C# Solution 64ms
IntuitionI will modify this post later.ApproachComplexity
Time complexity:

Space complexity:
Code
# Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
I will modify this post later.
# Approach
<!-- Describe your approach to solving the problem. -->

# Complexity
- Time complexity:
<!-- Add your time complexity here, e.g. $$O(n)$$ -->

- Space complexity:
<!-- Add your space complexity here, e.g. $$O(n)$$ -->

# Code
```csharp []
public class Solution {
    public int MinTimeToReach(int[][] moveTime) {
        return AnotherVersionBFSSolution(moveTime);    
    }

    private static int M;
    private static int N;
    private static bool[,] Seen;
    private static int[,] Directions = new int[,]{{0, 1}, {1, 0}, {0, -1}, {-1, 0}};
    public static int AnotherVersionBFSSolution(int[][] moveTime)
    {
        if(moveTime == null || moveTime.Length == 0 || moveTime[0] == null || moveTime[0].Length == 0)
        {
            return 0;
        }

        M = moveTime.Length;
        N = moveTime[0].Length;
        if(M == 1 && N == 1)
        {
            return 0;
        }

        Seen = new bool[M, N];
        // It could be proved take different routes to the same node will end using same odd or even edges
        // Because different routes will either share edges or formating circles on the grid
        // Circles on the grid always have even number of edges
        // Route Length = shareEdge + edges on circle
        // Thus final step to one node's odd/even parity is determined
        // Thus we only need to check every node once
        int[,] finalStep = new int[M, N];
        Seen[0, 0] = true;
        finalStep[0, 0] = 2;
        var timeToPoints = new Dictionary<int, HashSet<(int X, int Y)>>();
        PriorityQueue<int, int> timePQ = new PriorityQueue<int, int>();
        var end = (M - 1, N - 1);
        timePQ.Enqueue(0, 0);
        timeToPoints[0] = new HashSet<(int X, int Y)>();
        timeToPoints[0].Add((0, 0));
        while(timePQ.Count > 0)
        {
            int currentTime = timePQ.Dequeue();
            HashSet<(int X, int Y)> points = timeToPoints[currentTime];
            timeToPoints.Remove(currentTime);
            foreach(var point in points)
            {
                int x = point.X;
                int y = point.Y;
                int nextStep = 3 - finalStep[x, y];
                for(int k = 0; k < 4; ++k)
                {
                    int i = x + Directions[k, 0];
                    int j = y + Directions[k, 1];
                    if(CouldBeVisited(i, j))
                    {
                        Seen[i, j] = true;
                        finalStep[i, j] = nextStep;
                        int nextTime = Math.Max(currentTime + nextStep, moveTime[i][j] + nextStep);
                        if((i, j).Equals(end))
                        {
                            return nextTime;
                        }

                        if(!timeToPoints.ContainsKey(nextTime))
                        {
                            timePQ.Enqueue(nextTime, nextTime);
                            timeToPoints[nextTime] = new HashSet<(int X, int Y)>();
                        }

                        timeToPoints[nextTime].Add((i, j));
                    }
                }
            }
        }

        return -1;
    }

    private static bool CouldBeVisited(int i, int j)
    {
        if(i < 0 || i > M -1 || j < 0 || j > N - 1)
        {
            return false;
        }

        return !Seen[i, j];
    }
}
```
['C#']