Why this fails

1. Infinite Cycles: Because we can move back and forth to "waste" time, a standard traversal without strict state management (visited set based on time and position) will enter infinite loops.
2. State Explosion: The state space includes (row, col, time). Since time can grow up to $10^5$, the number of states is massive, leading to Time Limit Exceeded (TLE) errors.
3. Inefficiency: Standard BFS finds the shortest path in an unweighted graph. Here, the "cost" to enter a node is dynamic (dependent on the wait time), effectively making this a weighted graph problem where a simple queue is insufficient.

Core Insight for Minimum Time to Visit a Cell In a Grid

The key challenge in LeetCode 2577 is handling the time constraints. If we are at a cell at time t and want to move to a neighbor with a requirement grid[nr][nc], we have two scenarios:

1. Direct Entry: If t + 1 >= grid[nr][nc], we simply move there. The new time is t + 1.
2. Forced Delay: If t + 1 < grid[nr][nc], we are too early. We cannot stay still; we must move. To "wait," we move to an adjacent cell and immediately return. This round trip takes 2 seconds.

The Parity Observation: Because "waiting" consumes time in multiples of 2, the parity (even/odd nature) of our arrival time is locked once we decide to wait.

• If the difference between the target time grid[nr][nc] and our current arrival time t + 1 is even, we can burn exactly that amount of time moving back and forth and arrive exactly at grid[nr][nc].
• If the difference is odd, we cannot arrive exactly at grid[nr][nc]. We must waste one extra second (total wait time must be even), so we arrive at grid[nr][nc] + 1.

Visualizing the Priority Queue: Imagine the grid as a terrain where the "height" of a cell is its time requirement. Dijkstra's algorithm acts like water flooding the terrain. The Priority Queue ensures we always expand the "water front" from the cell with the lowest current time. When the water hits a "cliff" (a cell with a high time requirement), it rises (waits) until it matches the height (plus parity adjustment) before flowing in.

Algorithm Strategy: Graph - Shortest Path (Dijkstra's Algorithm)

1. Initialization: Use a Priority Queue (Min-Heap) to store states (time, row, col). Start at (0, 0, 0). Maintain a visited matrix to record the minimum time we reached each cell to avoid redundant processing.

2. Initial Check: Check if it is possible to leave the start. If both neighbors of (0,0) (i.e., (0,1) and (1,0)) have values greater than 1, we are trapped immediately because we have no space to "ping-pong" back and forth. Return -1.

3. Processing:

   • Extract the cell with the smallest time from the Min-Heap.
   • For each neighbor (nr, nc):
     • Calculate the wait time.
     • If grid[nr][nc] <= time + 1, the new arrival time is simply time + 1.
     • If grid[nr][nc] > time + 1, calculate the difference diff = grid[nr][nc] - (time + 1).
       • If diff is even, we can arrive exactly at grid[nr][nc].
       • If diff is odd, we arrive at grid[nr][nc] + 1.
   • If this new arrival time is better than what we've seen for (nr, nc), update the visited matrix and push to the heap.

4. Termination: The first time we pop the bottom-right cell (m-1, n-1) from the heap, the associated time is the global minimum.

Execution Flow

Let's trace a simplified scenario. Current state: time=1 at (0,1). Target neighbor: (0,2) with grid[0][2] = 4.

1. Pop State: Remove (1, 0, 1) from Priority Queue.
2. Identify Neighbor: We look at (0, 2).
3. Check Constraint:
   • Current time t = 1.
   • Direct move would result in t = 2.
   • Constraint is 4. Since 2 < 4, we are early.
4. Calculate Wait:
   • Difference diff = 4 - 2 = 2.
   • Since diff (2) is even, we can waste exactly 2 seconds (move back to (0,0) and return to (0,1)).
   • Arrival time at (0,2) will be exactly 4.
5. Update: Push (4, 0, 2) into Priority Queue. Mark (0, 2) as visited.

Proof of Correctness

Dijkstra's algorithm is proven to find the shortest path in graphs with non-negative edge weights. In this problem:

1. Non-negative Weights: The cost to move is always at least 1 (time always increases), satisfying the non-negative weight requirement.
2. Optimal Substructure: If the shortest path to cell B goes through A, then the path from start to A must be the shortest path to A (adjusted for parity constraints).
3. Parity Handling: The formula (grid[nr][nc] - (time + 1)) % 2 correctly calculates the minimal waiting time required to satisfy the grid constraint, ensuring that the edge weight calculated is the true minimum cost to traverse that edge.

Thus, the first time the destination node is extracted from the priority queue, it is guaranteed to be the minimum time.

