LeetCode 1223. Dice Roll Simulation

标签：therefore   etc   least   and   this   other   not   constrain   simulator   

原题链接在这里：https://leetcode.com/problems/dice-roll-simulation/

题目：

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

题解：

Let dp[i][j][k] denotes number of distinct sequences up to i dice, ending with number j with k repeating j at the end.

For previous p, if j != p, then dp[i][j][1] = sum(dp[i - 1][p][k]), sum of previous ending with p, with k repeating p at the end.

Else if j == p, we need to make sure k + 1 <= rollMax[j], dp[i][j][k + 1] = dp[i - 1][j][k]. otherwise it would be 0. 

Time Complexity: O(n*m*m*rMax). m = 6. rMax = max(rollMax[i]).

Space: O(n*m*rMax).

AC Java:

 1 class Solution {
 2     public int dieSimulator(int n, int[] rollMax) {
 3         int mod = 1000000007;
 4         int rMax = 15;
 5         
 6         int [][][] dp = new int[n + 1][6][rMax + 1];
 7         for(int i = 0; i < 6; i++){
 8             dp[1][i][1] = 1;
 9         }
10         
11         for(int i = 2; i <= n; i++){
12             for(int j = 0; j < 6; j++){
13                 for(int p = 0; p < 6; p++){
14                     for(int k = 1; k <= rMax; k++){
15                         if(j != p){
16                             dp[i][j][1] = (dp[i][j][1] + dp[i - 1][p][k]) % mod;
17                         }else if(k < rollMax[j]){ // same number, make sure k + 1 <= rollMax[j]
18                             dp[i][j][k + 1] = dp[i - 1][j][k];
19                         }
20                     }
21                 }
22             }
23         }
24         
25         int res = 0;
26         for(int j = 0; j < 6; j++){
27             for(int k = 1; k <= rMax; k++){
28                 res = (res + dp[n][j][k]) % mod;
29             }
30         }
31         
32         return res;
33     }
34 }

LeetCode 1223. Dice Roll Simulation

标签：therefore   etc   least   and   this   other   not   constrain   simulator   

原文地址：https://www.cnblogs.com/Dylan-Java-NYC/p/12325988.html