Solution Explanation: Finding the Maximum Minimum Path in a Grid

The problem asks to find the maximum score of a path from the top-left corner (0,0) to the bottom-right corner (m-1, n-1) in a given grid. The score of a path is defined as the minimum value encountered along that path. This means we need to find the path where the smallest element is as large as possible.

The optimal approach is to use a combination of binary search and union-find.

1. Binary Search:

We can perform binary search on the range of possible minimum values in the grid (from 0 to the maximum value in the grid). For a given threshold, we check if there exists a path from (0,0) to (m-1, n-1) where all values along the path are greater than or equal to threshold. If such a path exists, it means we might be able to find a path with an even higher minimum value. Otherwise, we need to lower the threshold.

2. Union-Find:

To efficiently check the existence of a path with a minimum value greater than or equal to threshold, we use a union-find data structure.

• Initialization: We create a union-find structure with m*n elements, one for each cell in the grid.
• Iterative Processing: We process the grid cells in decreasing order of their values. For each cell with value greater than or equal to the threshold, we union it with its adjacent cells that also have values above the threshold.
• Path Existence Check: If, after processing all cells, the top-left cell (0,0) and the bottom-right cell (m-1, n-1) belong to the same set in the union-find structure, it indicates that there exists a path from (0,0) to (m-1, n-1) where all values are at least threshold.

Time Complexity Analysis:

• Binary Search: O(log(max_value)) where max_value is the maximum value in the grid.
• Union-Find: The union and find operations in a union-find data structure take amortized O(α(n)) time, where α is the inverse Ackermann function and n is the number of nodes (m*n in our case). In practice, α(n) is very small and can be considered constant. We perform union-find operations for each cell once in each iteration of the binary search.
• Overall: The overall time complexity is O(mnlog(max_value) + mnα(mn)), which simplifies to O(mnlog(max_value)) as α(mn) is negligible.

Space Complexity Analysis:

The space complexity is dominated by the union-find data structure and the grid itself, which is O(m*n).

Code Examples (Python):

import itertools
 
class UnionFind:
    def __init__(self, n):
        self.parent = list(range(n))
        self.rank = [0] * n
 
    def find(self, i):
        if self.parent[i] == i:
            return i
        self.parent[i] = self.find(self.parent[i])
        return self.parent[i]
 
    def union(self, i, j):
        root_i = self.find(i)
        root_j = self.find(j)
        if root_i != root_j:
            if self.rank[root_i] < self.rank[root_j]:
                self.parent[root_i] = root_j
            elif self.rank[root_i] > self.rank[root_j]:
                self.parent[root_j] = root_i
            else:
                self.parent[root_j] = root_i
                self.rank[root_i] += 1
            return True
        return False
 
def maximumMinimumPath(grid):
    m, n = len(grid), len(grid[0])
    cells = sorted([(grid[i][j], i, j) for i, j in itertools.product(range(m), range(n))], reverse=True)
    low, high = 0, max(max(row) for row in grid)
    ans = 0
    while low <= high:
        mid = (low + high) // 2
        uf = UnionFind(m * n)
        for val, i, j in cells:
            if val >= mid:
                for x, y in [(i - 1, j), (i + 1, j), (i, j - 1), (i, j + 1)]:
                    if 0 <= x < m and 0 <= y < n and grid[x][y] >= mid:
                        uf.union(i * n + j, x * n + y)
        if uf.find(0) == uf.find(m * n - 1):
            ans = mid
            low = mid + 1
        else:
            high = mid - 1
    return ans
 

This Python code efficiently solves the problem using binary search and union-find, providing a clear and concise solution with optimal time complexity. Other languages (Java, C++, Go, TypeScript, Rust) would follow similar algorithmic structures, although the specific syntax and data structures might vary.