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  4. Snakes and Ladders: The Quickest Way Up
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Snakes and Ladders: The Quickest Way Up

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  • zoyashah41306
     + 0 comments

    Locksmithlocal services ensure your security needs are met, while Snakes and Ladders offers a fun metaphor for life's ups and downs. In the game, ladders represent opportunities that quickly elevate your position, while snakes symbolize setbacks that can bring you down. Just like in life, it’s about navigating challenges and seizing opportunities. The quickest way up involves taking calculated risks and making the most of each climb, no matter the obstacles.

  • zoyashah41306
     + 0 comments

    Branded merchandise can make your Snakes and Ladders game even more exciting! Just like the game, where players take the quickest route up the ladder or face setbacks with snakes, these items add a fun twist to your promotional strategy. Whether it's a custom board or unique game pieces, incorporating your brand ensures that every move is remembered and creates lasting impressions. Elevate your branding with a playful, engaging touch!

  • mah_shamim
     + 0 comments

    "Snakes and Ladders: The Quickest Way Up" problem, we can use the Breadth-First Search (BFS) algorithm. Below is the PHP solution for this problem:

    <?php

function quickestWayUp($ladders, $snakes) {
    $board = array_fill(1, 100, 0);

    // Set up ladders
    foreach ($ladders as $ladder) {
        $board[$ladder[0]] = $ladder[1] - $ladder[0];
    }

    // Set up snakes
    foreach ($snakes as $snake) {
        $board[$snake[0]] = $snake[1] - $snake[0];
    }

    // BFS to find the shortest path
    $queue = [[1, 0]];
    $visited = array_fill(1, 100, false);
    $visited[1] = true;

    while (!empty($queue)) {
        list($current, $moves) = array_shift($queue);

        for ($dice = 1; $dice <= 6; $dice++) {
            $next = $current + $dice;
            if ($next <= 100) {
                $next += $board[$next];

                if ($next == 100) {
                    return $moves + 1;
                }

                if (!$visited[$next]) {
                    $visited[$next] = true;
                    array_push($queue, [$next, $moves + 1]);
                }
            }
        }
    }

    return -1;
}

?>

    Explanation:

    1. Initialization:

      • The board array stores the effect of ladders and snakes on each cell.
      • queue is used for BFS, initialized with the starting position (1, 0) representing the start of the board and 0 moves.
      • visited array keeps track of visited cells to avoid revisiting them.

    2. Setup Ladders and Snakes:

      • For ladders, the board at the start cell of the ladder has the value of the end cell minus the start cell.
      • Similarly, for snakes, the board at the start cell of the snake has the value of the end cell minus the start cell.

    3. BFS to Find Shortest Path:

      • Dequeue the current position and number of moves.
      • For each dice roll (1 to 6), calculate the next position.
      • Apply any ladder or snake effect.
      • Check if the next position is 100 (end of the board).
      • If not visited, mark it as visited and enqueue it with incremented move count.

    4. Handling Input:

      • The readInput function reads the input from standard input and processes it.
      • For each test case, read the number of ladders and snakes, and their respective positions.
      • Call the quickestWayUp function for each test case and print the result.

  • wgcv92
     + 0 comments

    Pthon

    BFS

    def bfs(node, squares):
    visited = set()
    q = []
    q.append(node)
    sentinel = None
    q.append(sentinel)
    movement = 0
    while(len(q)>0):
        n = q.pop(0)
       
        if(n is sentinel):
            if(q):
                movement += 1
                q.append(sentinel)
                continue
            else:
                return -1
        if (n.finish):
            return movement
        if(n in visited):
            continue
        visited.add(n)
        for i in n.edges:
            q.append(squares[i])
            
def quickestWayUp(ladders, snakes):
    squares = {}
    dladders = {}
    dsnakes = {}
    for i in ladders:
        s1 = i[0]
        s2 = i[1]
        dladders[s1] = s2
    for i in snakes:
        s1 = i[0]
        s2 = i[1]
        dsnakes[s1] = s2
    for i in range(100, 0, -1):
        n = Node(i)
        squares[i] = n
        for j in range(i+1, i+7):
            if (j<=100):
                if(j in dladders):
                    n.edges.append(dladders[j])
                elif (j in dsnakes):
                    n.edges.append(dsnakes[j])
                else:
                    n.edges.append(j)  
    # for i in range(1,101):
    #     print("n", squares[i].val)
    #     print(squares[i].edges)
    squares[100].finish=True
    return bfs(squares[1], squares)

  • marcioreisjr
     + 0 comments

    Rust is generally a bit verbose. The execution time is nice, though. time ./target/release/rust < input_1.txt (test case #1) real 0m0.016s user 0m0.006s sys 0m0.008s

    use std::collections::HashMap;
const BOARD_LEN: usize = 100;

fn quickest_way_up_dp(ii: usize, memo: &mut HashMap<usize, i32>, st: &Vec<i32>) -> Option<i32> {
    if memo.contains_key(&ii) {
        return Some(*memo.get(&ii).unwrap());
    }
    if ii == BOARD_LEN {
        return Some(0);
    }
    if ii > BOARD_LEN || st[ii] > (BOARD_LEN as i32) || st[ii] < 0 || st[ii] == ii as i32 {
        return None;
    }
    if st[ii] != 0 {
        return quickest_way_up_dp(st[ii] as usize, memo, st);
    }
    memo.insert(ii, i32::MAX);
    let mut min = i32::MAX;
    for dice in 1..=6 {
        let next = ii + dice;
        if let Some(res) = quickest_way_up_dp(next, memo, st) {
            min = min.min(res);
        }
    }
    if min != i32::MAX {
        memo.insert(ii, 1 + min);
        return Some(*memo.get(&ii).unwrap());
    }
    None
}

fn quickestWayUp(ladders: &[Vec<i32>], snakes: &[Vec<i32>]) -> i32 {
    let mut steps: Vec<i32> = vec![0; 101];
    for v in ladders {
        steps[v[0] as usize] = v[1];
    }
    for v in snakes {
        steps[v[0] as usize] = v[1];
    }
    quickest_way_up_dp(1, &mut HashMap::new(), &steps).unwrap_or(-1)
}