920. Number of Music Playlists

Problem

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
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

class Solution {
    public int numMusicPlaylists(int n, int l, int k) {
        long[][] dp = new long[l][n + 1];
        for (int i = 0; i < l; i++) {
            Arrays.fill(dp[i], -1);
        }
        return (int) helper(0, l, 0, n, k, dp);
    }

    private long helper(int songNumber, int l, int usedSong, int n, int k, long[][] dp) {
        if (songNumber == l) {
            return usedSong == n ? 1 : 0;
        }
        if (dp[songNumber][usedSong] != -1) {
            return dp[songNumber][usedSong];
        }
        long ans;
        if (songNumber < k) {
            ans = (n - usedSong) * helper(songNumber + 1, l, usedSong + 1, n, k, dp);
        } else if (usedSong == n) {
            ans = (usedSong - k) * helper(songNumber + 1, l, usedSong, n, k, dp);
        } else {
            ans =
                    (n - usedSong) * helper(songNumber + 1, l, usedSong + 1, n, k, dp)
                            + (usedSong - k) * helper(songNumber + 1, l, usedSong, n, k, dp);
        }
        int mod = (int) 1e9 + 7;
        dp[songNumber][usedSong] = ans % mod;
        return ans % mod;
    }
}

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).