Snakes and Ladders

class Solution {
public:
    int boardGame(vector<vector<int>> &board) {
        int size = board.size();
        vector<pair<int, int>> slots(size * size + 1);
        int count = 1;
        vector<int> col_order(size);
        iota(col_order.begin(), col_order.end(), 0);
        for (int i = size - 1; i >= 0; i--) {
            for (int col : col_order) {
                slots[count++] = {i, col};
            }
            reverse(col_order.begin(), col_order.end());
        }
        vector<int> distance(size * size + 1, -1);
        priority_queue<pair<int, int>, vector<pair<int, int>>, greater<pair<int, int>>> queue;
        distance[1] = 0;
        queue.emplace(0, 1);
        while (!queue.empty()) {
            auto [cost, pos] = queue.top();
            queue.pop();
            if (cost != distance[pos]) {
                continue;
            }
            for (int move = pos + 1; move <= min(pos + 6, size * size); move++) {
                auto [r, c] = slots[move];
                int target = board[r][c] != -1 ? board[r][c] : move;
                if (distance[target] == -1 || distance[pos] + 1 < distance[target]) {
                    distance[target] = distance[pos] + 1;
                    queue.emplace(distance[target], target);
                }
            }
        }
        return distance[size * size];
    }
};

The given C++ solution efficiently tackles the "Snakes and Ladders" board game problem using a breadth-first search approach represented in a more optimized manner with the help of a priority queue and vector data structures for state management.

• Begin by understanding the board mapping. The code maps the 2D board game into a linear array of slots to simplify moving from one square to another.

• Defining the board matrix as vector<vector<int>> allows us to handle the board in standard memory-efficient way. Each cell in the matrix indicates either a regular cell (-1) or a shortcut/snake which jumps directly to another cell.

• Use a vector, slots, to keep track of coordinates of each board cell linearly. The vector col_order supports in initializing slots with alternating order rows to represent the zigzag movement of normal game play.

• An array, distance, keeps track of the shortest distance from the start to any cell (initialized to -1, meaning unvisited). The priority queue helps in exploring the minimum cost path first ensuring that the solution is reaching closer to less weighted paths first.

• The algorithm applies a typical breadth-first search configured by rolling a dice where:

  1. Loop through every possible "roll" outcome (1-6).
  2. Calculate the potential new position considering board shortcuts.
  3. Push the new states in the priority queue, if visiting them offers a shorter path.

• This implementation ensures that the gameplay dynamics such as snakes and ladders (which cause jumps in the board position) are efficiently handled by directly moving to the targeted cell rather than moving one cell at a time.

Use distance[size * size] to retrieve the shortest number of moves required to reach the end of the board, given optimal plays. The function returns this value as the outcome.

Ensure an understanding of priority queue usage, vector operations, and typical breadth-first search modifications to grasp the complete functionality of this game solution. This approach is effective as it minimizes unnecessary computations and efficiently manages the exploration of possible moves.

class Solution {
    public int computeMinThrows(int[][] grid) {
        int size = grid.length;
        Pair<Integer, Integer>[] positions = new Pair[size * size + 1];
        int count = 1;
        Integer[] rows = new Integer[size];
        for (int k = 0; k < size; k++) {
            rows[k] = k;
        }
        for (int i = size - 1; i >= 0; i--) {
            for (int j : rows) {
                positions[count++] = new Pair<>(i, j);
            }
            Collections.reverse(Arrays.asList(rows));
        }
        int[] moves = new int[size * size + 1];
        Arrays.fill(moves, -1);
        class Node implements Comparable<Node> {
            int movesTaken, position;

            public Node(int movesTaken, int position) {
                this.movesTaken = movesTaken;
                this.position = position;
            }

            public int compareTo(Node n) {
                return this.movesTaken - n.movesTaken;
            }
        }
        PriorityQueue<Node> queue = new PriorityQueue<Node>();
        moves[1] = 0;
        queue.add(new Node(0, 1));
        while (!queue.isEmpty()) {
            Node current = queue.poll();
            int dist = current.movesTaken, spot = current.position;
            if (dist != moves[spot]) {
                continue;
            }
            for (int nextSpot = spot + 1; nextSpot <= Math.min(spot + 6, size * size); nextSpot++) {
                int r = positions[nextSpot].getKey(), c = positions[nextSpot].getValue();
                int target = grid[r][c] != -1 ? grid[r][c] : nextSpot;
                if (moves[target] == -1 || moves[spot] + 1 < moves[target]) {
                    moves[target] = moves[spot] + 1;
                    queue.add(new Node(moves[target], target));
                }
            }
        }
        return moves[size * size];
    }
}

