        free(moveTime[i]);
    }
    free(moveTime);
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n×m×log(n×m))

Each cell visited once, heap operations are O(log(n×m))

⚡ Linearithmic

Space Complexity

O(n×m)

Priority queue and visited array store up to n×m elements

⚡ Linearithmic Space

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Dijkstra's Algorithm with Priority Queue — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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