What is the time complexity of Number of Music Playlists?

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

What data structures are used to solve Number of Music Playlists?

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

What are the key insights for solving Number of Music Playlists?

Insight 1: Dynamic Programming is suitable because the number of playlists of length `i` depends on the number of playlists of length `i-1` with different numbers of unique songs. Insight 2: The state of our DP table is defined by the length of the playlist and the number of unique songs used so far. `dp[i][j]` represents the number of playlists of length `i` using `j` unique songs.

Which companies ask LeetCode 920. Number of Music Playlists in interviews?

This problem is frequently asked at Coursera.

JavaScript is required for the full interactive experience. However, all content is accessible without JavaScript.

Number of Music Playlists - LeetCode 920 Solution

Q: What is the space complexity of Number of Music Playlists?

The space complexity is O(goal*n).

Number of Music Playlists - Complete Solution Guide

Number of Music Playlists is LeetCode problem 920, 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

Your music player contains n different songs. You want to listen to goal songs (not necessarily different) during your trip. To avoid boredom, you will create a playlist so that: Every song is played at least once . A song can only be played again only if k other songs have been played. Given n , goal , and k , return the number of possible playlists that you can create . Since the answer can be very large, return it modulo 10 9 + 7 . Example 1: Input: n = 3, goal = 3, k = 1 Output: 6 Explanatio

Detailed Explanation

The problem asks us to find the number of possible music playlists of a given length (`goal`) from a set of `n` distinct songs, with the constraints that each song must be played at least once, and a song can only be repeated if `k` other songs have been played since its last appearance. We need to return the result modulo 10^9 + 7.

Solution Approach

The provided solution uses a 2D dynamic programming approach. We build a table `dp` where `dp[i][j]` represents the number of playlists of length `i` with `j` distinct songs. The base case is `dp[0][0] = 1` (an empty playlist with zero unique songs). We iterate through the `dp` table, calculating each entry based on two possibilities: the last song added was a new song, or it was a repetition of a previously played song. When adding a new song, we have `n - (j - 1)` choices (total songs minus already used songs). When repeating a song, we have `j - k` choices (unique songs minus those on 'cooldown'). We take the modulo at each step to prevent integer overflow.

Step-by-Step Algorithm

Step 1: Initialize a 2D array `dp` of size `(goal + 1) x (n + 1)` with all values set to 0. `dp[i][j]` will store the number of playlists of length `i` with `j` unique songs.
Step 2: Set the base case `dp[0][0] = 1`. An empty playlist (length 0) with 0 unique songs has only 1 possibility.
Step 3: Iterate through the `dp` array, starting from `i = 1` to `goal` and `j = 1` to `n`.
Step 4: For each cell `dp[i][j]`, calculate the value based on two cases:
- Case 1: The last song is a new song. We can add a new song if we have a playlist of length `i-1` with `j-1` unique songs. The number of choices for the new song is `n - (j - 1)`. So, `term1 = dp[i-1][j-1] * (n - j + 1)`.
- Case 2: The last song is a repetition of an existing song. We can repeat a song if we have a playlist of length `i-1` with `j` unique songs, and the number of available songs to repeat is `j - k` (we can't repeat a song that was played in the last k songs). So, `term2 = dp[i-1][j] * (j - k)` if `j > k`, otherwise `term2 = 0`.
Step 5: Calculate `dp[i][j] = (term1 + term2) % MOD`, where `MOD = 10^9 + 7`.
Step 6: After filling the `dp` array, return `dp[goal][n]`, which represents the number of playlists of length `goal` with `n` unique songs.

Key Insights

Insight 1: Dynamic Programming is suitable because the number of playlists of length `i` depends on the number of playlists of length `i-1` with different numbers of unique songs.
Insight 2: The state of our DP table is defined by the length of the playlist and the number of unique songs used so far. `dp[i][j]` represents the number of playlists of length `i` using `j` unique songs.
Insight 3: We need to consider two cases when building our playlist: adding a new song or repeating a previously played song. The constraint on repeating a song (k other songs played) determines how many songs are available for repetition.

Complexity Analysis

Time Complexity: O(goal*n)

Space Complexity: O(goal*n)

Topics

This problem involves: Math, Dynamic Programming, Combinatorics.

Companies

Asked at: Coursera.

Number of Music Playlists | LeetCode 920

Hard

Math Dynamic Programming Combinatorics

Time: O(goal*n)Space: O(goal*n)

Solution Code

Available in:

class Solution {
    public int numMusicPlaylists(int n, int goal, int k) {
        int MOD = (int)1e9 + 7;
        
        long[][] dp = new long[goal + 1][n + 1];
        dp[0][0] = 1;
        
        for (int i = 1; i <= goal; i++) {
            for (int j = 1; j <= n; j++) {
                // Case 1: The last song is a new song.
                // We have a playlist of length i-1 with j-1 unique songs.
                // We choose one of the n - (j-1) new songs.
                long term1 = dp[i-1][j-1] * (n - j + 1);
                
                // Case 2: The last song is a repetition of an existing song.
                // We have a playlist of length i-1 with j unique songs.
                // A song can be repeated if k other songs have been played since its last appearance.
                // This means we can choose from j songs, but k of them are on "cooldown".
                // The number of available songs to repeat is j - k. This is valid only if j > k.
                long term2 = 0;
                if (j > k) {
                    term2 = dp[i-1][j] * (j - k);
                }
                
                dp[i][j] = (term1 + term2) % MOD;
            }
        }
        
        return (int)dp[goal][n];
    }
}

View Full Solution on GitHubGitHub Solution View on LeetCodeLeetCode

Problem Info

Description

Problem 920: Number of Music Playlists - Detailed explanation and solution approaches.

