What is the time complexity of Number of Distinct Roll Sequences?

The optimal solution has a time complexity of O(n).

What data structures are used to solve Number of Distinct Roll Sequences?

This problem uses Dynamic Programming, Memoization. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Number of Distinct Roll Sequences?

Insight 1: Dynamic programming is a natural fit because the number of valid sequences for 'n' rolls depends on the valid sequences of 'n-1' and 'n-2' rolls. We need to keep track of the last two rolls to enforce both GCD and separation constraints. Insight 2: We can represent the state of the DP with two variables: the value of the (i-1)th roll and the value of the ith roll.

Which companies ask LeetCode 2318. Number of Distinct Roll Sequences in interviews?

This problem is frequently asked at ServiceNow.

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Number of Distinct Roll Sequences - LeetCode 2318 Solution

Number of Distinct Roll Sequences - Complete Solution Guide

Number of Distinct Roll Sequences is LeetCode problem 2318, a Hard level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

You are given an integer n . You roll a fair 6-sided dice n times. Determine the total number of distinct sequences of rolls possible such that the following conditions are satisfied: The greatest common divisor of any adjacent values in the sequence is equal to 1 . There is at least a gap of 2 rolls between equal valued rolls. More formally, if the value of the i th roll is equal to the value of the j th roll, then abs(i - j) > 2 . Return the total number of distinct sequences possible . Since

Detailed Explanation

The problem asks us to find the number of distinct sequences of rolling a 6-sided die 'n' times, subject to two constraints: 1) The greatest common divisor (GCD) of any two adjacent rolls must be 1, meaning they are coprime. 2) Equal values must be separated by at least two rolls. The input is an integer 'n', and the output is the number of valid sequences modulo 10^9 + 7.

Solution Approach

The solution uses dynamic programming with a 2D array `dp[i][j]` to store the number of valid sequences of length 'k' ending with the sequence `(i+1, j+1)`, where 'i' and 'j' are die face indices (0 to 5). The solution iterates from k=3 to n. For each length k, it calculates a new `dp` array based on the previous `dp` array, ensuring the two constraints are met: GCD of adjacent rolls is 1, and there's a separation of at least 2 rolls between equal values.

Step-by-Step Algorithm

Step 1: Initialize `gcd_ok` array: Create a boolean 2D array to store whether the GCD of two die faces (i+1, j+1) is 1. Pre-compute this for all possible pairs of die faces.
Step 2: Initialize base case `dp` array: Create a 2D array `dp[i][j]` and set `dp[i][j] = 1` if the GCD of `i+1` and `j+1` is 1 and `i != j`. This represents the number of valid sequences of length 2.
Step 3: Iterate from 3 to n: For each 'k' from 3 to 'n', create a `new_dp` array to store the number of valid sequences of length 'k'.
Step 4: Calculate sums for efficient computation: For each possible last roll `p2`, compute the sum of all valid sequences ending with `p2` in the previous iteration. This sum represents the total number of valid sequences of length k-1 that can have 'p2' as the (k-1)-th roll.
Step 5: Update `new_dp` based on previous `dp`: For each possible second-to-last roll `p2` and last roll `p3`, if `p2 != p3` and the GCD of `p2+1` and `p3+1` is 1, then the number of valid sequences of length 'k' ending with `(p2+1, p3+1)` is the sum of all valid sequences of length `k-1` ending with `p2` minus the number of sequences that end with `(p3+1, p2+1)`, preventing a sequence like `..., x, y, x` where x and y are adjacent and identical after reduction.
Step 6: Update `dp` array: Copy the values from `new_dp` to `dp` for the next iteration.
Step 7: Calculate total count: After iterating through all lengths up to 'n', sum all the values in the final `dp` array to get the total number of distinct sequences.
Step 8: Return result modulo 10^9 + 7: Return the total count modulo 10^9 + 7.

