zoukankan      html  css  js  c++  java
  • 576. Out of Boundary Paths

    There is an m by n grid with a ball. Given the start coordinate (i,j) of the ball, you can move the ball to adjacent cell or cross the grid boundary in four directions (up, down, left, right). However, you can at most move N times. Find out the number of paths to move the ball out of grid boundary. The answer may be very large, return it after mod 109 + 7.

    Example 1:

    Input: m = 2, n = 2, N = 2, i = 0, j = 0
    Output: 6
    Explanation:
    

    Example 2:

    Input: m = 1, n = 3, N = 3, i = 0, j = 1
    Output: 12
    Explanation:
    

    Note:

    1. Once you move the ball out of boundary, you cannot move it back.
    2. The length and height of the grid is in range [1,50].
    3. N is in range [0,50].
     

    Approach #1: DP. [C++]

    class Solution {
    public:
        int findPaths(int m, int n, int N, int i, int j) {
            const int mod = 1000000007;
            vector<vector<vector<int>>> dp(N+1, vector<vector<int>>(m, vector<int>(n, 0)));
            vector<int> dirs = {1, 0, -1, 0, 1};
            for (int s = 1; s <= N; ++s) {
                for (int x = 0; x < m; ++x) {
                    for (int y = 0; y < n; ++y) {
                        for (int k = 0; k < 4; ++k) {
                            int dx = x + dirs[k];
                            int dy = y + dirs[k+1];
                            if (dx < 0 || dy < 0 || dx >= m || dy >= n) 
                                dp[s][x][y] += 1;
                            else 
                                dp[s][x][y] = (dp[s][x][y] + dp[s-1][dx][dy]) % mod;
                        }
                    }
                }
            }
            return dp[N][i][j];
        }
    };
    

      

    Analysis:

    Observation:

    Number of paths start from (i, j) to out of boundary <=> Number of paths start from out of boundary to (i, j).

    dp[N][i][j] : Number of paths start from out of boundary to (i, j) by moving N steps.

    dp[*][y][x] = 1, if (x, y) are out of boundary

    dp[s][i][j] = dp[s-1][i+1][j] + dp[s-1][i-1][j] + dp[s-1][i][j+1] + dp[s-1][i][j-1]

    Ans: dp[N][i][j]

    Time complexity: O(N*m*n)

    Space complexity: O(N*m*n) -> O(m*n)

    Reference:

    http://zxi.mytechroad.com/blog/dynamic-programming/leetcode-576-out-of-boundary-paths/

    永远渴望,大智若愚(stay hungry, stay foolish)
  • 相关阅读:
    第十四周课程总结&实验报告(简单记事本的实现)
    第十三周课程总结
    第十二周课程总结
    第十一周课程总结
    第十周课程总结
    第九周课程总结&实验报告(七)
    第八周课程总结&实验报告(六)
    第七周课程总结&试验报告(五)
    基于C的
    RMQ 区间最值问题
  • 原文地址:https://www.cnblogs.com/h-hkai/p/10498491.html
Copyright © 2011-2022 走看看