What is the time complexity of Dice Roll Simulation?

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

What data structures are used to solve Dice Roll Simulation?

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

What are the key insights for solving Dice Roll Simulation?

Insight 1: The core idea is to use dynamic programming to build up the solution incrementally. We need to keep track of the number of ways to reach a particular number of rolls and the last number rolled and the number of times it has been rolled consecutively. Insight 2: We can use a 2D DP array where `dp[i][j]` represents the number of sequences of length `i` that end with the number `j+1`. Then we accumulate the total number of distinct sequences of length i into the `S[i]` array.

Which companies ask LeetCode 1223. Dice Roll Simulation in interviews?

This problem is frequently asked at Akuna Capital.

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Dice Roll Simulation - LeetCode 1223 Solution

Dice Roll Simulation - Complete Solution Guide

Dice Roll Simulation is LeetCode problem 1223, 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

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

Detailed Explanation

The problem asks us to find the number of distinct sequences of dice rolls that can be obtained with exactly `n` rolls, given a constraint. The constraint is defined by the `rollMax` array, where `rollMax[i]` represents the maximum number of consecutive times the number `i+1` can be rolled. We need to return the number of such sequences modulo 10^9 + 7.

Solution Approach

The provided code uses dynamic programming to solve the problem. It maintains two arrays: `dp` and `S`. `dp[i][j]` stores the number of distinct sequences of length `i` ending with the number `j+1`. `S[i]` stores the total number of distinct sequences of length `i`. The code iterates through the possible lengths (from 1 to n) and the possible die values (from 1 to 6). For each `i` and `j`, it calculates `dp[i][j]` based on `S[i-1]` (the total number of ways to reach length `i-1`). It then subtracts any 'bad' sequences (those where rolling the number `j+1` would exceed its `rollMax` limit) from `S[i-1]` to get the number of valid sequences that can be extended by rolling `j+1`. Finally, it sums all the `dp[i][j]` values to get `S[i]`, representing the total number of distinct sequences of length `i`.

Step-by-Step Algorithm

Step 1: Initialize `dp` and `S` arrays. `S[0] = 1` as there is one way to have 0 rolls (an empty sequence).
Step 2: Iterate through the number of rolls `i` from 1 to `n`.
Step 3: For each roll `i`, iterate through each possible face of the die `j` from 0 to 5.
Step 4: Calculate the number of total ways `total_ways` to reach length `i` if there were no constraints, which is just `S[i-1]`.
Step 5: Determine the maximum consecutive rolls allowed for the face `j+1` i.e. `k=rollMax[j]`.
Step 6: Check if subtracting `k+1` rolls from `i` is valid by `i - k - 1 >= 0`. If valid, calculate `bad_sequences`. These are the sequences which have `rollMax[j]` consecutive `j+1`'s ending at roll `i`
Step 7: Subtract these `bad_sequences` from `total_ways` to get valid sequences that can reach `dp[i][j]`.
Step 8: Accumulate the total sequences for the length `i` using `S[i] = (S[i] + dp[i][j]) % MOD`
Step 9: Return the value of `S[n]`.

Key Insights

Insight 1: The core idea is to use dynamic programming to build up the solution incrementally. We need to keep track of the number of ways to reach a particular number of rolls and the last number rolled and the number of times it has been rolled consecutively.
Insight 2: We can use a 2D DP array where `dp[i][j]` represents the number of sequences of length `i` that end with the number `j+1`. Then we accumulate the total number of distinct sequences of length i into the `S[i]` array.
Insight 3: Be careful to handle the cases where rolling a specific number is constrained by `rollMax`. This requires subtracting the number of 'bad' sequences, those where a number is rolled consecutively more than allowed, from the total.
Insight 4: Modulo operation is important to prevent integer overflow.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(n)

Topics

This problem involves: Array, Dynamic Programming.

Companies

Asked at: Akuna Capital.

Dice Roll Simulation | LeetCode 1223

Hard

Array Dynamic Programming

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

Solution Code

Available in:

class Solution {
    public int dieSimulator(int n, int[] rollMax) {
        int MOD = 1000000007;

        int[][] dp = new int[n + 1][6];
        int[] S = new int[n + 1];
        S[0] = 1;

        for (int i = 1; i <= n; i++) {
            int current_total_sum = 0;
            for (int j = 0; j < 6; j++) {
                long total_ways = S[i - 1];

                int k = rollMax[j];

                if (i - k - 1 >= 0) {
                    long bad_sequences = (S[i - k - 1] - dp[i - k - 1][j] + (long)MOD) % MOD;
                    total_ways = (total_ways - bad_sequences + (long)MOD) % MOD;
                }

                dp[i][j] = (int)total_ways;
                current_total_sum = (current_total_sum + dp[i][j]) % MOD;
            }
            S[i] = current_total_sum;
        }

        return S[n];
    }
}

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Problem Info

Description

Problem 1223: Dice Roll Simulation - Detailed explanation and solution approaches.