Key Insights

Insight 1: Dynamic programming is a natural fit because the number of valid sequences for 'n' rolls depends on the valid sequences of 'n-1' and 'n-2' rolls. We need to keep track of the last two rolls to enforce both GCD and separation constraints.
Insight 2: We can represent the state of the DP with two variables: the value of the (i-1)th roll and the value of the ith roll.
Insight 3: Pre-computing the GCD relationships between die faces avoids repeated calculations and simplifies the logic.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(1)

Topics

This problem involves: Dynamic Programming, Memoization.

Companies

Asked at: ServiceNow.

Number of Distinct Roll Sequences | LeetCode 2318

Hard

Dynamic Programming Memoization

Time: O(n)Space: O(1)

Solution Code

Available in:

class Solution {
    public int distinctSequences(int n) {
        int MOD = (int)1e9 + 7;

        if (n == 1) {
            return 6;
        }

        boolean[][] gcd_ok = new boolean[6][6];
        for (int i = 0; i < 6; i++) {
            for (int j = 0; j < 6; j++) {
                if (gcd(i + 1, j + 1) == 1) {
                    gcd_ok[i][j] = true;
                }
            }
        }

        int[][] dp = new int[6][6];

        for (int p1 = 0; p1 < 6; p1++) {
            for (int p2 = 0; p2 < 6; p2++) {
                if (p1 != p2 && gcd_ok[p1][p2]) {
                    dp[p1][p2] = 1;
                }
            }
        }
        
        for (int _ = 3; _ <= n; _++) {
            int[][] new_dp = new int[6][6];
            
            int[] sums = new int[6];
            for (int p2 = 0; p2 < 6; p2++) {
                int current_sum = 0;
                for (int p1 = 0; p1 < 6; p1++) {
                    current_sum = (int)(((long)current_sum + dp[p1][p2]) % MOD);
                }
                sums[p2] = current_sum;
            }
            
            for (int p2 = 0; p2 < 6; p2++) {
                for (int p3 = 0; p3 < 6; p3++) {
                    if (p2 != p3 && gcd_ok[p2][p3]) {
                        long count = ((long)sums[p2] - dp[p3][p2] + MOD) % MOD;
                        new_dp[p2][p3] = (int)count;
                    }
                }
            }
            dp = new_dp;
            
        }

        int total_count = 0;
        for (int p1 = 0; p1 < 6; p1++) {
            for (int p2 = 0; p2 < 6; p2++) {
                total_count = (int)(((long)total_count + dp[p1][p2]) % MOD);
            }
        }
        
        return total_count;
    }

    private int gcd(int a, int b) {
        if (b == 0) {
            return a;
        }
        return gcd(b, a % b);
    }
}

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Problem Info

Description

Problem 2318: Number of Distinct Roll Sequences - Detailed explanation and solution approaches.

Topics

Dynamic Programming Memoization

Companies

ServiceNow

Stats

Difficulty:Hard

Languages:4

Companies:1

Problem #:2318

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

8 steps

1
Step 1: Initialize `gcd_ok` array: Create a boolean 2D array to store whether the GCD of two die faces (i+1, j+1) is 1. Pre-compute this for all possible pairs of die faces.
2
Step 2: Initialize base case `dp` array: Create a 2D array `dp[i][j]` and set `dp[i][j] = 1` if the GCD of `i+1` and `j+1` is 1 and `i != j`. This represents the number of valid sequences of length 2.
3
Step 3: Iterate from 3 to n: For each 'k' from 3 to 'n', create a `new_dp` array to store the number of valid sequences of length 'k'.
4
Step 4: Calculate sums for efficient computation: For each possible last roll `p2`, compute the sum of all valid sequences ending with `p2` in the previous iteration. This sum represents the total number of valid sequences of length k-1 that can have 'p2' as the (k-1)-th roll.
5
Step 5: Update `new_dp` based on previous `dp`: For each possible second-to-last roll `p2` and last roll `p3`, if `p2 != p3` and the GCD of `p2+1` and `p3+1` is 1, then the number of valid sequences of length 'k' ending with `(p2+1, p3+1)` is the sum of all valid sequences of length `k-1` ending with `p2` minus the number of sequences that end with `(p3+1, p2+1)`, preventing a sequence like `..., x, y, x` where x and y are adjacent and identical after reduction.
6
Step 6: Update `dp` array: Copy the values from `new_dp` to `dp` for the next iteration.
7
Step 7: Calculate total count: After iterating through all lengths up to 'n', sum all the values in the final `dp` array to get the total number of distinct sequences.
8
Step 8: Return result modulo 10^9 + 7: Return the total count modulo 10^9 + 7.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n)

👍 Linear - Good! Visits each element once.

