Path With Minimum Effort

class Solution {
public:
    int findMinEffort(vector<vector<int>> &terrain) {
        int low = 0;
        int high = 1000000;
        int smallest = high;
        while (low <= high) {
            int mid = (low + high) / 2;
            if (canTraverse(terrain, mid)) {
                smallest = min(smallest, mid);
                high = mid - 1;
            } else {
                low = mid + 1;
            }
        }
        return smallest;
    }
    
    int move[4][2] = {{1, 0}, {-1, 0}, {0, 1}, {0, -1}};
    bool canTraverse(vector<vector<int>> &terrain, int mid) {
        int numRows = terrain.size();
        int numCols = terrain[0].size();
        vector<vector<bool>> seen(numRows, vector<bool>(numCols, false));
        return reachable(0, 0, terrain, seen, numRows, numCols, mid);
    }
    
    bool reachable(int x, int y, vector<vector<int>> &terrain,
                  vector<vector<bool>> &seen, int numRows, int numCols,
                  int mid) {
        if (x == numRows - 1 && y == numCols - 1) {
            return true;
        }
        seen[x][y] = true;
        for (auto moveDir : move) {
            int newX = x + moveDir[0];
            int newY = y + moveDir[1];
            if (cellIsValid(newX, newY, numRows, numCols) &&
                !seen[newX][newY]) {
                int diff = abs(terrain[newX][newY] - terrain[x][y]);
                if (diff <= mid) {
                    if (reachable(newX, newY, terrain, seen, numRows, numCols, mid))
                        return true;
                }
            }
        }
        return false;
    }
    
    bool cellIsValid(int x, int y, int numRows, int numCols) {
        return x >= 0 && x < numRows && y >= 0 && y < numCols;
    }
};

Implementing the "Path With Minimum Effort" solution in C++ involves a Binary Search technique combined with Depth-First Search (DFS) for path finding through a 2D grid/terrain. The main goal is to determine the least maximum effort required to traverse from the top-left to the bottom-right corner of the grid.

Solution Steps:

1. Define a Binary Search over potential minimum maximum efforts, ranging initially from 0 to a high value (e.g., 1,000,000). Use a variable smallest to keep track of the smallest traversable effort found.

2. In each Binary Search iteration, calculate the mid value as the average of low and high.

3. Check if it is possible to traverse the grid with the current mid value as the maximum allowed effort using a helper function canTraverse.

4. Inside the canTraverse function:

   • Initiate a matrix seen to keep track of visited cells.
   • Use a DFS approach starting from (0, 0) through the helper function reachable to determine if the bottom-right corner of the grid can be reached.
   • reachable function recursively tests all four possible directions (up, down, left, right).
   • Validate movement to a new cell based on boundary conditions and whether the absolute difference in elevation between adjacent cells is less than or equal to mid.

5. If canTraverse returns true, update smallest and adjust the high value to mid - 1. If false, adjust low to mid + 1.

6. After exiting the loop, return the value of smallest.

Key Functions:

• canTraverse: Checks if a path exists for a given effort level using DFS.
• reachable: A recursive DFS function to explore paths by moving to adjacent cells and checking conditions.
• cellIsValid: Returns true if a cell is within grid boundaries.

By the end of this implementation, the function will effectively identify the minimum effort required to traverse the terrain under the specified conditions. The utilization of Binary Search optimizes the effort checking process, while DFS ensures thorough pathfinding in the grid.

class PathFinder {
    public int findMinimumEffort(int[][] terrain) {
        int min = 0;
        int max = 1000000;
        int optimalEffort = max;
        while (min <= max) {
            int median = (min + max) / 2;
            if (canTraverse(terrain, median)) {
                optimalEffort = Math.min(optimalEffort, median);
                max = median - 1;
            } else {
                min = median + 1;
            }
        }
        return optimalEffort;
    }
    
    int[][] moveOptions = {{1, 0}, {-1, 0}, {0, 1}, {0, -1}};
    boolean canTraverse(int[][] terrain, int limitation) {
        int rows = terrain.length;
        int columns = terrain[0].length;
        boolean[][] visited = new boolean[rows][columns];
        return checkPath(0, 0, terrain, visited, rows, columns, limitation);
    }
    
    boolean checkPath(int i, int j, int[][] terrain,
                               boolean[][] visited, int rows, int columns, int limitation) {
        if (i == rows - 1 && j == columns - 1) {
            return true;
        }
        visited[i][j] = true;
        for (int[] move : moveOptions) {
            int nextI = i + move[0];
            int nextJ = j + move[1];
            if (isWithinBound(nextI, nextJ, rows, columns) && !visited[nextI][nextJ]) {
                int heightDiff = Math.abs(terrain[nextI][nextJ] - terrain[i][j]);
                if (heightDiff <= limitation) {
                    if (checkPath(nextI, nextJ, terrain, visited, rows, columns, limitation))
                        return true;
                }
            }
        }
        return false;
    }
    
    boolean isWithinBound(int i, int j, int rows, int columns) {
        return i >= 0 && i < rows && j >= 0 && j < columns;
    }
}

The given Java code introduces the PathFinder class to solve the problem of finding the minimum effort required to traverse a terrain represented by a 2D grid. The effort is quantified by the maximum difference in heights between adjacent cells that can be visited in the path from the top left corner to the bottom right corner of the grid.

