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Hard
2008
Weekly Contest 158 Q3
Array
Dynamic Programming

1223. Dice Roll Simulation

中文文档

Description

A die simulator generates a random number from 1 to 6 for each roll. You introduced a constraint to the generator such that it cannot roll the number i more than rollMax[i] (1-indexed) consecutive times.

Given an array of integers rollMax and an integer n, return the number of distinct sequences that can be obtained with exact n rolls. Since the answer may be too large, return it modulo 109 + 7.

Two sequences are considered different if at least one element differs from each other.

 

Example 1:

Input: n = 2, rollMax = [1,1,2,2,2,3]
Output: 34
Explanation: There will be 2 rolls of die, if there are no constraints on the die, there are 6 * 6 = 36 possible combinations. In this case, looking at rollMax array, the numbers 1 and 2 appear at most once consecutively, therefore sequences (1,1) and (2,2) cannot occur, so the final answer is 36-2 = 34.

Example 2:

Input: n = 2, rollMax = [1,1,1,1,1,1]
Output: 30

Example 3:

Input: n = 3, rollMax = [1,1,1,2,2,3]
Output: 181

 

Constraints:

  • 1 <= n <= 5000
  • rollMax.length == 6
  • 1 <= rollMax[i] <= 15

Solutions

Solution 1: Memoization Search

We can design a function $dfs(i, j, x)$ to represent the number of schemes starting from the $i$-th dice roll, with the current dice roll being $j$, and the number of consecutive times $j$ is rolled being $x$. The range of $j$ is $[1, 6]$, and the range of $x$ is $[1, rollMax[j - 1]]$. The answer is $dfs(0, 0, 0)$.

The calculation process of the function $dfs(i, j, x)$ is as follows:

  • If $i \ge n$, it means that $n$ dice have been rolled, return $1$.
  • Otherwise, we enumerate the number $k$ rolled next time. If $k \ne j$, we can directly roll $k$, and the number of consecutive times $j$ is rolled will be reset to $1$, so the number of schemes is $dfs(i + 1, k, 1)$. If $k = j$, we need to judge whether $x$ is less than $rollMax[j - 1]$. If it is less, we can continue to roll $j$, and the number of consecutive times $j$ is rolled will increase by $1$, so the number of schemes is $dfs(i + 1, j, x + 1)$. Finally, add all the scheme numbers to get the value of $dfs(i, j, x)$. Note that the answer may be very large, so we need to take the modulus of $10^9 + 7$.

During the process, we can use memoization search to avoid repeated calculations.

The time complexity is $O(n \times k^2 \times M)$, and the space complexity is $O(n \times k \times M)$. Here, $k$ is the range of dice points, and $M$ is the maximum number of times a certain point can be rolled consecutively.

Python3

class Solution:
    def dieSimulator(self, n: int, rollMax: List[int]) -> int:
        @cache
        def dfs(i, j, x):
            if i >= n:
                return 1
            ans = 0
            for k in range(1, 7):
                if k != j:
                    ans += dfs(i + 1, k, 1)
                elif x < rollMax[j - 1]:
                    ans += dfs(i + 1, j, x + 1)
            return ans % (10**9 + 7)

        return dfs(0, 0, 0)

Java

class Solution {
    private Integer[][][] f;
    private int[] rollMax;

    public int dieSimulator(int n, int[] rollMax) {
        f = new Integer[n][7][16];
        this.rollMax = rollMax;
        return dfs(0, 0, 0);
    }

    private int dfs(int i, int j, int x) {
        if (i >= f.length) {
            return 1;
        }
        if (f[i][j][x] != null) {
            return f[i][j][x];
        }
        long ans = 0;
        for (int k = 1; k <= 6; ++k) {
            if (k != j) {
                ans += dfs(i + 1, k, 1);
            } else if (x < rollMax[j - 1]) {
                ans += dfs(i + 1, j, x + 1);
            }
        }
        ans %= 1000000007;
        return f[i][j][x] = (int) ans;
    }
}

C++

class Solution {
public:
    int dieSimulator(int n, vector<int>& rollMax) {
        int f[n][7][16];
        memset(f, 0, sizeof f);
        const int mod = 1e9 + 7;
        function<int(int, int, int)> dfs = [&](int i, int j, int x) -> int {
            if (i >= n) {
                return 1;
            }
            if (f[i][j][x]) {
                return f[i][j][x];
            }
            long ans = 0;
            for (int k = 1; k <= 6; ++k) {
                if (k != j) {
                    ans += dfs(i + 1, k, 1);
                } else if (x < rollMax[j - 1]) {
                    ans += dfs(i + 1, j, x + 1);
                }
            }
            ans %= mod;
            return f[i][j][x] = ans;
        };
        return dfs(0, 0, 0);
    }
};

Go

func dieSimulator(n int, rollMax []int) int {
	f := make([][7][16]int, n)
	const mod = 1e9 + 7
	var dfs func(i, j, x int) int
	dfs = func(i, j, x int) int {
		if i >= n {
			return 1
		}
		if f[i][j][x] != 0 {
			return f[i][j][x]
		}
		ans := 0
		for k := 1; k <= 6; k++ {
			if k != j {
				ans += dfs(i+1, k, 1)
			} else if x < rollMax[j-1] {
				ans += dfs(i+1, j, x+1)
			}
		}
		f[i][j][x] = ans % mod
		return f[i][j][x]
	}
	return dfs(0, 0, 0)
}

Solution 2: Dynamic Programming

We can change the memoization search in Solution 1 to dynamic programming.

