Jobs

Interviews

Discussions

Number of Music Playlists

Hard

Coursera

3 views

Topics:

Dynamic Programming

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

Try coding it in LeetCode

Number of Music Playlists

Hard

Coursera

3 views

Topics:

Dynamic Programming

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

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What are the constraints on `n`, `goal`, and `k`? What are the maximum and minimum possible values for each?
Is it possible for `k` to be greater than or equal to `n`? If so, what is the expected behavior?
Should the result be returned modulo some specific number (e.g., 10^9 + 7)?
Can `n` or `goal` ever be zero?
Are we concerned about integer overflow during calculations, and if so, what's the appropriate way to handle it?

Brute Force Solution

Approach

The goal is to create a playlist with specific length and number of unique songs using a given set of songs. The brute force approach tries every single combination of songs to create a playlist and checks if that playlist is valid.

Here's how the algorithm would work step-by-step:

Start building the playlist with the first song.
Then, for each spot in the playlist, try every song in the entire song set.
Make sure you keep track of how many songs are in the playlist, and how many unique songs have been used so far.
If you reach the desired playlist length and have the right number of unique songs, then you've found one valid playlist.
If you reach the desired playlist length, but don't have the right number of unique songs, it's not a valid playlist, so discard it and try another way.
If the playlist is still too short, keep adding songs and checking for validity.
Repeat this process for all possible combinations of songs.
Once all possibilities are exhausted, count the total number of valid playlists you found.

Code Implementation

def number_of_music_playlists_brute_force(number_of_songs, playlist_length, number_of_unique_songs):
    number_of_valid_playlists = 0

    def generate_playlist(current_playlist, unique_songs_in_playlist):
        nonlocal number_of_valid_playlists

        if len(current_playlist) == playlist_length:
            # Check if the playlist meets the criteria
            if len(unique_songs_in_playlist) == number_of_unique_songs:
                number_of_valid_playlists += 1
            return

        # Try adding each song to the playlist
        for song in range(number_of_songs):
            new_playlist = current_playlist + [song]

            # Create the set of unique songs in this playlist
            new_unique_songs_in_playlist = set(unique_songs_in_playlist)

            # Update to include the song
            new_unique_songs_in_playlist.add(song)

            generate_playlist(new_playlist, new_unique_songs_in_playlist)

    # Start generating playlists from scratch
    generate_playlist([], set())

    return number_of_valid_playlists

Big(O) Analysis

Time Complexity

O(n^L) – The brute force approach explores all possible playlists of length L, where each position in the playlist can be filled with one of the n songs. Therefore, for each of the L positions, there are n choices, leading to n multiplied by itself L times, or n^L possible playlists. The algorithm iterates through each of these potential playlists to check if it meets the criteria of having exactly k unique songs, thus resulting in O(n^L) time complexity. The check for unique songs within each playlist adds a negligible amount of time compared to generating the playlists themselves.

Space Complexity

O(K) – The brute force approach described relies on recursion to explore all possible playlists. The recursion depth corresponds to the playlist length, K, as stated in the problem, since we add a song for each spot in the playlist using recursive calls. Each recursive call adds a new frame to the call stack to store the state of the playlist being built, the number of unique songs, and the current position. Therefore, the maximum depth of the call stack is K, leading to O(K) auxiliary space due to the recursion call stack.

Optimal Solution

Approach

This problem asks us to count the number of ways to create music playlists with specific constraints on the number of songs, unique songs, and a minimum gap between repeats. The optimal solution involves using dynamic programming to efficiently calculate the possibilities, avoiding redundant computations by storing and reusing intermediate results.

Here's how the algorithm would work step-by-step:

Understand that we are building playlists one song at a time, and need to keep track of how many songs are in the playlist and how many unique songs we've used so far.
Recognize that the number of ways to build a playlist of a certain length using a certain number of unique songs depends on the number of ways we could have built the playlist just before adding the last song.
Realize that the last song we add is either a song we've already used, or a brand-new unique song.
If we choose a song we've already used, we need to make sure we have played enough other songs since the last time we played this song to meet the minimum gap requirement.
If we choose a new, unique song, there are a limited number of choices because we only have a fixed number of unique songs to choose from.
Store the number of ways to build each playlist configuration (length, unique songs used) as we calculate it, so we don't have to recalculate it later. This is the key to efficiency.
Start with a base case: There's one way to have an empty playlist with no songs used.
Build up the solutions for larger playlists step-by-step, using the stored values and the rules about repeats and unique songs.
The final answer is the number of ways to create the desired playlist length using all the available unique songs.

