Count of N length Strings having S as a Subsequence

We are given a string of length S and given an another number n which represents the length of the strings which might contains S as the subsequence. We have to find the number of unique strings of length N which contains S as the subsequence, where a subsequence is the set of characters from the given string which may be all the characters or not and they not need to be continuous.

Sample Examples

Input

string str = "xyz"
int n = 3


Output

1


Explanation

There is only one string of length 3 that contains the given string as its subsequence.

Input

string str = "aback"
int n = 4


Output

0


Explanation

No string of length 4 can store the subsequence of size 5 or the size of given string is more than the given string so there is no solution.

Input

string str = "aba"
int n = 5


Output

6376


Approach

We have seen the example above to get an idea about the problem. In this approach we are going to use the concepts of the permutations and the combinations.

We will create a function which will take the current string and the given number N as the input and will return an integer representing the required number of strings.

In the function we will create an array which will store the factorial result for each number 0 to N both inclusive.

We will use the for loop to traverse over the range from the length of the given string and to the given number N.

To get the nCr value for the number of combinations we need to define another function that will take the n, r and factorial as the input and will return required value.

We will define another function getPower that will return the power of given number to the given value but the catch is it will return the value with the mod.

We will call to nCr function and from there we will call to the power function and get the required methods which we can see in the code.

Example

#include <bits/stdc++.h>
using namespace std;

int mod = 1e9 + 7; // mod value to take mod with
// function to get the power as the function of mod
long long getPower(long long num1, long long num2){
long long ans = 1;
num1 = num1 % mod;
if(num1 == 0){
return 0;
}
while(num2 > 0){
if(num2 & 1){
ans = (ans * num1) % mod;
}
num2 /= 2;
num1 = (num1 * num1) % mod;
}
return ans;
}
// get the value of the nCr
long long nCr(int n, int r, long long fact[]){
// base case
if(r > n){
return 0;
}
long long res = fact[n];
res *= getPower(fact[r], mod - 2);
res = res % mod;
res *= getPower(fact[n - r], mod - 2);
res = res % mod;
return res;
}
// function to get the required number of
int numberOfStrings(string str, int n){
int len = str.length(); // get the lenght of the given string
// if the n is less than than given stirng size
// then there will be no string present of given length
if(len > n){
return 0;
}
// array to store the value of the factorials
long long fact[n + 1];
fact[0] = 1; // initializing 0th index value
// calculating the result for the other indexes
for(int i = 1; i <= n; i++){
fact[i] = fact[i - 1] * i;
fact[i] %= mod;
}
// variable to store the result
long long res = 0;
// iterating over the all the possible positions
for (int i = len; i <= n; i++){
long long temp = nCr(i-1, len-1, fact);
temp *= getPower(26, n - i);
temp %= mod;
temp *= getPower(25, i - len);
temp %= mod;
res += temp;
res %= mod;
}
return res;
}
int main(){
int n = 5;
string str = "aba";
cout << "The number of strings of n length having str as a Subsequence are: "<< numberOfStrings(str, n)<<endl;
return 0;
}


Output

The number of strings of n length having str as a Subsequence are: 6376


Time and Space Complexity

The time complexity of the above code is O(N*log(N)), where N is the given limit of character of sequence.

The space complexity of the above code is O(N), as we are using an array to store the factorial upto length N.

Conclusion

In this tutorial, we have implemented a program to find the count of the number of strings of given length N that contains the given string S as the subsequence and as the count of string can be high so we need to take mod with the 1e9 + 7. We have implemented a approach using the factorial and combinations methods with the help of the power with the mod.

Updated on: 24-Aug-2023

36 Views