Define $f[i][j][x]$ as the number of schemes for the first $i$ dice rolls, with the $i$-th dice roll being $j$, and the number of consecutive times $j$ is rolled being $x$. Initially, $f[1][j][1] = 1$, where $1 \leq j \leq 6$. The answer is:

$$ \sum_{j=1}^6 \sum_{x=1}^{rollMax[j-1]} f[n][j][x] $$

We enumerate the last dice roll as $j$, and the number of consecutive times $j$ is rolled as $x$. The current dice roll can be $1, 2, \cdots, 6$. If the current dice roll is $k$, there are two cases:

  • If $k \neq j$, we can directly roll $k$, and the number of consecutive times $j$ is rolled will be reset to $1$. Therefore, the number of schemes $f[i][k][1]$ will increase by $f[i-1][j][x]$.
  • If $k = j$, we need to judge whether $x+1$ is less than or equal to $rollMax[j-1]$. If it is less than or equal to, we can continue to roll $j$, and the number of consecutive times $j$ is rolled will increase by $1$. Therefore, the number of schemes $f[i][j][x+1]$ will increase by $f[i-1][j][x]$.

The final answer is the sum of all $f[n][j][x]$.

The time complexity is $O(n \times k^2 \times M)$, and the space complexity is $O(n \times k \times M)$. Here, $k$ is the range of dice points, and $M$ is the maximum number of times a certain point can be rolled consecutively.

Python3

class Solution:
    def dieSimulator(self, n: int, rollMax: List[int]) -> int:
        f = [[[0] * 16 for _ in range(7)] for _ in range(n + 1)]
        for j in range(1, 7):
            f[1][j][1] = 1
        for i in range(2, n + 1):
            for j in range(1, 7):
                for x in range(1, rollMax[j - 1] + 1):
                    for k in range(1, 7):
                        if k != j:
                            f[i][k][1] += f[i - 1][j][x]
                        elif x + 1 <= rollMax[j - 1]:
                            f[i][j][x + 1] += f[i - 1][j][x]
        mod = 10**9 + 7
        ans = 0
        for j in range(1, 7):
            for x in range(1, rollMax[j - 1] + 1):
                ans = (ans + f[n][j][x]) % mod
        return ans

Java

class Solution {
    public int dieSimulator(int n, int[] rollMax) {
        int[][][] f = new int[n + 1][7][16];
        for (int j = 1; j <= 6; ++j) {
            f[1][j][1] = 1;
        }
        final int mod = (int) 1e9 + 7;
        for (int i = 2; i <= n; ++i) {
            for (int j = 1; j <= 6; ++j) {
                for (int x = 1; x <= rollMax[j - 1]; ++x) {
                    for (int k = 1; k <= 6; ++k) {
                        if (k != j) {
                            f[i][k][1] = (f[i][k][1] + f[i - 1][j][x]) % mod;
                        } else if (x + 1 <= rollMax[j - 1]) {
                            f[i][j][x + 1] = (f[i][j][x + 1] + f[i - 1][j][x]) % mod;
                        }
                    }
                }
            }
        }
        int ans = 0;
        for (int j = 1; j <= 6; ++j) {
            for (int x = 1; x <= rollMax[j - 1]; ++x) {
                ans = (ans + f[n][j][x]) % mod;
            }
        }
        return ans;
    }
}

C++

class Solution {
public:
    int dieSimulator(int n, vector<int>& rollMax) {
        int f[n + 1][7][16];
        memset(f, 0, sizeof f);
        for (int j = 1; j <= 6; ++j) {
            f[1][j][1] = 1;
        }
        const int mod = 1e9 + 7;
        for (int i = 2; i <= n; ++i) {
            for (int j = 1; j <= 6; ++j) {
                for (int x = 1; x <= rollMax[j - 1]; ++x) {
                    for (int k = 1; k <= 6; ++k) {
                        if (k != j) {
                            f[i][k][1] = (f[i][k][1] + f[i - 1][j][x]) % mod;
                        } else if (x + 1 <= rollMax[j - 1]) {
                            f[i][j][x + 1] = (f[i][j][x + 1] + f[i - 1][j][x]) % mod;
                        }
                    }
                }
            }
        }
        int ans = 0;
        for (int j = 1; j <= 6; ++j) {
            for (int x = 1; x <= rollMax[j - 1]; ++x) {
                ans = (ans + f[n][j][x]) % mod;
            }
        }
        return ans;
    }
};

Go

func dieSimulator(n int, rollMax []int) (ans int) {
	f := make([][7][16]int, n+1)
	for j := 1; j <= 6; j++ {
		f[1][j][1] = 1
	}
	const mod = 1e9 + 7
	for i := 2; i <= n; i++ {
		for j := 1; j <= 6; j++ {
			for x := 1; x <= rollMax[j-1]; x++ {
				for k := 1; k <= 6; k++ {
					if k != j {
						f[i][k][1] = (f[i][k][1] + f[i-1][j][x]) % mod
					} else if x+1 <= rollMax[j-1] {
						f[i][j][x+1] = (f[i][j][x+1] + f[i-1][j][x]) % mod
					}
				}
			}
		}
	}
	for j := 1; j <= 6; j++ {
		for x := 1; x <= rollMax[j-1]; x++ {
			ans = (ans + f[n][j][x]) % mod
		}
	}
	return
}