Leetcode 935.Knight Dialer

The chess knight has a unique movement, it may move two squares vertically and one square horizontally, or two squares horizontally and one square vertically (with both forming the shape of an L). The possible movements of chess knight are shown in this diagaram:

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

We have a chess knight and a phone pad as shown below, the knight can only stand on a numeric cell (i.e. blue cell).

Given an integer n, return how many distinct phone numbers of length n we can dial.

You are allowed to place the knight on any numeric cell initially and then you should perform n - 1 jumps to dial a number of length n. All jumps should be valid knight jumps.

As the answer may be very large, return the answer modulo 109 + 7.

Example 1:

Input: n = 1
Output: 10
Explanation: We need to dial a number of length 1, so placing the knight over any numeric cell of the 10 cells is sufficient.

Example 2:

Input: n = 2
Output: 20
Explanation: All the valid number we can dial are [04, 06, 16, 18, 27, 29, 34, 38, 40, 43, 49, 60, 61, 67, 72, 76, 81, 83, 92, 94]

Example 3:

Input: n = 3
Output: 46

Example 4:

Input: n = 4
Output: 104

Example 5:

Input: n = 3131
Output: 136006598
Explanation: Please take care of the mod.

Constraints:

• 1 <= n <= 5000

Solution 1: DFS + memoization

public class Knight_Dialer_DFS {
    // Time O(10 * N) = O(N)
    public int knightDial(int N) {
        int count = 0; // count the result
        int[][] memo = new int[N][10]; // memoization
        for (int pos = 0; pos < 10; pos++) {
            memo[0][pos] = 1; // initialize the memo
        }
        for (int pos = 0; pos < 10; pos++) {
            count += knightDial(pos, N - 1, memo);
        }
        return count;
    }

    private int knightDial(int pos, int hops, int[][] memo) {
        if (memo[hops][pos] != 0) {
            return memo[hops][pos];
        }

        int count = 0;
        for (int next: neighbors[pos]) {
            count += knightDial(next, hops - 1, memo);
        }
        memo[hops][pos] = count;
        return count;
    }

    //private final int[][] neighbors = {4, 6}, {6, 8}, {7, 9}, {4, 8}, {0, 3, 9}, {}, {0, 1, 7}, {2, 6}, {1, 3}, {2, 4};

    public static void main(String[] args) {
        Knight_Dialer_DFS solution = new Knight_Dialer_DFS();

        // test case
        int a = solution.knightDial(10);
        System.out.println(a);
    }
}

Solution 2: DP

public class Knight_Dialer_DP {
    public int knightDial(int N) {
        // memo[hops][pos] represents the number of solutions ending at pos in hops
        int[][] memo = new int[N][10];
        for (int i = 0; i < 10; i++) {
            memo[0][i] = 1;
        }
        for (int i = 1; i < N; i++) {
            for (int j = 0; j < 10; j++) {
                for (int next: neighbors[j]) {
                    memo[i][j] += memo[i - 1][next];
                  	memo[i][j] %= Math.pow(10, 9) + 7; // 防止stack overflow
                }
            }
        }

        int sum = 0;
        for (int k = 0; k < 10; k++) {
            sum += memo[N - 1][k];
        }
        return (int)(sum % (Math.pow(10, 9) + 7));
    }

    // private final int[][] neighbors = {4, 6}, {6, 8}, {7, 9}, {4, 8}, {0, 3, 9}, {}, {0, 1, 7}, {2, 6}, {1, 3}, {2, 4};

    public static void main(String[] args) {
        DP.Knight_Dialer_DP solution = new DP.Knight_Dialer_DP();

        // test case
        int a = solution.knightDial(10);
        System.out.println(a);
    }