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Number of Music Playlists in C++
Suppose we have a music player, that contains N different songs and we want to listen to L songs during our trip. So we have to make a playlist so that it meets these conditions −
Every song is played at least once
A song can only be played again only if K other songs have been played.
We have to find the number of possible playlists. The answer can be very large, so we will return it modulo 10^9 + 7.
So, if the input is like N = 2, L = 3, K = 0, then the output will be 6, as there are 6 possible playlists [1,1,2], [1,2,1], [2,1,1], [2,2,1], [2,1,2], [1,2,2].
To solve this, we will follow these steps −
Define a function add(), this will take a, b,
return ((a mod m) + (b mod m)) mod m
Define a function sub(), this will take a, b,
return ((a mod m) - (b mod m) + m) mod m
Define a function mul(), this will take a, b,
return ((a mod m) * (b mod m)) mod m
From the main method, do the following −
Make one 2d array of size dp(L + 1) x (N + 1)
dp[0, 0] := 1
for initialize i := 1, when i <= L, update (increase i by 1), do −
for initialize j := 1, when j <= N, update (increase j by 1), do −
dp[i, j] := mul(dp[i - 1, j - 1], (N - (j - 1)))
if j > K, then −
dp[i, j] := add(dp[i, j], mul(dp[i - 1, j], j - K))
return dp[L, N]
Let us see the following implementation to get better understanding −
Example
#include <bits/stdc++.h>
using namespace std;
const int m = 1e9 + 7;
typedef long long int lli;
class Solution {
public:
int add(lli a, lli b){
return ((a % m) + (b % m)) % m;
}
int sub(lli a, lli b){
return ((a % m) - (b % m) + m) % m;
}
int mul(lli a, lli b){
return ((a % m) * (b % m)) % m;
}
int numMusicPlaylists(int N, int L, int K) {
vector < vector <int> > dp(L + 1, vector <int>(N + 1));
dp[0][0] = 1;
for(int i = 1; i <= L; i++){
for(int j = 1; j <= N; j++){
dp[i][j] = mul(dp[i - 1][j - 1], (N - (j - 1)));
if(j > K){
dp[i][j] = add(dp[i][j], mul(dp[i - 1][j], j -
K));
}
}
}
return dp[L][N];
}
};
main(){
Solution ob;
cout << (ob.numMusicPlaylists(2, 3, 0));
}Input
2,3,0
Output
6
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