This Java solution implements a method to compute the minimum number of dice throws required to win the game of Snakes and Ladders, based on an input grid representing the game board. The game board is a two-dimensional array where each cell either has a number indicating a shortcut (ladder or snake) or -1 implying a regular step. The grid dimension is size x size, representing a sequential structure flattened in a snake-like pattern.

• The solution initializes a mapping of board positions to grid coordinates (positions array) by zigzagging through the grid.

• Each position number references its respective row and column in the grid, carefully accounting for the reverse order in alternate rows to mimic the gameplay structure.

• A custom class Node, which encapsulates both the moves taken to reach a position and the position itself, helps in prioritizing exploration of paths with fewer moves.

• A priority queue efficiently manages nodes to explore, populating it initially with the starting node (the first square).

• The main computational logic iteratively explores potential moves from the current position:

  • For each valid dice roll move (1 to 6):
    • It calculates the resultant position on the grid.
    • If there exists a snake or ladder, it adjusts the target position accordingly.
    • If this new adjusted position offers a shorter path than previous recorded attempts, the solution updates the moves taken to reach this new position and adds the new position for further exploration.

• The process repeats until all possible positions are evaluated or the queue depletes, ultimately returning the minimum moves needed to reach the last square or -1 if not possible.

This approach ensures efficiency by leveraging a priority queue to always process the most promising next move first and dynamically adjusting paths using the game's shortcuts. Through an object-oriented approach and methodical variable management, the solution remains scalable and relatively easy to understand within a gaming context.

class Solution:
    def playSnakesLadders(self, grid: List[List[int]]) -> int:
        dimension = len(grid)
        positions = [None] * (dimension**2 + 1)
        idx = 1
        cols = list(range(0, dimension))
        for r in range(dimension - 1, -1, -1):
            for c in cols:
                positions[idx] = (r, c)
                idx += 1
            cols.reverse()
        steps = [-1] * (dimension * dimension + 1)
        steps[1] = 0
        queue = [(0, 1)]
        while queue:
            step, current = heapq.heappop(queue)
            if step != steps[current]:
                continue
            for next_cell in range(current + 1, min(current + 6, dimension**2) + 1):
                r, c = positions[next_cell]
                jump = (grid[r][c] if grid[r][c] != -1 else next_cell)
                if steps[jump] == -1 or steps[current] + 1 < steps[jump]:
                    steps[jump] = steps[current] + 1
                    heapq.heappush(queue, (steps[jump], jump))
        return steps[dimension * dimension]

This solution is an efficient Python3 implementation to solve the "Snakes and Ladders" problem using the standard Breadth-First Search (BFS) approach.

The approach follows these steps:

1. Instead of moving around in a matrix fashion which can be cumbersome due to the irregular movement in snakes and ladders, this solution flattens the board into a list for easy access. It maps each square of the board to a unique position in the flattened list.

2. The calculation of the board position considers if the row is odd or even, mirroring real movement on a physical Snakes and Ladders board.

3. BFS is used for finding the shortest path. The list steps keeps track of minimum steps needed to reach each position. The position starts at 1, following typical game rules. It uses a priority queue to ensure the smallest step counts are processed first.

4. For each cell within a dice throw range (1 to 6), the target cell is determined. If a ladder or snake is present, it directly modifies the target position accordingly.

5. If this new computed target position can be reached in fewer steps than currently recorded, the count is updated in the steps list, and the position is added back to the queue for further exploration.

The algorithm efficiently computes the minimum number of rolls needed to reach the end of the board, or indicates if the end is not reachable by returning a special value.

The logic ensures that all possible scenarios on the board are considered, from stepping on a ladder, snake, or ordinary cells, and calculates the minimum required moves efficiently. A priority queue (min-heap) is utilized to always process the next most promising position. The program deals with 1-based index for compatibility with common game configurations.