Code Implementation

def number_of_music_playlists(playlist_length, 
                               number_of_unique_songs, 
                               minimum_song_gap):

    modulo_value = 10**9 + 7

    # dp[i][j] is the number of playlists of length i using j unique songs
    dynamic_programming_table = [[0] * (number_of_unique_songs + 1) 
                                 for _ in range(playlist_length + 1)]

    # Base case: an empty playlist with no songs used
    dynamic_programming_table[0][0] = 1

    for playlist_index in range(1, playlist_length + 1):
        for unique_song_count in range(1, number_of_unique_songs + 1):

            # Add a new unique song
            # There are number_of_unique_songs - (unique_song_count - 1) choices
            dynamic_programming_table[playlist_index][unique_song_count] = 
                (dynamic_programming_table[playlist_index - 1][unique_song_count - 1] 
                 * (number_of_unique_songs - (unique_song_count - 1))) % modulo_value

            # Re-use an old song, but only if allowed
            if unique_song_count > minimum_song_gap:
                # Ensures we've played enough songs for repeats
                dynamic_programming_table[playlist_index][unique_song_count] += 
                    (dynamic_programming_table[playlist_index - 1][unique_song_count] 
                     * (unique_song_count - minimum_song_gap)) % modulo_value

                dynamic_programming_table[playlist_index][unique_song_count] %= modulo_value

    # The answer is the playlist with playlist_length using all unique songs.
    return dynamic_programming_table[playlist_length][number_of_unique_songs]

Big(O) Analysis

Time Complexity

O(n * goal) – The dynamic programming solution builds a 2D table (dp) where the dimensions are `n` (number of unique songs available) and `goal` (desired playlist length). The algorithm iterates through each cell of this table to compute the number of possible playlists for a given length (up to `goal`) using a given number of unique songs (up to `n`). Each cell computation involves a constant amount of work (checking for new songs vs. repeated songs and applying the modular arithmetic). Thus, the time complexity is determined by the size of the dp table, which is n * goal, so the overall time complexity is O(n * goal).

Space Complexity

O(N*K) – The solution utilizes dynamic programming, storing intermediate results in a table (likely a 2D array or matrix) to avoid redundant computations. The dimensions of this table are determined by the length of the playlist (N) and the number of unique songs (K). Therefore, the auxiliary space used is proportional to the product of N and K, where N is the desired playlist length and K is the number of unique songs. This space is used to store the number of ways to build each playlist configuration (length, unique songs used). Thus, the space complexity is O(N*K).

Edge Cases

Case	How to Handle
n = 0 (no songs), goal = 0, k = 0 (no restrictions)	Return 1, as there is one way to play 0 songs.
n = 1, goal = 1, k = 0 (only one song to play for one playlist slot)	Return 1 since there is one way to play that song.
n = 1, goal > 1, k = 0 (one song to play for multiple playlist slots)	Return 0, as we can only play that song once.
goal < n (playlist length shorter than total songs)	Return 0 since it is impossible to create a playlist with less slots than available songs.
k >= n (can replay all songs immediately)	The standard formula or DP approach might need modification, likely simplifying to a combination problem without reuse restrictions after initial playlist creation.
Large n, goal, or k that cause integer overflow during calculations	Use modulo operator during intermediate calculations to prevent overflow and return the result modulo 10^9 + 7.
n is very close to goal and k is small such that it pushes calculations to the limit of the modulo.	Ensure the modulo operations are applied correctly to intermediate results to prevent errors caused by overflowing calculations followed by modulo.
n and goal are large, consider memory requirements for a DP solution	Use space optimization techniques like rolling arrays to reduce the memory footprint of the DP table.

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What are the constraints on `n`, `goal`, and `k`? What are the maximum and minimum possible values for each?
Is it possible for `k` to be greater than or equal to `n`? If so, what is the expected behavior?
Should the result be returned modulo some specific number (e.g., 10^9 + 7)?
Can `n` or `goal` ever be zero?
Are we concerned about integer overflow during calculations, and if so, what's the appropriate way to handle it?

Brute Force Solution

Approach

Here's how the algorithm would work step-by-step:

Start building the playlist with the first song.
Then, for each spot in the playlist, try every song in the entire song set.
Make sure you keep track of how many songs are in the playlist, and how many unique songs have been used so far.
If you reach the desired playlist length and have the right number of unique songs, then you've found one valid playlist.
If you reach the desired playlist length, but don't have the right number of unique songs, it's not a valid playlist, so discard it and try another way.
If the playlist is still too short, keep adding songs and checking for validity.
Repeat this process for all possible combinations of songs.
Once all possibilities are exhausted, count the total number of valid playlists you found.

