Path With Minimum Effort

Minimum Effort Path

This problem requires finding the minimum effort path from the top-left to the bottom-right cell in a 2D grid, where effort is defined as the maximum absolute difference between consecutive cell heights along the path. Let's explore a couple of approaches.

1. Brute Force Approach (Not Efficient)

One naive approach is to explore all possible paths using Depth-First Search (DFS) or Breadth-First Search (BFS). For each path, calculate the maximum absolute difference between consecutive cells, and then find the minimum among all such paths.

Limitations:

• This approach has exponential time complexity because it explores all possible paths. It's not efficient for larger grids and will likely result in a Time Limit Exceeded (TLE) error.

2. Optimal Solution: Binary Search + DFS/BFS

A more efficient approach combines binary search with either Depth-First Search (DFS) or Breadth-First Search (BFS). The idea is to binary search the minimum effort required. For a given mid value (potential minimum effort), we check if a path exists from the top-left to the bottom-right cell such that the absolute difference between consecutive cells along the path is at most mid. This check can be done using DFS or BFS.

Algorithm:

1. Binary Search: Define a search space for the minimum effort. The lowest possible effort is 0, and the highest possible effort is the maximum absolute difference between any two adjacent cells in the grid.
2. isPossible(mid) Function: This function checks if a path exists with a maximum effort of mid. We can use DFS or BFS to traverse the grid.
   • Start from the top-left cell.
   • For each neighbor, check if the absolute difference in height is less than or equal to mid. If it is, explore that neighbor.
   • If we reach the bottom-right cell, return true.
   • If we explore all possible paths and don't reach the bottom-right cell, return false.
3. Binary Search Iteration:
   • If isPossible(mid) returns true, it means we can reach the destination with an effort of at most mid. Try a smaller effort by setting right = mid - 1.
   • If isPossible(mid) returns false, it means we need a larger effort. Set left = mid + 1.
4. Return left: After the binary search, left will be the minimum effort required.

Code Implementation (Java):

import java.util.LinkedList;
import java.util.Queue;

class Solution {
    public int minimumEffortPath(int[][] heights) {
        int rows = heights.length;
        int cols = heights[0].length;
        int left = 0;
        int right = 1000000; // Maximum possible effort
        int ans = 0;

        while (left <= right) {
            int mid = left + (right - left) / 2;
            if (isPossible(heights, mid)) {
                ans = mid;
                right = mid - 1;
            } else {
                left = mid + 1;
            }
        }

        return ans;
    }

    private boolean isPossible(int[][] heights, int effort) {
        int rows = heights.length;
        int cols = heights[0].length;
        boolean[][] visited = new boolean[rows][cols];
        Queue<int[]> queue = new LinkedList<>();
        queue.offer(new int[]{0, 0});
        visited[0][0] = true;

        int[][] directions = {{0, 1}, {0, -1}, {1, 0}, {-1, 0}};

        while (!queue.isEmpty()) {
            int[] cell = queue.poll();
            int row = cell[0];
            int col = cell[1];

            if (row == rows - 1 && col == cols - 1) {
                return true;
            }

            for (int[] dir : directions) {
                int newRow = row + dir[0];
                int newCol = col + dir[1];

                if (newRow >= 0 && newRow < rows && newCol >= 0 && newCol < cols && !visited[newRow][newCol]
                        && Math.abs(heights[row][col] - heights[newRow][newCol]) <= effort) {
                    queue.offer(new int[]{newRow, newCol});
                    visited[newRow][newCol] = true;
                }
            }
        }

        return false;
    }
}

Explanation:

• The minimumEffortPath function performs a binary search on the possible effort values.
• The isPossible function uses BFS to check if a path exists with the given effort limit.

Time Complexity: O(m * n * log(M)), where 'm' is the number of rows, 'n' is the number of columns, and 'M' is the maximum possible effort (maximum height difference). The binary search takes O(log(M)) time, and the BFS takes O(m * n) time in the worst case.

Space Complexity: O(m * n), due to the visited array and the queue used in BFS.

Edge Cases:

• Single cell grid: If the grid is 1x1, the effort is 0.
• No path exists: If there is no valid path from the start to the end, the binary search will still converge to a value, which represents the minimum effort needed to attempt to find a path. The isPossible function helps determine whether a path is actually possible. For cases where there is no path, the algorithm will still give a result, but this result will simply reflect the smallest effort that allows the algorithm to explore as much of the grid as possible.
• Heights are the same: If all the heights in the grid are the same, the minimum effort is 0.
• Large Height Differences: The initial range for the binary search (left, right) needs to be considered with respect to the maximum differences in the height.