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We define a dynamic programming approach where we use a 2D DP table where dp[i][j] means the number of ways to form a playlist of length j using i distinct songs. We will eventually find dp[n][goal].
The key recurrence relations are:
(n - i + 1) songs available, hence dp[i][j] += dp[i - 1][j - 1] * (n - i + 1)i songs, considering k songs must be between repeats, we have: dp[i][j] += dp[i][j - 1] * (i - k), where i > kThis draws from the constraints on repeats and expands the playlist iteratively.
Time Complexity: O(n * goal), since we iterate through each subproblem.
Space Complexity: O(n * goal), due to the size of the DP table.
1
2class Solution {
3 public int numMusicPlaylists(int n, int goal, int k) {
4 int MOD = 1000000007;
5 long[][] dp = new long[n + 1][goal + 1];
6 dp[0][0] = 1;
7 for (int i = 1; i <= n; ++i) {
8 for (int j = 1; j <= goal; ++j) {
9 dp[i][j] = dp[i - 1][j - 1] * (n - i + 1) % MOD;
10 if (i > k) {
11 dp[i][j] += dp[i][j - 1] * (i - k) % MOD;
12 }
13 dp[i][j] %= MOD;
14 }
15 }
16 return (int)dp[n][goal];
17 }
18}
19The Java solution mirrors the logic used in Python, employing a two-dimensional array to calculate the number of playlists that can be made.
This approach uses memoization to recursively compute the number of playlists. It calculates the playlists by considering whether a new song is added or an existing song is reused, accounting for k intervening songs.
Time Complexity: O(n * goal), due to memoization.
Space Complexity: O(n * goal), for memo storage.
1
2const numMusicPlaylists = (n, goal
The JavaScript solution implements a top-down dynamic programming approach using memoization to efficiently explore all possible playlists and caches intermediate results to optimize repeated subproblem solutions.