Minimum Time to Visit a Cell In a Grid - Practice Coding Problems

Minimum Time to Visit a Cell In a Grid - Problem

Array Breadth-First Search Graph Heap (Priority Queue) Matrix Shortest Path Hard

Journey Through Time: Navigate a time-constrained grid where each cell has a minimum arrival time requirement!

You begin your adventure at the top-left corner (0,0) of an m x n grid at time 0. Each cell grid[row][col] contains a minimum time requirement - you can only enter that cell when your current time is greater than or equal to this value.

Movement Rules:
• Move in 4 directions: up, down, left, right
• Each move takes exactly 1 second
• You can wait by moving back and forth between accessible cells

Goal: Find the minimum time needed to reach the bottom-right corner (m-1, n-1), or return -1 if impossible.

Key Insight: If you can't move from the starting position initially (both right and down cells have time > 1), the journey is impossible!

Input & Output

basic_grid.py — Python

$ Input: grid = [[0,1,3,2],[5,1,2,5],[4,3,8,6]]

› Output: 7

💡 Note: One possible path: (0,0) → (0,1) → (0,2) → wait until time 4 → (0,3) → (1,3) → (2,3). Total time = 7.

impossible_case.py — Python

$ Input: grid = [[0,1,99],[2,3,4]]

› Output: -1

💡 Note: Cannot move from (0,0) because both (0,1) requires time ≥ 1 and (1,0) requires time ≥ 2, but we can only move to time 1.

single_cell.py — Python

$ Input: grid = [[0]]

› Output: 0

💡 Note: Already at destination (0,0) at time 0.

Visualization

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Understanding the Visualization

Check Initial Movement

Verify you can leave the starting room

Use Priority Queue

Always process the cell reachable in minimum time

Calculate Wait Time

If arriving early, move back and forth to waste time

Find Optimal Path

Continue until reaching the destination

Key Takeaway

🎯 Key Insight: Use Dijkstra's algorithm with smart waiting - move back and forth between accessible cells to waste time when arriving too early at time-constrained cells.

Time & Space Complexity

Time Complexity

⏱️

O(4^(m*n))

Each cell can be visited multiple times with 4 directions, leading to exponential paths

✓ Linear Growth

Space Complexity

O(m*n)

Recursion stack depth limited by grid size

⚡ Linearithmic Space

Constraints

m == grid.length
n == grid[i].length
2 ≤ m, n ≤ 1000
4 ≤ m * n ≤ 10⁵
0 ≤ grid[i][j] ≤ 10⁵
grid[0][0] == 0

Asked in

a Amazon 25 G Google 18 ⊞ Microsoft 12 f Meta 8

The optimal solution uses Dijkstra's algorithm with a priority queue to find the shortest time path. Key insight: if both adjacent cells from start require time > 1, return -1 immediately. Otherwise, use the waiting technique - move back and forth between accessible cells to waste time when arriving too early.

Common Approaches

Approach	Time	Space	Notes
✓ Brute Force DFS with Backtracking	O(4^(m*n))	O(m*n)	Explore all possible paths using recursive DFS
Dijkstra's Algorithm with Priority Queue	O(mnlog(m*n))	O(m*n)	Use modified Dijkstra's algorithm to find shortest time path

Brute Force DFS with Backtracking — Algorithm Steps

Start DFS from (0,0) with time = 0
For each cell, try all 4 directions
Calculate arrival time considering wait time if needed
Recursively explore valid moves
Backtrack and try other paths
Return minimum time found

Visualization

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Step-by-Step Walkthrough

Start DFS

Begin at (0,0) with time 0

Explore Neighbors

Try all 4 directions from current cell

Calculate Wait Time

If arrival time < required time, calculate wait

Recursive Call

Continue DFS from valid neighbors

Backtrack

Return and try other paths

Code -

solution.c — C

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

int dfs(int** grid, int gridSize, int* gridColSize, int row, int col, int time, bool** visited) {
    int m = gridSize, n = gridColSize[0];
    int directions[4][2] = {{0,1}, {1,0}, {0,-1}, {-1,0}};
    
    if (row == m-1 && col == n-1) {
        return (time > grid[row][col]) ? time : grid[row][col];
    }
    
    int minTime = INT_MAX;
    visited[row][col] = true;
    
    for (int i = 0; i < 4; i++) {
        int nr = row + directions[i][0], nc = col + directions[i][1];
        if (nr >= 0 && nr < m && nc >= 0 && nc < n && !visited[nr][nc]) {
            int enterTime = time + 1;
            if (enterTime < grid[nr][nc]) {
                int diff = grid[nr][nc] - enterTime;
                enterTime = grid[nr][nc] + (diff % 2);
            }
            
            int result = dfs(grid, gridSize, gridColSize, nr, nc, enterTime, visited);
            if (result < minTime) minTime = result;
        }
    }
    
    visited[row][col] = false;
    return minTime;
}

int minimumTime(int** grid, int gridSize, int* gridColSize, int* returnSize) {
    if (grid[0][1] > 1 && grid[1][0] > 1) {
        return -1;
    }
    
    bool** visited = (bool**)malloc(gridSize * sizeof(bool*));
    for (int i = 0; i < gridSize; i++) {
        visited[i] = (bool*)calloc(gridColSize[i], sizeof(bool));
    }
    
    int result = dfs(grid, gridSize, gridColSize, 0, 0, 0, visited);
    
    for (int i = 0; i < gridSize; i++) {
        free(visited[i]);
    }
    free(visited);
    
    return (result == INT_MAX) ? -1 : result;
}

Time & Space Complexity

Time Complexity

⏱️

O(4^(m*n))

Each cell can be visited multiple times with 4 directions, leading to exponential paths

✓ Linear Growth

Space Complexity

O(m*n)

Recursion stack depth limited by grid size

⚡ Linearithmic Space

Constraints

m == grid.length
n == grid[i].length
2 ≤ m, n ≤ 1000
4 ≤ m * n ≤ 10⁵
0 ≤ grid[i][j] ≤ 10⁵
grid[0][0] == 0

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Input & Output

Visualization

Time & Space Complexity

Related Problems

Constraints

Common Approaches

Brute Force DFS with Backtracking — Algorithm Steps

Visualization

Code -

Time & Space Complexity

Constraints

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