How do you solve Number of Distinct Roll Sequences on LeetCode?

Number of Distinct Roll Sequences (LeetCode #2318) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Understanding GCD properties helps in filtering valid sequences quickly.

What is the brute force solution for Number of Distinct Roll Sequences?

The brute force approach for Number of Distinct Roll Sequences is Brute Force, with O(n²) time complexity and O(1) space complexity. We can generate all possible sequences of dice rolls and filter out the invalid ones based on the given conditions. This approach is straightforward but inefficient due to the exponential growth of possibilities as n increases. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Number of Distinct Roll Sequences?

Number of Distinct Roll Sequences uses the following concepts: Dynamic Programming, Memoization. The recommended patterns to study are: Dynamic Programming, Memoization.

What are the interview tips for Number of Distinct Roll Sequences?

Always clarify the problem constraints and edge cases before jumping into coding. Discuss your thought process and potential approaches with the interviewer. Be prepared to optimize your initial solution and explain your reasoning.

What are common mistakes when solving Number of Distinct Roll Sequences?

Not considering the GCD condition properly, leading to invalid sequences being counted. Failing to account for the gap condition when checking for duplicates.

#2318

Number of Distinct Roll Sequences

Hard

Dynamic ProgrammingMemoizationDynamic ProgrammingMemoization

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

We can use dynamic programming to build valid sequences incrementally while keeping track of the last two rolls. This way, we can efficiently calculate the number of valid sequences without generating all possibilities.

⚙️

Algorithm

4 steps

1Step 1: Create a DP array where dp[i][j] represents the number of valid sequences of length i ending with the value j.
2Step 2: Initialize base cases for sequences of length 1.
3Step 3: For each length from 2 to n, calculate the number of valid sequences by considering the last roll and ensuring the conditions are met.
4Step 4: Sum up all valid sequences of length n and return the result modulo 10^9 + 7.

solution.py23 lines

1# Full working Python code
2MOD = 10**9 + 7
3
4def count_sequences(n):
5    dp = [[0] * 7 for _ in range(n + 1)]
6    for j in range(1, 7):
7        dp[1][j] = 1
8
9    for i in range(2, n + 1):
10        for j in range(1, 7):
11            for k in range(1, 7):
12                if gcd(j, k) == 1:
13                    dp[i][j] = (dp[i][j] + dp[i - 1][k]) % MOD
14            if i > 2:
15                for k in range(1, 7):
16                    if j != k:
17                        dp[i][j] = (dp[i][j] + dp[i - 2][k]) % MOD
18
19    result = sum(dp[n]) % MOD
20    return result
21
22# Example usage
23print(count_sequences(4))  # Output: 184

ℹ

Complexity note: The time complexity is O(n) because we iterate through the sequence lengths and the possible last rolls, while the space complexity is O(n) due to the DP table storing results for each roll length.

1Understanding GCD properties helps in filtering valid sequences quickly.
2Dynamic programming can significantly reduce the complexity by avoiding redundant calculations.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.