Golang solution with the explanation
Intuition\nThe problem requires finding the minimum time to travel from the top-left to the bottom-right of the grid while adhering to the constraints of cell a
# Intuition\nThe problem requires finding the minimum time to travel from the top-left to the bottom-right of the grid while adhering to the constraints of cell availability. A shortest-path algorithm, similar to Dijkstra\'s, is well-suited for this problem because we are minimizing the time to reach each cell.\n\nKey insights:\n1. Use a priority queue to always process the cell with the smallest time first.\n2. Handle cell availability by calculating the required wait time if the current time does not satisfy the cell\'s constraints.\n\n# Approach\n1. **Priority Queue**:\n   - Use a priority queue (min-heap) to process cells in increasing order of time.\n   - Each entry in the queue contains the current time, row, and column of a cell.\n\n2. **Edge Case Handling**:\n   - If the initial neighbors of the starting cell are blocked (i.e., `grid[1][0] > 1` and `grid[0][1] > 1`), return `-1` as the destination is unreachable.\n\n3. **Time Matrix**:\n   - Maintain a 2D matrix `time` to store the minimum time to reach each cell. Initialize all cells with infinity (`INT_MAX`) except the starting cell, which is `0`.\n\n4. **Neighbor Traversal**:\n   - For each cell, check its four neighbors (up, down, left, right).\n   - Compute the earliest possible time to move to the neighbor based on the constraints:\n     - If the current time plus 1 is less than the neighbor\'s grid constraint, calculate the wait time required to synchronize the timing (adjust for even/odd constraints).\n   - If the new time to reach the neighbor is smaller than its recorded time, update it and push it into the priority queue.\n\n5. **Termination**:\n   - If the bottom-right cell is reached, return the time. If the queue is empty and the destination is not reached, return `-1`.\n\n# Complexity\n- **Time Complexity**:  \n  \\(O(n \\cdot m \\cdot \\log(n \\cdot m))\\), where \\(n\\) and \\(m\\) are the number of rows and columns in the grid.  \n  - Each cell can be pushed into the priority queue at most once.\n  - Priority queue operations (push and pop) take \\(O(\\log(n \\cdot m))\\).\n\n- **Space Complexity**:  \n  \\(O(n \\cdot m)\\), for the `time` matrix and priority queue storage.\n\n# Code\n```golang []\ntype Info struct {\n\ttime, row, col int\n}\n\ntype MinHeap []Info\n\nfunc (h MinHeap) Len() int            { return len(h) }\nfunc (h MinHeap) Less(i, j int) bool  { return h[i].time < h[j].time }\nfunc (h MinHeap) Swap(i, j int)       { h[i], h[j] = h[j], h[i] }\nfunc (h *MinHeap) Push(x interface{}) { *h = append(*h, x.(Info)) }\nfunc (h *MinHeap) Pop() interface{} {\n\told := *h\n\tn := len(old)\n\titem := old[n-1]\n\t*h = old[:n-1]\n\treturn item\n}\n\nvar directions = [5]int{0, 1, 0, -1, 0}\n\nfunc isOutside(row, col, rows, cols int) bool {\n\treturn row < 0 || row >= rows || col < 0 || col >= cols\n}\n\nfunc minimumTime(grid [][]int) int {\n\trows := len(grid)\n\tcols := len(grid[0])\n\n\t// Edge case: if the immediate neighbors are blocked\n\tif grid[1][0] > 1 && grid[0][1] > 1 {\n\t\treturn -1\n\t}\n\n\ttime := make([][]int, rows)\n\tfor i := range time {\n\t\ttime[i] = make([]int, cols)\n\t\tfor j := range time[i] {\n\t\t\ttime[i][j] = math.MaxInt32\n\t\t}\n\t}\n\n\tpq := &MinHeap{}\n\theap.Push(pq, Info{time: 0, row: 0, col: 0})\n\ttime[0][0] = 0\n\n\tfor pq.Len() > 0 {\n\t\tcurr := heap.Pop(pq).(Info)\n\t\tcurrTime, currRow, currCol := curr.time, curr.row, curr.col\n\n\t\t// If we reached the bottom-right cell\n\t\tif currRow == rows-1 && currCol == cols-1 {\n\t\t\treturn currTime\n\t\t}\n\n\t\t// Explore neighbors\n\t\tfor d := 0; d < 4; d++ {\n\t\t\tnextRow := currRow + directions[d]\n\t\t\tnextCol := currCol + directions[d+1]\n\n\t\t\tif isOutside(nextRow, nextCol, rows, cols) {\n\t\t\t\tcontinue\n\t\t\t}\n\n\t\t\t// Calculate next time\n\t\t\twait := 0\n\t\t\tif (grid[nextRow][nextCol]-currTime)&1 == 0 {\n\t\t\t\twait = 1\n\t\t\t}\n\t\t\tnextTime := max(currTime+1, grid[nextRow][nextCol]+wait)\n\n\t\t\t// Update if the path is quicker\n\t\t\tif nextTime < time[nextRow][nextCol] {\n\t\t\t\ttime[nextRow][nextCol] = nextTime\n\t\t\t\theap.Push(pq, Info{time: nextTime, row: nextRow, col: nextCol})\n\t\t\t}\n\t\t}\n\t}\n\n\treturn -1\n}\n\nfunc max(a, b int) int {\n\tif a > b {\n\t\treturn a\n\t}\n\treturn b\n}\n```
['Go']