# 1223. Dice Roll Simulation

## 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.

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.

• 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;
}
}

• 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);
}
};

• 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)


• 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)
}