Reference Implementation

C++ Solution for LeetCode 2577

cpp
#include <vector>
#include <queue>
#include <tuple>

using namespace std;

class Solution {
public:
    int minimumTime(vector<vector<int>>& grid) {
        // Edge case: If we cannot even make the first move
        if (grid[0][1] > 1 && grid[1][0] > 1) {
            return -1;
        }

        int m = grid.size();
        int n = grid[0].size();
        
        // Min-heap stores {time, row, col}
        // Ordered by time ascending
        priority_queue<tuple<int, int, int>, vector<tuple<int, int, int>>, greater<tuple<int, int, int>>> pq;
        
        // Visited array to keep track of minimum time to reach each cell
        vector<vector<bool>> visited(m, vector<bool>(n, false));
        
        pq.push({0, 0, 0});
        visited[0][0] = true;
        
        int dr[] = {0, 0, 1, -1};
        int dc[] = {1, -1, 0, 0};
        
        while (!pq.empty()) {
            auto [t, r, c] = pq.top();
            pq.pop();
            
            // If we reached the bottom-right cell
            if (r == m - 1 && c == n - 1) {
                return t;
            }
            
            for (int i = 0; i < 4; i++) {
                int nr = r + dr[i];
                int nc = c + dc[i];
                
                if (nr >= 0 && nr < m && nc >= 0 && nc < n && !visited[nr][nc]) {
                    int nextTime = 0;
                    int diff = grid[nr][nc] - t;
                    
                    if (diff <= 1) {
                        // Standard move
                        nextTime = t + 1;
                    } else {
                        // Need to wait (ping-pong)
                        // Wait time logic:
                        // If (grid[nr][nc] - t) is odd, we can arrive exactly at grid[nr][nc]
                        // Note: Our current time is t. We arrive at t+1. 
                        // If grid[nr][nc] > t+1, we wait.
                        // The parity check is simpler: 
                        // If (grid[nr][nc] - t) is odd, wait = 0 (arrive at grid[nr][nc])
                        // If (grid[nr][nc] - t) is even, wait = 1 (arrive at grid[nr][nc] + 1)
                        if (diff % 2 == 1) {
                            nextTime = grid[nr][nc];
                        } else {
                            nextTime = grid[nr][nc] + 1;
                        }
                    }
                    
                    visited[nr][nc] = true;
                    pq.push({nextTime, nr, nc});
                }
            }
        }
        
        return -1;
    }
};

Java Solution for LeetCode 2577

java
import java.util.PriorityQueue;

class Solution {
    public int minimumTime(int[][] grid) {
        // Edge case: blocked at start
        if (grid[0][1] > 1 && grid[1][0] > 1) {
            return -1;
        }

        int m = grid.length;
        int n = grid[0].length;
        
        // Priority Queue stores int[] {time, row, col}
        PriorityQueue<int[]> pq = new PriorityQueue<>((a, b) -> Integer.compare(a[0], b[0]));
        boolean[][] visited = new boolean[m][n];
        
        pq.offer(new int[]{0, 0, 0});
        visited[0][0] = true;
        
        int[] dr = {0, 0, 1, -1};
        int[] dc = {1, -1, 0, 0};
        
        while (!pq.isEmpty()) {
            int[] curr = pq.poll();
            int time = curr[0];
            int r = curr[1];
            int c = curr[2];
            
            if (r == m - 1 && c == n - 1) {
                return time;
            }
            
            for (int i = 0; i < 4; i++) {
                int nr = r + dr[i];
                int nc = c + dc[i];
                
                if (nr >= 0 && nr < m && nc >= 0 && nc < n && !visited[nr][nc]) {
                    int nextTime = 0;
                    
                    if (grid[nr][nc] <= time + 1) {
                        nextTime = time + 1;
                    } else {
                        // Calculate wait time based on parity
                        int diff = grid[nr][nc] - time;
                        if (diff % 2 == 1) {
                            nextTime = grid[nr][nc];
                        } else {
                            nextTime = grid[nr][nc] + 1;
                        }
                    }
                    
                    visited[nr][nc] = true;
                    pq.offer(new int[]{nextTime, nr, nc});
                }
            }
        }
        
        return -1;
    }
}

Python Solution for LeetCode 2577

python
import heapq

class Solution:
    def minimumTime(self, grid: list[list[int]]) -> int:
        # Edge case: blocked at start
        if grid[0][1] > 1 and grid[1][0] > 1:
            return -1
            
        m, n = len(grid), len(grid[0])
        # Min-heap stores (time, row, col)
        pq = [(0, 0, 0)]
        visited = set()
        visited.add((0, 0))
        
        dirs = [(0, 1), (0, -1), (1, 0), (-1, 0)]
        
        while pq:
            t, r, c = heapq.heappop(pq)
            
            if r == m - 1 and c == n - 1:
                return t
            
            for dr, dc in dirs:
                nr, nc = r + dr, c + dc
                
                if 0 <= nr < m and 0 <= nc < n and (nr, nc) not in visited:
                    next_time = 0
                    
                    if grid[nr][nc] <= t + 1:
                        next_time = t + 1
                    else:
                        # Wait logic
                        diff = grid[nr][nc] - t
                        if diff % 2 == 1:
                            next_time = grid[nr][nc]
                        else:
                            next_time = grid[nr][nc] + 1
                            
                    visited.add((nr, nc))
                    heapq.heappush(pq, (next_time, nr, nc))
                    
        return -1