Topics

Array Dynamic Programming

Companies

Akuna Capital

Stats

Difficulty:Hard

Languages:4

Companies:1

Problem #:1223

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

9 steps

1
Step 1: Initialize `dp` and `S` arrays. `S[0] = 1` as there is one way to have 0 rolls (an empty sequence).
2
Step 2: Iterate through the number of rolls `i` from 1 to `n`.
3
Step 3: For each roll `i`, iterate through each possible face of the die `j` from 0 to 5.
4
Step 4: Calculate the number of total ways `total_ways` to reach length `i` if there were no constraints, which is just `S[i-1]`.
5
Step 5: Determine the maximum consecutive rolls allowed for the face `j+1` i.e. `k=rollMax[j]`.
6
Step 6: Check if subtracting `k+1` rolls from `i` is valid by `i - k - 1 >= 0`. If valid, calculate `bad_sequences`. These are the sequences which have `rollMax[j]` consecutive `j+1`'s ending at roll `i`
7
Step 7: Subtract these `bad_sequences` from `total_ways` to get valid sequences that can reach `dp[i][j]`.
8
Step 8: Accumulate the total sequences for the length `i` using `S[i] = (S[i] + dp[i][j]) % MOD`
9
Step 9: Return the value of `S[n]`.

💡

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(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The nested loops determine the time complexity. The outer loop runs `n` times, and the inner loop runs 6 times. The space complexity is determined by the size of the `dp` array, which is `(n+1) * 6`, and the `S` array which is `(n+1)`. Therefore, the space complexity is O(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

Recursive approach with memoization

A recursive approach can also solve this problem. You define a function that calculates the number of sequences for a given length, last die rolled, and the count of consecutive rolls for that die. Memoization can be used to store the calculated values and avoid redundant computations.

Time: O(n*rollMax_max*6) where rollMax_max is maximum value from rollMax.Space: O(n*rollMax_max*6)

Trade-offs: The recursive approach can be more intuitive to understand, but it may be slower due to function call overhead. However, for larger n, recursion might lead to stack overflow if not handled correctly. It becomes especially useful if some subproblems are never visited.

✨

Key Insights

1Insight 1: The core idea is to use dynamic programming to build up the solution incrementally. We need to keep track of the number of ways to reach a particular number of rolls and the last number rolled and the number of times it has been rolled consecutively.
2Insight 2: We can use a 2D DP array where `dp[i][j]` represents the number of sequences of length `i` that end with the number `j+1`. Then we accumulate the total number of distinct sequences of length i into the `S[i]` array.
3Insight 3: Be careful to handle the cases where rolling a specific number is constrained by `rollMax`. This requires subtracting the number of 'bad' sequences, those where a number is rolled consecutively more than allowed, from the total.
4Insight 4: Modulo operation is important to prevent integer overflow.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Dynamic Programming

Breaking complex problems into smaller subproblems and storing results.

Edge Cases

Edge case 1: n = 1. The code correctly initializes S[0] to 1, which represents the empty sequence before any rolls. The loops then handle the cases for the first roll correctly, accounting for the rollMax constraints.
Edge case 2: rollMax[i] = 1 for all i. This means no two consecutive rolls can be the same. The DP transitions will correctly handle the subtraction of the sequences with consecutive same numbers.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not taking the modulo at each step. The number of sequences can grow very large, so it's important to take the modulo at each step to prevent integer overflow.

✗Common mistake 2: Incorrectly calculating the 'bad' sequences. Make sure to only subtract sequences where the number of consecutive rolls exceeds the `rollMax` limit.

✅

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: Designing a pseudo-random number generator with constraints on consecutive values. This kind of constraint can be useful to reduce bias.

🏦Real-world scenario 2: Simulating events with frequency limitations. For instance, in network simulations, limiting the rate at which certain types of packets can be sent consecutively.

💡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

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

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

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

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: October 22, 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 14

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

@TechLeadJul 7

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jun 30

❓

Frequently Asked Questions

When should I use in-place modification vs creating a new array?

Use in-place modification when space complexity matters (O(1) space requirement) and the original array can be modified. Create a new array when you need to preserve the original data or when the problem involves significant restructuring that would complicate in-place logic.

Should I use top-down (memoization) or bottom-up (tabulation)?

Start with top-down (recursion + memoization) as it directly maps to the recursive thinking and handles only needed subproblems. Convert to bottom-up for better space efficiency and to avoid stack overflow. Bottom-up is often faster due to no recursion overhead.

Is it okay to not fully solve hard problems in interviews?

Yes! Interviewers care more about your problem-solving process than the final answer. Demonstrate: clear communication, systematic approach, ability to recognize when to pivot strategies, and graceful handling of hints. Partial progress with good reasoning is valued.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Dice Roll Simulation"?

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

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