How do you solve Number of Music Playlists on LeetCode?

Number of Music Playlists (LeetCode #920) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n * goal) time complexity and O(n * goal) space complexity. Key insight: Dynamic programming is key for counting combinations efficiently.

What is the brute force solution for Number of Music Playlists?

The brute force approach for Number of Music Playlists is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach generates all possible playlists and checks if they meet the criteria. This is simple but inefficient, as it doesn't leverage any patterns or properties of the problem. The optimal Optimal Solution approach improves this to O(n * goal).

What algorithm or data structure is used to solve Number of Music Playlists?

Number of Music Playlists uses the following concepts: Math, Dynamic Programming, Combinatorics. The recommended patterns to study are: Dynamic Programming, Combinatorics.

What are the interview tips for Number of Music Playlists?

Explain your thought process clearly while coding. Discuss edge cases and constraints before diving into coding. Practice writing clean and efficient code.

What are common mistakes when solving Number of Music Playlists?

Not initializing the DP table correctly. Forgetting to apply the modulo operation.

#920

Number of Music Playlists

Hard

MathDynamic ProgrammingCombinatoricsDynamic ProgrammingCombinatorics

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

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

Using dynamic programming, we can efficiently count valid playlists by building on smaller subproblems. This approach avoids generating all permutations and instead focuses on valid combinations.

⚙️

Algorithm

5 steps

1Step 1: Create a DP table where dp[i][j] represents the number of ways to create a playlist of length 'i' using 'j' different songs.
2Step 2: Initialize dp[0][0] = 1 (1 way to create an empty playlist).
3Step 3: Iterate through lengths from 1 to 'goal' and number of songs from 1 to 'n'.
4Step 4: For each combination, calculate the number of ways to add a new song or reuse an existing song based on the 'k' constraint.
5Step 5: Return dp[goal][n] as the result.

solution.py12 lines

1# Full working Python code
2MOD = 10**9 + 7
3
4def count_playlists(n, goal, k):
5    dp = [[0] * (n + 1) for _ in range(goal + 1)]
6    dp[0][0] = 1
7    for i in range(1, goal + 1):
8        for j in range(1, n + 1):
9            dp[i][j] = (dp[i - 1][j - 1] * (n - (j - 1))) % MOD  # Add a new song
10            if j > k:
11                dp[i][j] = (dp[i][j] + dp[i - 1][j] * (j - k)) % MOD  # Reuse a song
12    return dp[goal][n]

ℹ

Complexity note: The time complexity is O(n * goal) because we fill a 2D DP table of size (goal+1) x (n+1). Each cell is computed based on previously computed values, making it efficient.

1Dynamic programming is key for counting combinations efficiently.
2Understanding constraints helps in formulating the DP transitions.

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