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The time complexity is dominated by the outer loop that runs 'n' times. The inner computations are constant time operations as they only depend on the number of die faces (6). The space complexity is constant because the arrays used to store intermediate results have fixed dimensions, irrespective of the input value 'n'.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Recursion with Memoization

A recursive approach could be used along with memoization to avoid recomputing the same subproblems. The recursive function would take the current length of the sequence and the values of the last two rolls as input. The base cases are n = 1 and n = 2. The recursive step would explore all possible rolls and recursively call itself to build longer valid sequences.

Time: O(n)Space: O(n)

Trade-offs: While conceptually similar, the recursive approach with memoization might incur a larger space complexity due to the call stack. In this case, the iterative DP approach is slightly more efficient and easier to implement.

✨

Key Insights

1Insight 1: Dynamic programming is a natural fit because the number of valid sequences for 'n' rolls depends on the valid sequences of 'n-1' and 'n-2' rolls. We need to keep track of the last two rolls to enforce both GCD and separation constraints.
2Insight 2: We can represent the state of the DP with two variables: the value of the (i-1)th roll and the value of the ith roll.
3Insight 3: Pre-computing the GCD relationships between die faces avoids repeated calculations and simplifies the logic.

📚

Key Terms Explained

For Students

Dynamic Programming

Breaking complex problems into smaller subproblems and storing results.

Edge Cases

Edge case 1: n = 1: If n is 1, there are 6 possible sequences (1, 2, 3, 4, 5, 6). The code handles this case with an explicit check and returns 6.
Edge case 2: The modulus operation: The intermediate sums can become very large, potentially exceeding the maximum integer value. The solution uses the modulo operator (%) to prevent integer overflow and ensures that the final result is within the specified range.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting the modulus operator: Failing to apply the modulo operator at each step can lead to integer overflow and incorrect results.

✗Common mistake 2: Incorrect GCD calculation: Making errors in the GCD calculation or using the wrong indices can lead to incorrect validation of valid sequences.

✗Common mistake 3: Not enforcing the separation of equal values: Failing to ensure that identical roll values are separated by at least two rolls will lead to overcounting and incorrect results.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Generating sequences with constraints: This algorithm can be adapted to generate sequences in other domains where certain constraints need to be satisfied between adjacent elements, such as generating valid code sequences, musical melodies, or protein structures.

🏦Real-world scenario 2: Resource allocation: Consider allocating different types of resources (represented by the die faces) over time, where some resource combinations are incompatible (GCD != 1) or a resource cannot be used too soon after its previous use. The algorithm can model the number of valid resource allocation schedules.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

Start with recursion, then memoize. The hardest part is identifying the state. Ask yourself: "What information do I need to make the current decision?"

🎯 Interview Tip

These problems test your ability to optimize. Start with a brute force solution, then improve.

⚠️ Watch Out

Forgetting base cases or initializing the DP array incorrectly. Start by solving the smallest subproblems.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: November 6, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 13

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 6

❓

Frequently Asked Questions

How do I identify if a problem needs dynamic programming?

DP problems typically have: 1) Overlapping subproblems (same calculation repeated), 2) Optimal substructure (optimal solution uses optimal solutions to subproblems). Look for keywords like "minimum/maximum," "count ways," "is it possible," or "longest/shortest."

How do I approach hard algorithm problems?

Hard problems often require: 1) Breaking down into subproblems, 2) Combining multiple advanced techniques, 3) Careful optimization of brute force. Start with a working solution (even O(n²) or worse), then optimize. Draw examples, identify patterns, and think about what makes this problem unique.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Number of Distinct Roll Sequences"?

For this hard-level Dynamic Programming problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether memoization techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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