1223. Dice Roll Simulation

Problem Description

In this problem, we are given a virtual die that can roll numbers from 1 to 6. However, there is an additional constraint on the die: each number i cannot be rolled more than rollMax[i] consecutive times, where rollMax is an array given as input and is defined for each number from 1 to 6 (1-indexed).

The task is to calculate the number of distinct sequences that can be obtained with exactly n rolls under this constraint, where n is a given integer. Since the number of sequences could be very large, we are required to return the result modulo (10^9 + 7).

A sequence is considered distinct from another if at least one element in the sequence is different. This means the order and the number rolled matters for the distinctness of sequences.

Intuition

The solution to this problem uses dynamic programming and depth-first search (DFS) to explore all possible combinations of rolls under the given constraints.

The intuition behind using DFS is that, for each roll, we have 6 choices (numbers from 1 to 6), but we need to respect the rollMax constraints for consecutive rolls. We can define a function that takes the current position in the sequence (i), the last number rolled (j), and the count of how many times that number has been rolled consecutively (x).

We recursively call this function until we reach n rolls. While exploring each possibility, we conditionally add to our total count based on two criteria:

1. If the next number we are trying to roll is different from the last (k != j), then we can reset the consecutive count and continue from the next position.
2. If the next number is the same and we have not exceeded the maximum allowed consecutive rolls for this number (x < rollMax[j - 1]), then we increment the consecutive count and move to the next position.

By using a @cache decorator (assuming from Python's functools), we cache the results of subproblems and avoid recomputation, thus saving time and optimizing performance.

We repeatedly take the modulus of the result to keep the number within the required limit (10^9 + 7) at each step, as the final number can be quite large.

With this approach, we will be able to cover all paths of sequences without violating the given constraints, and we'll incrementally build up the total number of distinct sequences modulo (10^9 + 7).

Learn more about Dynamic Programming patterns.

Solution Approach

The implementation uses a DFS approach combined with memoization to efficiently explore all valid sequences. The key components of the solution are:

• A recursive function dfs that takes three parameters: i, which represents the current position in the sequence (or the current roll number); j, the last number rolled; and x, the current streak of the last number rolled (how many times it has been rolled consecutively).
• The base case for the dfs function occurs when i equals n, which means that we've successfully generated a sequence of n rolls without violating the constraints. Whenever this happens, the function returns 1 as it represents a distinct sequence.
• To utilize memoization, the @cache decorator is used. This stores the results of the dfs function calls with particular arguments, so when the same state is encountered again, the function can return the cached result instead of recalculating it.
• The dfs function iterates over the possible numbers to be rolled next (from 1 to 6), and for each choice, it checks whether the choice is different from the last rolled number (k != j). If it is, the function resets the consecutive count (x) to 1 and calls dfs for the next position (i + 1).
• If the next number equals the last rolled number, and the consecutive roll count x has not yet breached the maximum allowed by rollMax, it increments the roll count (x + 1) and recurses to i + 1.
• To handle the modulus operation, the sum is taken modulo (10^9 + 7) after each increment.

Given this recursive structure, the algorithm starts by calling dfs(0, 0, 0), meaning it starts with the first roll (position 0), with no last number rolled (0 in this context is not a valid die number and acts as a placeholder) and no consecutive count.

The function then branches out into all possible sequences while adhering to the constraints. The memoization ensures that previously encountered states contribute to the solution without further computational overhead. Finally, the answer is a sum of all branches modulo (10^9 + 7).

Example Walkthrough

Let's walk through a smaller example to illustrate the solution approach. Suppose we are given n = 2 rolls, and the rollMax array is [1, 1, 2, 2, 2, 2], which means we can roll the number 1 and 2 only once consecutively, but we can roll 3, 4, 5, and 6 up to twice consecutively.

We start with the call dfs(0, 0, 0). Here i equals 0 because we've made no rolls yet, j equals 0 because there is no last number rolled, and x equals 0 because we do not have a consecutive count yet.

From this starting point, we try all numbers from 1 to 6:

1. If we roll a 1, we call dfs(1, 1, 1) (one roll made, last number rolled is 1, consecutive count for the number 1 is 1).
   • On the next roll, we cannot roll a 1 again since rollMax[0] is 1, so we explore other numbers:
     • dfs(2, 2, 1): since we rolled a 2, a valid sequence [1, 2] is formed. This call returns 1 as it represents a distinct sequence.
     • Calls for numbers 3 to 6 follow the same logic as rolling a 2, each return 1 for their respective distinct sequences: [1, 3], [1, 4], [1, 5], [1, 6]. Therefore, rolling a 1 first contributes to 5 distinct sequences.
2. If we roll a 2, similarly to rolling a 1, we follow the same process and find we can't roll a 2 again, but we can roll numbers 1, 3, 4, 5, and 6, which gives us another 5 distinct sequences.

For numbers 3 to 6, since we can roll these numbers up to twice consecutively, we will have a few more possibilities:

3. If we roll a 3, we call dfs(1, 3, 1):
   • We could roll a 3 again (since rollMax[2] is 2), which is dfs(2, 3, 2), resulting in sequence [3, 3] contributing to 1 sequence.
   • We can also roll any other number, which would be similar to the previous cases and yield 5 distinct sequences for each initial 3: [3, 1], [3, 2], [3, 4], [3, 5], [3, 6].

Following the same logic for initial rolls of 4, 5, and 6, we end up with 5 distinct sequences for each of their alternative rolls (as they can only be followed by 5 other distinct numbers due to the rollMax constraint).

By summing up all the possible distinct sequences:

• When starting with a 1 or 2, we get 5 sequences each.
• When starting with 3, 4, 5, or 6, we obtain 6 sequences each because we can roll the same number twice or roll a different number (5 other possibilities).

The total number of sequences for n = 2 would be the sum of all these results, which is (2 * 5 + 4 * 6 = 10 + 24 = 34) distinct sequences.

This example clearly illustrates how the dfs function explores all possible combinations, while @cache stores the intermediate results, preventing recalculations and improving performance. The final step would involve taking this total (34 for this example) and returning it modulo (10^9 + 7).

Solution Implementation

Time and Space Complexity

The given code defines a function dieSimulator which uses depth-first search (DFS) with memoization to count the number of distinct die sequences that can be rolled.

Time Complexity

The time complexity of the algorithm is O(n * 6 * max(rollMax)).

• n is the number of dice rolls.
• 6 represents each possible die face value.
• max(rollMax) represents the maximum constraint for consecutive rolls of the same face.

For each state in our DFS, we have at most 6 choices of the die face to consider. Since we also consider the number of consecutive rolls of the same face (bounded by max(rollMax)), the time complexity includes this factor. DFS will run for each roll, so we multiply by n. Memoization ensures each state is calculated once, thus reducing the time complexity.

Space Complexity

The space complexity of the algorithm is O(n * max(rollMax) * 6).

• The cache potentially stores results for every combination of:
  • The number of dice left to roll (n).
  • The current face being rolled (6 faces).
  • The number of times the current face has been rolled consecutively, which is at most max(rollMax).

Each state requires a constant amount of space, and since the space is used to store the combination of states mentioned above, the space complexity is a product of these factors.

Learn more about how to find time and space complexity quickly using problem constraints.