Understand the core functionality of the code:

• Binary Search Implementation: The findMinimumEffort method uses a binary search approach to optimize the effort. min and max represent the search boundaries, initially set to 0 and 1,000,000, respectively. The search narrows down to find the minimum effort (optimalEffort) where the path is still feasible.

• Feasibility Check: The canTraverse method checks whether a path exists with a given limitation (maximum allowable height difference). It utilizes a helper method checkPath using recursion to explore possible paths in the grid, respecting the height difference limitation and making sure not to revisit cells.

• Directional Movement:

  • The 2D array moveOptions describes potential movements in the grid: right, left, up, and down.

• Path Checking: The checkPath function moves recursively through the grid, using the moveOptions for navigating through cells. It checks if moving to an adjacent cell complies with the height difference limitation and if the bottom right corner of the grid can be reached without exceeding this limitation.

• Boundary Checks: The isWithinBound method guarantees that the movement stays within the grid limits, preventing any index out of bounds errors.

In summary, this code effectively applies binary search combined with depth-first search (DFS) techniques to determine the minimal traversal effort across a matrix based on height differences, ensuring that each possible path adheres to the given constraints. The solution emphasizes optimization and efficiency in traversing complex grid-based data structures.

class Solution:
    def findLowestEffort(self, heightMap: List[List[int]]) -> int:
        totalRows = len(heightMap)
        totalCols = len(heightMap[0])
    
        def canFinish(x, y, limit):
            if x == totalRows-1 and y == totalCols-1:
                return True
            visited[x][y] = True
            for deltaX, deltaY in [[0, 1], [1, 0], [0, -1], [-1, 0]]:
                neighborX = x + deltaX
                neighborY = y + deltaY
                if 0 <= neighborX < totalRows and 0 <= neighborY < totalCols and not visited[neighborX][neighborY]:
                    diff = abs(heightMap[neighborX][neighborY] - heightMap[x][y])
                    if diff <= limit:
                        visited[neighborX][neighborY] = True
                        if canFinish(neighborX, neighborY, limit):
                            return True
        low = 0
        high = 10000000
        while low < high:
            mid = (low + high) // 2
            visited = [[False] * totalCols for _ in range(totalRows)]
            if canFinish(0, 0, mid):
                high = mid
            else:
                low = mid + 1
        return low

The Path With Minimum Effort problem in Python3 involves finding the minimum "effort" required to move from the top-left corner to the bottom-right corner of a matrix representing a height map. Here's a brief solution outline:

1. Initialization of Dimensions: Determine the total number of rows (totalRows) and columns (totalCols) in the height map.

2. Binary Search Setup: Utilize binary search to find the minimal effort. Set the initial search range with low as 0 and high as 10000000.

3. Depth-First Search (DFS) Helper Function (canFinish):

   • If the current position (x, y) is the bottom-right corner, return True.
   • Mark the current position as visited.
   • Iterate through the four possible movements (up, down, left, right) to adjacent cells.
   • Check conditionally:
     • Are the coordinates valid (within the grid bounds)?
     • Is the position unvisited?
     • Does the difference in height between the current cell and the neighboring cell not exceed the current limit being checked?
   • If all conditions are met, recursively continue the DFS from the neighboring cell.
   • If reaching the end is possible via any path, return True. Otherwise, return False after all possibilities are exhausted.

4. Binary Search Execution:

   • Calculate the mid value of the current search range.
   • Run the canFinish starting from the top-left corner (0,0) with the current mid as the effort limit.
   • If a path is successfully found (i.e., canFinish returns True), adjust the upper bound (high) to mid to explore potentially smaller effort limits.
   • If no path is found with the current limit, increase the lower bound (low) to mid + 1 to try with a higher limit.
   • Continue adjusting the search range until low meets high.

5. Return Result: The low value at the end of the binary search loop represents the minimum effort required to traverse from the first to the last cell under the given conditions.

This approach efficiently narrows down to the minimal required effort using a combination of DFS for checking feasible paths and binary search for minimizing the effort value.