Problem Description

You have a music player with n different songs. You want to create a playlist to listen to exactly goal songs during your trip (songs can repeat in the playlist). To keep things interesting and avoid boredom, your playlist must follow these rules:

1. Every song must be played at least once - All n songs must appear in your playlist at least one time.

2. Replay restriction - If you want to play a song again, you must wait until at least k other different songs have been played since the last time you played it.

For example, if k = 2 and you play song A, you need to play at least 2 other different songs before you can play song A again.

Given three integers:

• n: the number of different songs available
• goal: the total length of the playlist (total number of songs to play)
• k: the minimum number of different songs that must be played before repeating a song

Return the number of different valid playlists you can create. Since this number can be very large, return the answer modulo 10^9 + 7.

The problem uses dynamic programming where f[i][j] represents the number of ways to create a playlist of length i using exactly j different songs. The transition considers two cases:

• Adding a new song that hasn't been played yet: there are (n - j + 1) choices
• Repeating a song that was already played: there are (j - k) choices (only if j > k)

The final answer is f[goal][n], representing playlists of length goal that use all n songs.

Intuition

The key insight is to think about building the playlist step by step, keeping track of how many songs we've used so far. When constructing a playlist, at each position we need to make a choice: either play a new song we haven't played before, or replay a song we've already used.

Let's think about what state we need to track. At any point in building our playlist, we care about:

• How many songs we've added to the playlist so far (let's call this i)
• How many different songs we've used (let's call this j)

This naturally leads to a dynamic programming approach where f[i][j] represents the number of ways to create a playlist of length i using exactly j different songs.

Now, when we're at position i in our playlist and have used j different songs, we have two choices:

Choice 1: Play a new song

• We can only do this if j < n (we haven't used all songs yet)
• We had j - 1 different songs before, and now we're adding the j-th one
• How many new songs can we choose from? We have n total songs, and we've already used j - 1, so we have n - (j - 1) = n - j + 1 options
• This contributes f[i - 1][j - 1] * (n - j + 1) to our count

Choice 2: Replay an existing song

• We can only do this if we've already used some songs (j > 0)
• Due to the constraint, we can only replay songs that were played more than k songs ago
• If we have j different songs in our rotation and need to skip the last k songs, we have j - k songs available to replay
• This only works if j > k (otherwise we don't have enough songs to satisfy the gap requirement)
• This contributes f[i - 1][j] * (j - k) to our count

The base case is f[0][0] = 1 - there's exactly one way to create an empty playlist with zero songs.

Since we need all n songs to appear at least once in our goal-length playlist, our answer is f[goal][n].

Learn more about Math, Dynamic Programming and Combinatorics patterns.

Solution Approach

We implement the dynamic programming solution using a 2D array f where f[i][j] represents the number of ways to create a playlist of length i using exactly j different songs.

Initialization:

• Create a 2D array f of size (goal + 1) × (n + 1) initialized with zeros
• Set the base case f[0][0] = 1 (one way to create an empty playlist)

State Transition: For each position i from 1 to goal and for each number of different songs j from 1 to n, we calculate f[i][j] based on two scenarios:

1. Adding a new song:

   • Previous state: f[i - 1][j - 1] (playlist of length i - 1 with j - 1 different songs)
   • Number of choices: (n - j + 1) new songs available
   • Contribution: f[i][j] += f[i - 1][j - 1] * (n - j + 1)

2. Replaying an existing song:

   • Previous state: f[i - 1][j] (playlist of length i - 1 with j different songs)
   • Condition: Only possible if j > k (we have enough songs to satisfy the gap requirement)
   • Number of choices: (j - k) songs that can be replayed
   • Contribution: f[i][j] += f[i - 1][j] * (j - k)

Implementation Details:

for i in range(1, goal + 1):
    for j in range(1, n + 1):
        # Add a new song
        f[i][j] = f[i - 1][j - 1] * (n - j + 1)
      
        # Replay an existing song (if possible)
        if j > k:
            f[i][j] += f[i - 1][j] * (j - k)
      
        # Apply modulo to prevent overflow
        f[i][j] %= mod

Time Complexity: O(goal × n) as we iterate through all states in our DP table.

Space Complexity: O(goal × n) for storing the DP table.

The final answer is f[goal][n], which represents the number of playlists of length goal that include all n songs at least once.

Example Walkthrough

Let's walk through a small example with n = 3 songs, goal = 3 songs in playlist, and k = 1 (must play 1 different song before repeating).

We'll build our DP table f[i][j] where i is playlist length and j is number of different songs used.

