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Number of Music Playlists

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 109 + 7.

 

Example 1:

Input: n = 3, goal = 3, k = 1
Output: 6
Explanation: 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
Explanation: 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
Explanation: There are 2 possible playlists: [1, 2, 1] and [2, 1, 2].

 

Constraints:

  • 0 <= k < n <= goal <= 100

Solution Explanation for Number of Music Playlists

This problem asks to find the number of playlists that can be created given n songs, goal number of songs to play, and k as the minimum number of other songs that must be played before a song can be repeated.

The most efficient approach is using dynamic programming. We'll explore two versions: one with a larger space complexity and a space-optimized version.

Solution 1: Dynamic Programming (Detailed Explanation)

This solution uses a 2D DP array f[i][j], where f[i][j] represents the number of playlists of length i using exactly j different songs.

State Transition:

To understand the state transition, consider building a playlist of length i using j different songs. There are two possibilities for the i-th song:

  1. New Song: The i-th song is a song that hasn't been played yet. In this case, we have n - (j - 1) choices for the i-th song (since j-1 songs have already been played), and the number of ways to create the playlist of length i-1 using j-1 songs is f[i-1][j-1]. Thus, the number of playlists in this case is f[i-1][j-1] * (n - (j - 1)).

  2. Repeated Song: The i-th song is a song that has already been played. To avoid boredom, at least k other songs must have been played before repeating a song. This means we must have j > k. We have j choices for the i-th song (any of the already played songs). The number of ways to create the playlist of length i-1 using j songs is f[i-1][j]. Therefore, the number of playlists in this case is f[i-1][j] * (j - k).

Combining both cases, we get the recurrence relation:

f[i][j] = f[i-1][j-1] * (n - (j - 1)) + (j > k ? f[i-1][j] * (j - k) : 0)

Base Case:

f[0][0] = 1 (an empty playlist is one valid playlist).

Final Answer:

The final answer is f[goal][n] (playlists of length goal using all n songs).

Time and Space Complexity:

  • Time Complexity: O(goal * n) - We iterate through the DP array.
  • Space Complexity: O(goal * n) - Size of the DP array.

Solution 2: Dynamic Programming (Space Optimization)

Notice that in the recurrence relation, f[i][j] only depends on f[i-1][j-1] and f[i-1][j]. Therefore, we can optimize the space complexity to O(n) by using only two 1D arrays (or even just one if you're clever). The logic remains the same; only the DP array structure changes.

Time and Space Complexity:

  • Time Complexity: O(goal * n) - Same as the previous solution.
  • Space Complexity: O(n) - We now only use a 1D array of size n + 1.

The provided code snippets in multiple languages implement both solutions. The space-optimized versions are generally preferred for larger input sizes due to their reduced memory footprint. Remember that modulo operation (% mod) is used throughout the code to prevent integer overflow.