
Problem Statement
In this problem, you're tasked with generating playlists from a collection of n
distinct songs. Your objective is to create a playlist that consists of exactly goal
songs, tackling the challenge under specific conditions to enhance variety and listener experience. Each song in your playlist must be played at least once, and a song can only be replayed if at least k
other songs have been played since its last play. This rule helps in making sure that songs are not played too repetitively. The problem requires calculating the number of different playlists you can form under these rules. Since the potential number of playlists can be extremely high, you should provide the result modulo 10^9 + 7
to manage large numbers efficiently.
Examples
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
Approach and Intuition
In solving the problem of generating playlists that meet the specified criteria, the strategy revolves around dynamic programming due to the recursive nature of the constraints. Here's a breakdown of how we might think about and solve this:
Identifying Base Cases:
- When
goal
is equal ton
, every song must be unique, as each song has to be played at least once with no repetitions allowed. This simplifies to a permutation problem ofn
songs. - If
k
is 0, each song can be replayed immediately, increasing the potential combinations extensively.
- When
Recursive Formula Development:
- Consider
dp[i][j]
, wherei
is the number of songs in the playlist so far, andj
is the number of unique songs used. We need to manage transitions between states:- Adding a new song: This can be done in
(n-j)
ways if there are unused songs left. - Replaying a song: This can happen if at least
k
other songs have been played since its last play, providing(j-k)
options ifj>k
.
- Adding a new song: This can be done in
- Consider
Building the DP Table:
- Start by initializing the table where
dp[0][0] = 1
, representing an empty playlist. - Fill the table using the developed recursive relations, always taking care to apply the modulo operation to manage large numbers.
- Start by initializing the table where
Result Extraction:
- The answer to the problem will be found in
dp[goal][n]
, representing the number of ways to have exactlygoal
songs in the playlist with alln
distinct songs utilized.
- The answer to the problem will be found in
By modular arithmetic (mod 10^9 + 7
), the approach ensures computational feasiblity even with large numbers that result from combinatorial calculations involved in playlist generation. Understanding and applying these steps allow us to exactly quantify the combinations under the given constraints, providing the needed count of valid playlists.
Solutions
- C++
#include <vector>
class PlaylistCalculator {
private:
static const long long MODULO = 1000000007; // 10^9 + 7
std::vector<long long> fact; // Store factorials
std::vector<long long> invFact; // Store inverse factorials
// Function to perform modular exponentiation
long long modExp(long long base, int exp) {
long long res = 1;
while (exp > 0) {
if (exp % 2 == 1) {
res = (res * base) % MODULO;
}
base = (base * base) % MODULO;
exp /= 2;
}
return res;
}
// Precompute factorials and their modular inverses
void prepareFactorials(int maxN) {
fact.resize(maxN + 1);
invFact.resize(maxN + 1);
fact[0] = invFact[0] = 1;
for (int i = 1; i <= maxN; i++) {
fact[i] = (fact[i - 1] * i) % MODULO;
invFact[i] = modExp(fact[i], MODULO - 2); // Using Fermat's Little Theorem
}
}
public:
int countPlaylists(int numSongs, int playlistLength, int repeatLimit) {
prepareFactorials(numSongs);
long long result = 0;
int sign = 1;
for (int i = numSongs; i >= repeatLimit; i--) {
long long temp = modExp(i - repeatLimit, playlistLength - repeatLimit);
temp = temp * invFact[numSongs - i] % MODULO;
temp = temp * invFact[i - repeatLimit] % MODULO;
result = (result + sign * temp + MODULO) % MODULO;
sign *= -1;
}
return (fact[numSongs] * result) % MODULO;
}
};
Calculate the number of music playlists that meet specific conditions using C++, focusing on modular arithmetic for computational efficiency. Here's a detailed rundown of the implemented solution in the provided C++ code:
The system employs a
PlaylistCalculator
class that keeps all functional methods encapsulated:fact
andinvFact
vectors store the precomputed factorials and their modular inverses, respectively.
Key functions in the class:
modExp
: Handles modular exponentiation, critical for operations under a large prime modulus (MODULO
being (10^9 + 7)), ensuring the calculation does not overflow.prepareFactorials
: Computes all factorials and their inverses up to the required numbermaxN
using Fermat's Little Theorem for inverses.
The constructor,
PlaylistCalculator
:- Initializes factorials up to the desired maximum required for the computations, setting up the base for the main function.
The primary function,
countPlaylists
:- Sets up and calculates the number of valid playlists using combinatorial formulas adjusted for the modulo operations.
- Implements an inclusion-exclusion principle to accurately calculate the potential playlists by iterating from the total number of songs down to the minimum that may repeat (given by
repeatLimit
).
The outlined approach ensures:
- Efficient computation using modular arithmetic to handle large numbers.
- Precise calculations with inclusion-exclusion to accommodate for playlists restrictions.
- Use of precomputed values for factorial to optimize multiple factorial calculations, especially within the combinatorial context.
This ensures an optimal implementation to solve for the number of viable playlists adhering to both the length and repetition limitations, leveraging combinatorial mathematics and efficient use of modern C++ data structures and algorithms.