Code Implementation

def number_of_music_playlists_brute_force(number_of_songs, playlist_length, number_of_unique_songs):
    number_of_valid_playlists = 0

    def generate_playlist(current_playlist, unique_songs_in_playlist):
        nonlocal number_of_valid_playlists

        if len(current_playlist) == playlist_length:
            # Check if the playlist meets the criteria
            if len(unique_songs_in_playlist) == number_of_unique_songs:
                number_of_valid_playlists += 1
            return

        # Try adding each song to the playlist
        for song in range(number_of_songs):
            new_playlist = current_playlist + [song]

            # Create the set of unique songs in this playlist
            new_unique_songs_in_playlist = set(unique_songs_in_playlist)

            # Update to include the song
            new_unique_songs_in_playlist.add(song)

            generate_playlist(new_playlist, new_unique_songs_in_playlist)

    # Start generating playlists from scratch
    generate_playlist([], set())

    return number_of_valid_playlists

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

Here's how the algorithm would work step-by-step:

Understand that we are building playlists one song at a time, and need to keep track of how many songs are in the playlist and how many unique songs we've used so far.
Recognize that the number of ways to build a playlist of a certain length using a certain number of unique songs depends on the number of ways we could have built the playlist just before adding the last song.
Realize that the last song we add is either a song we've already used, or a brand-new unique song.
If we choose a song we've already used, we need to make sure we have played enough other songs since the last time we played this song to meet the minimum gap requirement.
If we choose a new, unique song, there are a limited number of choices because we only have a fixed number of unique songs to choose from.
Store the number of ways to build each playlist configuration (length, unique songs used) as we calculate it, so we don't have to recalculate it later. This is the key to efficiency.
Start with a base case: There's one way to have an empty playlist with no songs used.
Build up the solutions for larger playlists step-by-step, using the stored values and the rules about repeats and unique songs.
The final answer is the number of ways to create the desired playlist length using all the available unique songs.

Code Implementation

def number_of_music_playlists(playlist_length, 
                               number_of_unique_songs, 
                               minimum_song_gap):

    modulo_value = 10**9 + 7

    # dp[i][j] is the number of playlists of length i using j unique songs
    dynamic_programming_table = [[0] * (number_of_unique_songs + 1) 
                                 for _ in range(playlist_length + 1)]

    # Base case: an empty playlist with no songs used
    dynamic_programming_table[0][0] = 1

    for playlist_index in range(1, playlist_length + 1):
        for unique_song_count in range(1, number_of_unique_songs + 1):

            # Add a new unique song
            # There are number_of_unique_songs - (unique_song_count - 1) choices
            dynamic_programming_table[playlist_index][unique_song_count] = 
                (dynamic_programming_table[playlist_index - 1][unique_song_count - 1] 
                 * (number_of_unique_songs - (unique_song_count - 1))) % modulo_value

            # Re-use an old song, but only if allowed
            if unique_song_count > minimum_song_gap:
                # Ensures we've played enough songs for repeats
                dynamic_programming_table[playlist_index][unique_song_count] += 
                    (dynamic_programming_table[playlist_index - 1][unique_song_count] 
                     * (unique_song_count - minimum_song_gap)) % modulo_value

                dynamic_programming_table[playlist_index][unique_song_count] %= modulo_value

    # The answer is the playlist with playlist_length using all unique songs.
    return dynamic_programming_table[playlist_length][number_of_unique_songs]

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
n = 0 (no songs), goal = 0, k = 0 (no restrictions)	Return 1, as there is one way to play 0 songs.
n = 1, goal = 1, k = 0 (only one song to play for one playlist slot)	Return 1 since there is one way to play that song.
n = 1, goal > 1, k = 0 (one song to play for multiple playlist slots)	Return 0, as we can only play that song once.
goal < n (playlist length shorter than total songs)	Return 0 since it is impossible to create a playlist with less slots than available songs.
k >= n (can replay all songs immediately)	The standard formula or DP approach might need modification, likely simplifying to a combination problem without reuse restrictions after initial playlist creation.
Large n, goal, or k that cause integer overflow during calculations	Use modulo operator during intermediate calculations to prevent overflow and return the result modulo 10^9 + 7.
n is very close to goal and k is small such that it pushes calculations to the limit of the modulo.	Ensure the modulo operations are applied correctly to intermediate results to prevent errors caused by overflowing calculations followed by modulo.
n and goal are large, consider memory requirements for a DP solution	Use space optimization techniques like rolling arrays to reduce the memory footprint of the DP table.