LeetCode 1368: Minimum Cost to Make at Least One Valid Path in a Grid Solution

Master LeetCode problem 1368 (Minimum Cost to Make at Least One Valid Path in a Grid), a hard 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.

1368. Minimum Cost to Make at Least One Valid Path in a Grid

Hard

Array Breadth-First Search Graph Heap (Priority Queue)Matrix Shortest Path

Problem Explanation

Explanation:

To solve this problem, we can use Dijkstra's algorithm to find the shortest path from the start point (0, 0) to the end point (m - 1, n - 1) with some modifications. We create a priority queue to store the minimum cost to reach each cell. At the beginning, all costs are set to infinity except for the start cell which is set to 0. We iterate through the grid cells, updating the costs based on the direction of the arrows. If we encounter a cell pointing outside the grid, we can replace the arrow with a cost of 1. Finally, we return the cost to reach the end cell.

Algorithm:

Initialize a priority queue (min heap) to store cells with their costs.
Initialize the cost grid with all cells set to infinity except the start cell set to 0.
Add the start cell to the priority queue.
While the priority queue is not empty:
- Pop the cell with the minimum cost.
- Update the costs of neighboring cells based on the direction of the arrows.
- If a cell points outside the grid, update the arrow with a cost of 1.
- Add the updated cells to the priority queue.
Return the cost to reach the end cell.

Time Complexity: O(m * n * log(m * n))
Space Complexity: O(m * n)

Solution Code

import java.util.*;

class Solution {
    public int minCost(int[][] grid) {
        int m = grid.length;
        int n = grid[0].length;
        
        PriorityQueue<int[]> pq = new PriorityQueue<>(Comparator.comparingInt(a -> a[2]));
        int[][] costs = new int[m][n];
        
        for (int[] row : costs) {
            Arrays.fill(row, Integer.MAX_VALUE);
        }
        costs[0][0] = 0;
        
        pq.offer(new int[]{0, 0, 0});
        
        int[][] dirs = {{0, 1}, {0, -1}, {1, 0}, {-1, 0}};
        
        while (!pq.isEmpty()) {
            int[] cell = pq.poll();
            int r = cell[0];
            int c = cell[1];
            int cost = cell[2];
            
            if (r == m - 1 && c == n - 1) {
                return cost;
            }
            
            for (int i = 0; i < 4; i++) {
                int nr = r + dirs[i][0];
                int nc = c + dirs[i][1];
                
                if (nr >= 0 && nr < m && nc >= 0 && nc < n) {
                    int newCost = cost + (i + 1 == grid[r][c] ? 0 : 1);
                    
                    if (newCost < costs[nr][nc]) {
                        costs[nr][nc] = newCost;
                        pq.offer(new int[]{nr, nc, newCost});
                    }
                }
            }
        }
        
        return -1; // Should not reach here
    }
}

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