- Java
public class PlaylistCalculator {
private static final long MODULO = 1_000_000_007;
// Arrays for factorials and inverses
private long[] fact;
private long[] invFact;
// Calculates number of possible playlists
public int calculatePlaylists(int songs, int length, int k) {
// Prepare factorials and their inverses
setupFactorials(songs);
long total = 0;
int sig = 1;
// Calculate possibilities
for (int i = songs; i >= k; i--) {
long result = modPower(i - k, length - k);
result = (result * invFact[songs - i]) % MODULO;
result = (result * invFact[i - k]) % MODULO;
total = (total + sig * result + MODULO) % MODULO;
sig *= -1;
}
return (int) ((fact[songs] * total) % MODULO);
}
// Setup factorial values and their modular inverses
private void setupFactorials(int upTo) {
fact = new long[upTo + 1];
invFact = new long[upTo + 1];
fact[0] = invFact[0] = 1;
for (int i = 1; i <= upTo; i++) {
fact[i] = (fact[i - 1] * i) % MODULO;
invFact[i] = modPower(fact[i], (int) (MODULO - 2));
}
}
// Efficient modular exponentiation
private long modPower(long base, int exp) {
long res = 1L;
while (exp > 0) {
if ((exp & 1) == 1) {
res = (res * base) % MODULO;
}
exp >>= 1;
base = (base * base) % MODULO;
}
return res;
}
}
To solve the problem of calculating the number of music playlists, a Java class named PlaylistCalculator
is utilized. This class includes methods for calculating factorial values, their modular inverses, and efficiently computing powers with a modulo operation.
Within the calculatePlaylists
method:
- First, factorial arrays are initialized by invoking the
setupFactorials
method for a given number of songs. - The total number of combinations is calculated using a formula derived from combinatorial mathematics and properties of permutations and combinatorial arrangements. Alternating signs are managed using a variable
sig
that flips between each iteration to handle inclusion-exclusion principles. - Each iteration computes possible configurations based on available songs and spots in the playlist, adjusting for songs that cannot immediately repeat within
k
positions using modular arithmetic to ensure computations stay within bounds of the given MODULO (1,000,000,007
).
The array fact
is used to store precomputed factorial values, and invFact
holds the modular inverses of these factorial values. The inverses aid in efficient division under modulo properties.
The setupFactorials
method sets up the factorial and inverse arrays. Each factorial is computed sequentially, and inverses are calculated using the modPower
method:
- The
modPower
method performs modular exponentiation which is essential for finding factorial inverses via Fermat's Little Theorem. This process helps manage large numbers effectively without overflow, crucial for handling large factorial values within the modulo.
This solution ensures a linear-time operation for setting up factorials and a potentially O(L) complexity for playlist calculation, where L is the playlist length, making it efficient for the computation of very large numbers in combinatorial scenarios.
- Python
class Solution:
MODULUS = 1000000007
def exponentiation(self, base, exp):
result = 1
while exp > 0:
if exp % 2 == 1:
result = (result * base) % self.MODULUS
exp //= 2
base = (base * base) % self.MODULUS
return result
def precompute_factorials(self, n):
self.fact = [1] * (n + 1)
self.inv_fact = [1] * (n + 1)
for i in range(1, n + 1):
self.fact[i] = (self.fact[i - 1] * i) % self.MODULUS
self.inv_fact[i] = self.exponentiation(self.fact[i], self.MODULUS - 2)
def number_of_playlists(self, n, goal, k):
self.precompute_factorials(n)
sign = 1
total = 0
for i in range(n, k - 1, -1):
temp_result = self.exponentiation(i - k, goal - k)
temp_result = (temp_result * self.inv_fact[n - i]) % self.MODULUS
temp_result = (temp_result * self.inv_fact[i - k]) % self.MODULUS
total = (total + sign * temp_result + self.MODULUS) % self.MODULUS
sign *= -1
return (self.fact[n] * total) % self.MODULUS
The Python program provided solves the problem of determining the number of unique playlists that can be generated given specific constraints, using a modular arithmetic approach to address issues related to large numbers.
- The class
Solution
includes methods for modular exponentiation, precomputation of factorials and their modular inverses, and the main methodnumber_of_playlists
that calculates the number of ways to create a playlist. - Modular exponentiation, implemented in the
exponentiation
method, efficiently calculates powers of numbers under a modulus, a key operation given constraints often involve large values. - Factorials and their inverses modulo a large number (10^9 + 7) are precomputed in the
precompute_factorials
method. This precomputation speeds up the calculation in the main function as factorials are used repetitively. number_of_playlists
uses a dynamic programming approach with inclusion-exclusion to calculate the result.- It iterates backwards calculating contribution of each possible number of distinct songs to total playlists, adjusting for overcounts using alternating signs (managed by
sign
), mimicking the inclusion-exclusion principle.
The roles of major variables:
n
is the number of distinct songs available.goal
represents the target number of positions (length of the playlist) to be filled.k
sets a condition that a song cannot be repeated within 'k' subsequent positions once it is played.
Through these algorithms and number theory principles, the program efficiently handles calculating permutations with constraints by modular arithmetic, avoiding overflow problems and delivering results for large inputs expediently. This solution leverages mathematical properties to handle potentially vast numbers that could result from combinatorial calculations, crucial in cases where direct computation is not feasible due to time or space constraints.
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