#920

Hard

auto_awesomeState transition dynamic programming

LeetCode Problem Workspace

Number of Music Playlists

Solve the Number of Music Playlists problem with dynamic programming, focusing on state transitions and combinatorics to calculate possible playlists.

math dynamic programming combinatorics

Problem Statement

You have n different songs and you want to create a playlist with exactly goal songs. The playlist must follow the rule that no song is repeated within k consecutive songs. The challenge is to determine the number of possible playlists that can be formed under these conditions, and the result should be returned modulo 10^9 + 7.

Given n, goal, and k, you need to count how many different playlists are possible where songs may repeat but must respect the constraint on consecutive repetitions. The task is solved efficiently using dynamic programming techniques with state transitions.

Examples

Example 1

Input: n = 3, goal = 3, k = 1

Output: 6

There are 6 possible playlists: [1, 2, 3], [1, 3, 2], [2, 1, 3], [2, 3, 1], [3, 1, 2], and [3, 2, 1].

Example 2

Input: n = 2, goal = 3, k = 0

Output: 6

There are 6 possible playlists: [1, 1, 2], [1, 2, 1], [2, 1, 1], [2, 2, 1], [2, 1, 2], and [1, 2, 2].

Example 3

Input: n = 2, goal = 3, k = 1

Output: 2

There are 2 possible playlists: [1, 2, 1] and [2, 1, 2].

Constraints

0 <= k < n <= goal <= 100

Solution Approach

State Transition Dynamic Programming

The core approach to solving this problem is using dynamic programming with state transitions. Define dp[i][j] as the number of ways to create a playlist of length j using i distinct songs. The recursive formula updates the state by either adding a new song or repeating a song based on the given constraint k.

Modular Arithmetic

Since the result can be very large, all calculations are performed modulo 10^9 + 7 to ensure that the result remains within the integer range. This helps to manage large numbers and ensures that the solution remains feasible for the input constraints.

Optimization with Combinatorics

In addition to dynamic programming, combinatorics is used to count how new songs and repeated songs can be chosen. For each transition, new songs are added to the playlist, while ensuring that repeated songs do not violate the k-consecutive restriction.

Complexity Analysis

Metric	Value
Time	O(n \log \text{goal})
Space	O(n)

The time complexity of this solution is O(n log goal), which arises due to the dynamic programming table updates and the required state transitions for each song. The space complexity is O(n) because we only need to store the state for the number of songs being used, as the playlist size grows.

What Interviewers Usually Probe

The candidate demonstrates an understanding of dynamic programming and modular arithmetic.
The candidate can clearly explain the logic behind state transitions and playlist constraints.
The candidate is able to optimize a combinatorial counting problem using dynamic programming techniques.

Common Pitfalls or Variants

Common pitfalls

Misunderstanding the modular arithmetic and failing to apply modulo 10^9 + 7 correctly.
Incorrectly handling the case where no songs can be repeated within k songs, especially when constructing the DP table.
Overlooking the fact that playlist lengths must exactly equal goal, which can lead to edge case errors when calculating the result.

Follow-up variants

Handling larger values of n and goal, which increases the complexity of the state transitions and the time required to compute the result.
Adapting the problem to include additional constraints such as a maximum number of distinct songs in the playlist.
Solving the problem in a distributed manner where the DP states can be computed concurrently.

FAQ

What is the primary algorithmic pattern used to solve the Number of Music Playlists problem?

The primary pattern used is state transition dynamic programming, where the problem is broken into smaller subproblems and solved recursively.

How do you ensure the result is within limits for large numbers in this problem?

We use modular arithmetic with modulo 10^9 + 7 to prevent overflow and keep the numbers manageable during the computation process.

Why is combinatorics important in solving the Number of Music Playlists problem?

Combinatorics helps count how songs are added or repeated in the playlist, respecting the constraint on consecutive repetitions of songs.

How can dynamic programming be optimized in this problem?

The dynamic programming table can be optimized by reducing space complexity to O(n) by only storing the states for the current number of distinct songs used.

What edge cases should be considered for the Number of Music Playlists problem?

Consider edge cases such as when k = 0, where no song can repeat, or when n = goal, where the playlist must use every song exactly once.

terminal

Solution

Solution 1: Dynamic Programming

We define $f[i][j]$ to be the number of playlists that can be made from $i$ songs with exactly $j$ different songs. We have $f[0][0] = 1$ and the answer is $f[goal][n]$.

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class Solution:
    def numMusicPlaylists(self, n: int, goal: int, k: int) -> int:
        mod = 10**9 + 7
        f = [[0] * (n + 1) for _ in range(goal + 1)]
        f[0][0] = 1
        for i in range(1, goal + 1):
            for j in range(1, n + 1):
                f[i][j] = f[i - 1][j - 1] * (n - j + 1)
                if j > k:
                    f[i][j] += f[i - 1][j] * (j - k)
                f[i][j] %= mod
        return f[goal][n]

Solution 2: Dynamic Programming (Space Optimization)

We notice that $f[i][j]$ is only related to $f[i - 1][j - 1]$ and $f[i - 1][j]$. Therefore, we can use a rolling array to optimize the space complexity, reducing the space complexity to $O(n)$.

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class Solution:
    def numMusicPlaylists(self, n: int, goal: int, k: int) -> int:
        mod = 10**9 + 7
        f = [[0] * (n + 1) for _ in range(goal + 1)]
        f[0][0] = 1
        for i in range(1, goal + 1):
            for j in range(1, n + 1):
                f[i][j] = f[i - 1][j - 1] * (n - j + 1)
                if j > k:
                    f[i][j] += f[i - 1][j] * (j - k)
                f[i][j] %= mod
        return f[goal][n]

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