Introduction

In this blog post, we will explore how to solve the LeetCode problem "920" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

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

Explanation

Here's the breakdown of the solution:

• Dynamic Programming: We use dynamic programming to solve this problem. dp[i][j] stores the number of playlists of length i using j distinct songs.
• Base Case and Transitions: The base case is dp[0][0] = 1. The transitions consider two possibilities: either we add a new song or we replay an existing song.

• Modulo Arithmetic: We perform modulo operations to prevent integer overflow.

• Runtime Complexity: O(goal n), Storage Complexity: O(goal n)

Code

    def num_playlists(n: int, goal: int, k: int) -> int:
    """
    Calculates the number of possible playlists that can be created given the constraints.

    Args:
        n: The number of different songs.
        goal: The desired length of the playlist.
        k: The minimum number of other songs that must be played before a song can be replayed.

    Returns:
        The number of possible playlists modulo 10^9 + 7.
    """
    MOD = 10**9 + 7

    # dp[i][j] stores the number of playlists of length i using j distinct songs
    dp = [[0] * (n + 1) for _ in range(goal + 1)]

    # Base case: a playlist of length 0 with 0 distinct songs is 1
    dp[0][0] = 1

    # Iterate through the possible lengths of the playlist
    for i in range(1, goal + 1):
        # Iterate through the possible number of distinct songs used
        for j in range(1, n + 1):
            # If we add a new song
            dp[i][j] = (dp[i][j] + dp[i - 1][j - 1] * (n - j + 1)) % MOD

            # If we replay an existing song
            if j > k:
                dp[i][j] = (dp[i][j] + dp[i - 1][j] * (j - k)) % MOD

    return dp[goal][n]