Topics

Math Dynamic Programming Combinatorics

Companies

Coursera

Stats

Difficulty:Hard

Languages:4

Companies:1

Problem #:920

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 a 2D array `dp` of size `(goal + 1) x (n + 1)` with all values set to 0. `dp[i][j]` will store the number of playlists of length `i` with `j` unique songs.
2
Step 2: Set the base case `dp[0][0] = 1`. An empty playlist (length 0) with 0 unique songs has only 1 possibility.
3
Step 3: Iterate through the `dp` array, starting from `i = 1` to `goal` and `j = 1` to `n`.
4
Step 4: For each cell `dp[i][j]`, calculate the value based on two cases:
5
- Case 1: The last song is a new song. We can add a new song if we have a playlist of length `i-1` with `j-1` unique songs. The number of choices for the new song is `n - (j - 1)`. So, `term1 = dp[i-1][j-1] * (n - j + 1)`.
6
- Case 2: The last song is a repetition of an existing song. We can repeat a song if we have a playlist of length `i-1` with `j` unique songs, and the number of available songs to repeat is `j - k` (we can't repeat a song that was played in the last k songs). So, `term2 = dp[i-1][j] * (j - k)` if `j > k`, otherwise `term2 = 0`.
7
Step 5: Calculate `dp[i][j] = (term1 + term2) % MOD`, where `MOD = 10^9 + 7`.
8
Step 6: After filling the `dp` array, return `dp[goal][n]`, which represents the number of playlists of length `goal` with `n` unique songs.

💡

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(goal*n)

💾

Space Complexity

O(goal*n)

📝

Why?

The algorithm iterates through a 2D DP table of size `(goal + 1) x (n + 1)`. Each cell is calculated in O(1) time. Therefore, the time complexity is O(goal * n). The algorithm uses a 2D array of size `(goal + 1) x (n + 1)` to store the DP table, so the space complexity is O(goal * 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

We can solve this problem using recursion with memoization. The recursive function would take the current playlist length and the number of distinct songs used as input, and return the number of ways to complete the playlist. Memoization would store the results of already computed subproblems to avoid redundant calculations.

Time: O(goal * n)Space: O(goal * n)

Trade-offs: Recursion with memoization offers a more intuitive implementation for some, closely mirroring the recursive nature of the problem. However, it might be slightly slower due to function call overhead, and it can potentially lead to stack overflow issues for very large input values, although in this constrained problem it would not be a problem.

✨

Key Insights

1Insight 1: Dynamic Programming is suitable because the number of playlists of length `i` depends on the number of playlists of length `i-1` with different numbers of unique songs.
2Insight 2: The state of our DP table is defined by the length of the playlist and the number of unique songs used so far. `dp[i][j]` represents the number of playlists of length `i` using `j` unique songs.
3Insight 3: We need to consider two cases when building our playlist: adding a new song or repeating a previously played song. The constraint on repeating a song (k other songs played) determines how many songs are available for repetition.

📚

Key Terms Explained

For Students

Dynamic Programming

Breaking complex problems into smaller subproblems and storing results.

Edge Cases

Edge case 1: `k >= n`. In this case, no song can be repeated, so `term2` will always be 0. This scenario is handled correctly because the condition `j > k` ensures that `term2` remains 0 when `k >= n` and `j <= n`.
Edge case 2: `goal < n`. It is impossible to create a playlist of length goal if goal is less than n, since we need each of n songs to be played at least once. While the code itself does not explicitly handle cases where goal < n, the dynamic programming approach will correctly calculate a result of 0 for `dp[goal][n]` if it's impossible to have a valid playlist.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not taking the modulo at each step can lead to integer overflow, resulting in incorrect answers. It is crucial to apply the modulo operator after each addition and multiplication.

✗Common mistake 2: Incorrectly calculating the number of available songs for repetition. The condition `j > k` is crucial to ensure that we don't try to repeat a song if `k` other songs haven't been played since its last appearance. Forgetting the +1 or -1 in calculation `n-j+1` when choosing a new song.

✅

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: Optimizing content streaming. Similar logic can be applied to design a streaming service's playlist algorithm to ensure a diversity of content (like different genres of music or types of movies) and avoid immediate repetition based on user preferences and constraints.

🏦Real-world scenario 2: Advertising campaign scheduling. Consider an advertising campaign that wants to show 'n' different ads over a 'goal' number of slots, with the constraint that an ad can't be shown more than once in 'k' slots. This problem translates directly to the playlist problem.

💡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

Always test your solution with edge cases: empty input, single element, and boundary values.

👨‍💻 From Experience

I encountered a variation of this problem in a system design interview. The key was recognizing the underlying pattern.

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: June 21, 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

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

@DevExpert92Jul 13

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJul 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 Music Playlists"?

For this hard-level Math 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.

🤖 Need help with this problem?