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LeetCode 1102: Path With Maximum Minimum Value Solution

Master LeetCode problem 1102 (Path With Maximum Minimum Value), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

Problem Explanation

Explanation:

This problem can be solved using Dijkstra's algorithm. The idea is to start from the top-left cell and explore all possible paths while keeping track of the minimum value encountered so far in each path. We can use a priority queue to prioritize paths with higher minimum values. This way, we always select the path with the maximum minimum value at each step.

  1. Create a priority queue to store paths based on their minimum value.
  2. Initialize the priority queue with the starting cell (0, 0) with a minimum value of grid[0][0].
  3. While the priority queue is not empty, pop the path with the maximum minimum value.
  4. Explore all four directions from the current cell and update the minimum value for each path.
  5. If the current cell is the bottom-right cell, return the minimum value for this path.
  6. Continue this process until we reach the bottom-right cell. :

Solution Code

class Solution {
    public int maximumMinimumPath(int[][] A) {
        int m = A.length, n = A[0].length;
        int[][] dirs = {{0, 1}, {1, 0}, {0, -1}, {-1, 0}};
        
        PriorityQueue<int[]> pq = new PriorityQueue<>((a, b) -> A[b[0]][b[1]] - A[a[0]][a[1]]);
        pq.offer(new int[]{0, 0});
        
        boolean[][] visited = new boolean[m][n];
        visited[0][0] = true;
        
        while (!pq.isEmpty()) {
            int[] curr = pq.poll();
            int x = curr[0], y = curr[1];
            
            if (x == m - 1 && y == n - 1) {
                return A[x][y];
            }
            
            for (int[] dir : dirs) {
                int newX = x + dir[0];
                int newY = y + dir[1];
                
                if (newX >= 0 && newX < m && newY >= 0 && newY < n && !visited[newX][newY]) {
                    pq.offer(new int[]{newX, newY});
                    visited[newX][newY] = true;
                    A[newX][newY] = Math.min(A[x][y], A[newX][newY]);
                }
            }
        }
        
        return -1;
    }
}

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