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.

Two sequences are considered different if at least one element differs from each other. Since the answer may be too large, return it modulo 10^9 + 7.

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

Related Topics:
Dynamic Programming

Let dp[i][j] be the number of distinct sequences ending with j obtained using i rolles.

dp[1][j] = 1       1 <= j <= 6

Ignoring the rollMax, we have dp[i][j] = sum( dp[i-1][k] | 1 <= k <= 6 ).

If we pick j at i roll, we need to exclude the cases where there are already rollMax[j] j numbers right in front of ith roll.

The number of such cases is dp[i-rollMax[j]-1][k] where 1 <= k <= 6 and k != j.

// OJ: https://leetcode.com/problems/dice-roll-simulation/
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(N)
// Ref: https://leetcode.com/problems/dice-roll-simulation/discuss/403756/Java-Share-my-DP-solution
class Solution {
public:
    int dieSimulator(int n, vector<int>& A) {
        long mod = 1e9+7;
        vector<vector<long>> dp(n, vector<long>(7));
        for (int i = 0; i < 6; ++i) dp[0][i] = 1;
        dp[0][6] = 6;
        for (int i = 1; i < n; ++i) {
            long sum = 0;
            for (int j = 0; j < 6; ++j) {
                dp[i][j] = dp[i - 1][6];
                if (i < A[j]) sum = (sum + dp[i][j]) % mod;
                else {
                    if (i - A[j] - 1 >= 0) dp[i][j] = (dp[i][j] - (dp[i - A[j] - 1][6] - dp[i - A[j] - 1][j]) + mod) % mod;
                    else dp[i][j] = (dp[i][j] - 1) % mod;
                    sum = (sum + dp[i][j]) % mod;
                }
            }
            dp[i][6] = sum;
        }
        return dp[n - 1][6];
    }
};