Initial State:

f[0][0] = 1 (one way to make empty playlist)
All other entries start at 0

Building the DP Table:

For i = 1 (playlist of length 1):

• f[1][1]: Use 1 different song
  • Add new song: f[0][0] × (3 - 1 + 1) = 1 × 3 = 3
  • Can't replay (j ≤ k)
  • Result: f[1][1] = 3

For i = 2 (playlist of length 2):

• f[2][1]: Use 1 different song (must repeat the same song)

  • Add new song: f[1][0] × 3 = 0 (f[1][0] doesn't exist/is 0)
  • Can't replay (j = 1 ≤ k = 1)
  • Result: f[2][1] = 0

• f[2][2]: Use 2 different songs

  • Add new song: f[1][1] × (3 - 2 + 1) = 3 × 2 = 6
  • Replay existing: f[1][2] × (2 - 1) = 0 (f[1][2] is 0)
  • Result: f[2][2] = 6

For i = 3 (playlist of length 3):

• f[3][1]: Use 1 different song

  • Add new song: f[2][0] × 3 = 0
  • Can't replay (j ≤ k)
  • Result: f[3][1] = 0

• f[3][2]: Use 2 different songs

  • Add new song: f[2][1] × 2 = 0 × 2 = 0
  • Replay existing: f[2][2] × (2 - 1) = 6 × 1 = 6
  • Result: f[3][2] = 6

• f[3][3]: Use 3 different songs (all songs used)

  • Add new song: f[2][2] × (3 - 3 + 1) = 6 × 1 = 6
  • Replay existing: f[2][3] × (3 - 1) = 0 (f[2][3] is 0)
  • Result: f[3][3] = 6

Final DP Table:

     j=0  j=1  j=2  j=3
i=0   1    0    0    0
i=1   0    3    0    0
i=2   0    0    6    0
i=3   0    0    6    6

Answer: f[3][3] = 6

The 6 valid playlists are (using songs A, B, C):

1. [A, B, C] - each song played once
2. [A, C, B] - each song played once
3. [B, A, C] - each song played once
4. [B, C, A] - each song played once
5. [C, A, B] - each song played once
6. [C, B, A] - each song played once

Note that we can't have playlists like [A, A, B] because that would only use 2 different songs, not all 3 as required.

Solution Implementation

Time and Space Complexity

The time complexity is O(goal × n). The code uses two nested loops - the outer loop iterates from 1 to goal (which runs goal times), and the inner loop iterates from 1 to n (which runs n times). Inside the nested loops, all operations including arithmetic calculations, array access, and modulo operations take constant time O(1). Therefore, the overall time complexity is O(goal × n).

The space complexity is O(goal × n). The code creates a 2D array f with dimensions (goal + 1) × (n + 1), which requires O((goal + 1) × (n + 1)) space. This simplifies to O(goal × n) since we drop the constant additions when expressing big-O notation. No other data structures are created that would affect the space complexity, so the total space complexity remains O(goal × n).

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Off-by-One Error in Counting Available Songs

The Pitfall: When adding a new song that hasn't been played yet, developers often make mistakes in calculating how many songs are available to choose from. A common error is using (n - j) instead of (n - j + 1).

Why It Happens:

• If we have n total songs and have already used j - 1 different songs, we're now choosing the j-th unique song
• The number of remaining songs is n - (j - 1) = n - j + 1
• It's easy to incorrectly think "we have used j songs, so n - j remain"

Incorrect Code:

# Wrong: This assumes we've already used j songs
dp[i][j] = dp[i - 1][j - 1] * (n - j)  # ❌

Correct Code:

# Right: We're choosing the j-th song from (n - j + 1) available options
dp[i][j] = dp[i - 1][j - 1] * (n - j + 1)  # ✅

2. Integer Overflow from Not Applying Modulo Early

The Pitfall: Forgetting to apply the modulo operation after each calculation step, or only applying it at the very end, can cause integer overflow even in languages like Python that handle big integers.

Why It Happens:

• The intermediate values grow exponentially large
• Even though Python handles arbitrary precision integers, the computation becomes extremely slow
• In other languages like Java or C++, this causes actual overflow

Incorrect Code:

# Wrong: Accumulating large values without modulo
dp[i][j] = dp[i - 1][j - 1] * (n - j + 1)
if j > k:
    dp[i][j] += dp[i - 1][j] * (j - k)
# Only applying modulo at the end - values can get huge! ❌
dp[i][j] = dp[i][j] % MOD

Correct Code:

# Right: Apply modulo after each operation
dp[i][j] = (dp[i - 1][j - 1] * (n - j + 1)) % MOD
if j > k:
    dp[i][j] = (dp[i][j] + dp[i - 1][j] * (j - k)) % MOD  # ✅

3. Incorrect Base Case Initialization

The Pitfall: Setting dp[0][0] = 0 or forgetting to initialize the base case entirely, which propagates zeros throughout the entire DP table.

Why It Happens:

• It seems counterintuitive that there's "one way" to create an empty playlist
• Developers might think an empty playlist should have zero ways

Incorrect Code:

# Wrong: No valid playlists will be counted
dp[0][0] = 0  # ❌
# Or forgetting initialization entirely

Correct Code:

# Right: There's exactly one way to have an empty playlist
dp[0][0] = 1  # ✅

4. Misunderstanding the Replay Constraint

The Pitfall: Using j >= k instead of j > k when checking if we can replay a song, or miscalculating the number of replayable songs.

Why It Happens:

• The constraint states we need k OTHER songs between replays
• If we have exactly k unique songs, we can't replay any of them yet because we need all k as the "buffer"

Incorrect Code:

# Wrong: Allows replay when we only have k songs
if j >= k:  # ❌
    dp[i][j] += dp[i - 1][j] * (j - k)
    # When j = k, this gives 0 choices anyway, but the logic is wrong

Correct Code:

# Right: We need MORE than k songs to have replayable options
if j > k:  # ✅
    dp[i][j] += dp[i - 1][j] * (j - k)
    # When j = k+1, we have 1 song that can be replayed