Solution Explanation:

The problem asks for the minimum time required to reach the bottom-right corner (n-1, n-1) from the top-left corner (0,0) of an n x n grid, given that you can only move to adjacent cells with elevation less than or equal to the current time t. This is essentially a shortest path problem with a time-dependent constraint.

Several approaches can solve this, including binary search and union-find. The provided solutions use a Union-Find approach which is efficient for this problem.

Union-Find Approach:

The core idea is to use a Union-Find data structure to track connected components in the grid at different time steps.

1. Initialization:

   • Create a Union-Find data structure p to represent the connected components. Initially, each cell is its own component.
   • Create an array hi to map the elevation values to their corresponding cell indices (row * n + col). This allows efficient lookup of cell indices based on elevation.

2. Iterative Time Steps:

   • Iterate through time t from 0 to n*n -1 (maximum possible elevation).
   • At each time step, process cells with elevation <= t.
   • For each cell (i, j) with elevation <= t:
     • Check its four neighbors.
     • If a neighbor (x, y) also has elevation <= t, union the components of (i, j) and (x, y) using find and union operations.

3. Check Connectivity:

   • After processing all cells with elevation <= t, check if the top-left (0) and bottom-right (n*n - 1) cells belong to the same component using the find function.
   • If they are in the same component, it means a path exists from (0,0) to (n-1, n-1) at time t, so return t.

4. Return Result:

   • If the loop completes without finding a path, it implies there is an error (though this shouldn't happen with the given constraints).

Time Complexity Analysis:

• The outer loop iterates n*n times.
• Inside the loop, we iterate through at most 4*n*n neighbors for union operation.
• The find and union operations in the Union-Find structure have an amortized time complexity of almost O(1) due to path compression and union by rank (though it could be O(log n) in the worst case without optimizations).

Therefore, the overall time complexity is approximately O(n*n) (dominated by the outer loop and neighbor checks), which is efficient for the given constraints (1 <= n <= 50).

Space Complexity Analysis:

The space complexity is O(n*n) due to the p and hi arrays, which store information about the cells.

Code Explanation (Python3 as Example):

class Solution:
    def swimInWater(self, grid: List[List[int]]) -> int:
        def find(x):  # Find operation for Union-Find
            if p[x] != x:
                p[x] = find(p[x])
            return p[x]
 
        n = len(grid)
        p = list(range(n * n))  # Initialize Union-Find
        hi = [0] * (n * n)  # Mapping elevation to cell index
        for i, row in enumerate(grid):
            for j, h in enumerate(row):
                hi[h] = i * n + j  # Store cell index for each elevation
 
        for t in range(n * n):  # Iterate through time steps
            i, j = divmod(hi[t], n)  # Get cell coordinates from index
            for a, b in [(0, -1), (0, 1), (1, 0), (-1, 0)]: #Check neighbors
                x, y = i + a, j + b
                if 0 <= x < n and 0 <= y < n and grid[x][y] <= t:
                    p[find(x * n + y)] = find(hi[t]) #Union if neighbor is accessible
 
                if find(0) == find(n * n - 1):  #Check if connected
                    return t
 
        return -1  # Should not reach here with given constraints
 

The other languages (Java, C++, Go, TypeScript, Rust) follow a similar structure, implementing the Union-Find and the iterative time step process. The core logic remains the same across all the languages. The only difference lies in the syntax and standard library functions used.