花花酱 LeetCode 935. Knight Dialer

zxi — Sun, 04 Nov 2018 15:58:59 +0000

Problem

https://leetcode.com/problems/knight-dialer/description/

A chess knight can move as indicated in the chess diagram below:

This time, we place our chess knight on any numbered key of a phone pad (indicated above), and the knight makes N-1 hops. Each hop must be from one key to another numbered key.

Each time it lands on a key (including the initial placement of the knight), it presses the number of that key, pressing N digits total.

How many distinct numbers can you dial in this manner?

Since the answer may be large, output the answer modulo 10^9 + 7.

Example 1:

Input: 1
Output: 10

Example 2:

Input: 2
Output: 20

Example 3:

Input: 3
Output: 46

Note:

1 <= N <= 5000

Solution: DP

V1

We can define dp[k][i][j] as # of ways to dial and the last key is (j, i) after k steps

Note: dp[*][3][0], dp[*][3][2] are always zero for all the steps.

Init: dp[0][i][j] = 1

Transition: dp[k][i][j] = sum(dp[k – 1][i + dy][j + dx]) 8 ways of move from last step.

ans = sum(dp[k])

Time complexity: O(kmn) or O(k * 12 * 8) = O(k)

Space complexity: O(kmn) -> O(mn) or O(12*8) = O(1)

V2

define dp[k][i] as # of ways to dial and the last key is i after k steps

init: dp[0][0:10] = 1

transition: dp[k][i] = sum(dp[k-1][j]) that j can move to i

ans: sum(dp[k])

Time complexity: O(k * 10) = O(k)

Space complexity: O(k * 10) -> O(10) = O(1)

C++ V1

// Author: Huahua, 96 ms
class Solution {
public:
  int knightDialer(int N) {
    constexpr int kMod = 1e9 + 7;
    int dirs[8][2] = {{-2, -1}, {-2, 1}, {-1, -2}, {-1, 2}, {1, -2}, {1, 2}, {2, -1}, {2, 1}};
    vector> dp(4, vector(3, 1));
    dp[3][0] = dp[3][2] = 0;    
    for (int k = 1; k < N; ++k) {
      vector> tmp(4, vector(3));
      for (int i = 0; i < 4; ++i)
        for (int j = 0; j < 3; ++j) {
          if (i == 3 && j != 1) continue;
          for (int d = 0; d < 8; ++d) {           
            int tx = j + dirs[d][0];
            int ty = i + dirs[d][1];
            if (tx < 0 || ty < 0 || tx >= 3 || ty >= 4) continue;
            tmp[i][j] = (tmp[i][j] + dp[ty][tx]) % kMod;
          }          
        }
      dp.swap(tmp);
    }
    int ans = 0;
    for (int i = 0; i < 4; ++i)
      for (int j = 0; j < 3; ++j)
        ans = (ans + dp[i][j]) % kMod;
    return ans;
  }
};

C++ V2

// Author: Huahua, 24 ms
public:
  int knightDialer(int N) {
    constexpr int kMod = 1e9 + 7;
    vector> moves{{4,6},{8,6},{7,9},{4,8},{3,9,0},{},{1,7,0},{2,6},{1,3},{2,4}};
    vector dp(10, 1);
    for (int k = 1; k < N; ++k) {
      vector tmp(10);
      for (int i = 0; i < 10; ++i)
        for (int nxt : moves[i])
          tmp[nxt] = (tmp[nxt] + dp[i]) % kMod;        
      dp.swap(tmp);
    }
    int ans = 0;
    for (int i = 0; i < 10; ++i)
      ans = (ans + dp[i]) % kMod;
    return ans;
  }
};

花花酱 688. Knight Probability in Chessboard

zxi — Mon, 02 Oct 2017 06:18:42 +0000

https://leetcode.com/problems/knight-probability-in-chessboard/description/

Problem:

On an NxN chessboard, a knight starts at the r-th row and c-th column and attempts to make exactly Kmoves. The rows and columns are 0 indexed, so the top-left square is (0, 0), and the bottom-right square is (N-1, N-1).

A chess knight has 8 possible moves it can make, as illustrated below. Each move is two squares in a cardinal direction, then one square in an orthogonal direction.

Each time the knight is to move, it chooses one of eight possible moves uniformly at random (even if the piece would go off the chessboard) and moves there.

The knight continues moving until it has made exactly K moves or has moved off the chessboard. Return the probability that the knight remains on the board after it has stopped moving.

Example:

Input: 3, 2, 0, 0
Output: 0.0625
Explanation: There are two moves (to (1,2), (2,1)) that will keep the knight on the board.
From each of those positions, there are also two moves that will keep the knight on 
the board. The total probability the knight stays on the board is 0.0625.

Note:

N will be between 1 and 25.
K will be between 0 and 100.
The knight always initially starts on the board.

Idea:

Dynamic programming

Count the ways to reach (x, y) after k moves from start.

Time Complexity: O(k*n^2)

Space Complexity: O(n^2)

Solution:

// Author: Huahua
// Runtime: 6 ms
class Solution {
public:
    double knightProbability(int N, int K, int r, int c) {
        vector> dp0(N, vector(N, 0.0));
        dp0[r][c] = 1.0;
        int dirs[8][2] = {{1, 2}, {-1, -2}, {1, -2}, {-1, 2},
                          {2, 1}, {-2, -1}, {2, -1}, {-2, 1}};
        for (int k = 0; k < K; ++k) {            
            vector> dp1(N, vector(N, 0.0));
            for (int i = 0; i < N; ++i)
                for (int j = 0; j < N; ++j) 
                    for (int m = 0; m < 8; ++m) {
                        int x = j + dirs[m][0];
                        int y = i + dirs[m][1];
                        if (x < 0 || y < 0 || x >= N || y >= N) continue;
                        dp1[i][j] += dp0[y][x];
                    }
            std::swap(dp0, dp1);
        }
        
        double total = 0;
        for (int i = 0; i < N; ++i)
            for (int j = 0; j < N; ++j)
                total += dp0[i][j];
        
        return total / pow(8, K);
    }
};

Related problems:

The post 花花酱 688. Knight Probability in Chessboard appeared first on Huahua's Tech Road.

knight Archives - Huahua's Tech Road