# Number of ways to convert a character X to a string Y

Given a character X and a string Y of length N and the task is to find the number of ways to convert X to Y by appending characters to the left and the right ends of X. Note that any two ways are considered different if either the sequence of left and right appends are different or if the sequence is same, then characters appended are different i.e. a left append followed by a right append is different from a right append followed by a left append. Since the answer can be large print the final answer MOD (109 + 7).

Examples:

Input: X = ‘a’, Y = “xxay”
Output: 3
All possible ways are:

1. Left append ‘x’ (“xa”), left append ‘x’ (“xxa”), right append y(“xxay”).
2. Left append ‘x’ (“xa”), right append y(“xay”), left append ‘x’ (“xxay”).
3. Right append y(“ay”), left append ‘x’ (“xay”), left append ‘x’ (“xxay”).

Input: X = ‘a’, Y = “cd”
Output: 0

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Method 1: One way for solving this problem will be using dynamic programming.

• Initialise a variable ans = 0, mod = 1000000007.
• For all index ‘i’ such that Y[i] = X, update answer as ans = (ans + dp[i][i])%mod.

Here, dp[l][r] is the number of ways to make Y from the sub-string Y[l…r].
The recurrence relation will be:

dp[l][r] = (dp[l][r + 1] + dp[l – 1][r]) % mod

The time complexity for this approach will be O(N2).

Method 2:

• Initialise a variable ans = 0, mod = 1000000007.
• For all index i such that Y[i] = X, update answer as ans = (ans + F(i)) % mod where F(i) = (((N – 1)!) / (i! * (N – i – 1)!)) % mod.

Reason the above formula works: Just try to find the answer of the question, find the number of permutations of (p number of L) and (q number of R) where L and R the left append and the right append operations respectively.
The answer is (p + q)! / (p! * q!). For each valid i, just find the number of permutations of i Ls and N – i – 1 Rs.
The time complexity of this approach will be O(N).

Below is the implementation of the above approach:

## C++

 // C++ implementation of the approach #include using namespace std;    const int MOD = 1000000007;    // Function to find the modular-inverse long long modInv(long long a, long long p = MOD - 2) {     long long s = 1;        // While power > 1     while (p != 1) {            // Updating s and a         if (p % 2)             s = (s * a) % MOD;         a = (a * a) % MOD;            // Updating power         p /= 2;     }        // Return the final answer     return (a * s) % MOD; }    // Function to return the count of ways long long findCnt(char x, string y) {     // To store the final answer     long long ans = 0;        // To store pre-computed factorials     long long fact[y.size() + 1] = { 1 };        // Computing factorials     for (long long i = 1; i <= y.size(); i++)         fact[i] = (fact[i - 1] * i) % MOD;        // Loop to find the occurences of x     // and update the ans     for (long long i = 0; i < y.size(); i++) {         if (y[i] == x) {             ans += (modInv(fact[i])                     * modInv(fact[y.size() - i - 1]))                    % MOD;             ans %= MOD;         }     }        // Multiplying the answer by (n - 1)!     ans *= fact[(y.size() - 1)];     ans %= MOD;        // Return the final answer     return ans; }    // Driver code int main() {     char x = 'a';     string y = "xxayy";        cout << findCnt(x, y);        return 0; }

## Java

 // Java implementation of the approach     class GFG  {                final static int MOD = 1000000007;             // Function to find the modular-inverse      static long modInv(long a)      {          long p = MOD - 2;         long s = 1;                 // While power > 1          while (p != 1)          {                     // Updating s and a              if (p % 2 == 1)                  s = (s * a) % MOD;                                 a = (a * a) % MOD;                     // Updating power              p /= 2;          }                 // Return the final answer          return (a * s) % MOD;      }             // Function to return the count of ways      static long findCnt(char x, String y)      {          // To store the final answer          long ans = 0;                 // To store pre-computed factorials          long fact[] = new long[y.length() + 1];                    for(int i = 0; i < y.length() + 1; i++)             fact[i] = 1;                // Computing factorials          for (int i = 1; i <= y.length(); i++)              fact[i] = (fact[i - 1] * i) % MOD;                 // Loop to find the occurences of x          // and update the ans          for (int i = 0; i < y.length(); i++)          {              if (y.charAt(i) == x)             {                  ans += (modInv(fact[i])                      * modInv(fact[y.length() - i - 1])) % MOD;                                    ans %= MOD;              }          }                 // Multiplying the answer by (n - 1)!          ans *= fact[(y.length() - 1)];          ans %= MOD;                 // Return the final answer          return ans;      }             // Driver code      public static void main (String[] args)     {          char x = 'a';          String y = "xxayy";                 System.out.println(findCnt(x, y));             }  }    // This code is contributed by AnkitRai01

## Python3

 # Python3 implementation of the approach  MOD = 1000000007;     # Function to find the modular-inverse  def modInv(a, p = MOD - 2) :        s = 1;         # While power > 1      while (p != 1) :            # Updating s and a          if (p % 2) :             s = (s * a) % MOD;          a = (a * a) % MOD;             # Updating power          p //= 2;         # Return the final answer      return (a * s) % MOD;        # Function to return the count of ways  def findCnt(x, y) :        # To store the final answer      ans = 0;         # To store pre-computed factorials      fact = [1]*(len(y) + 1) ;         # Computing factorials      for i in range(1,len(y)) :         fact[i] = (fact[i - 1] * i) % MOD;         # Loop to find the occurences of x      # and update the ans      for i in range(len(y)) :         if (y[i] == x) :             ans += (modInv(fact[i]) *                      modInv(fact[len(y)- i - 1])) % MOD;              ans %= MOD;                     # Multiplying the answer by (n - 1)!      ans *= fact[(len(y) - 1)];      ans %= MOD;         # Return the final answer      return ans;     # Driver code  if __name__ == "__main__" :         x = 'a';      y = "xxayy";         print(findCnt(x, y));     # This code is contributed by AnkitRai01

## C#

 // C# implementation of the approach  using System;    class GFG  {                 static int MOD = 1000000007;             // Function to find the modular-inverse      static long modInv(long a)      {          long p = MOD - 2;          long s = 1;                 // While power > 1          while (p != 1)          {                     // Updating s and a              if (p % 2 == 1)                  s = (s * a) % MOD;                                 a = (a * a) % MOD;                     // Updating power              p /= 2;          }                 // Return the final answer          return (a * s) % MOD;      }             // Function to return the count of ways      static long findCnt(char x, String y)      {          // To store the final answer          long ans = 0;                 // To store pre-computed factorials          long []fact = new long[y.Length + 1];                     for(int i = 0; i < y.Length + 1; i++)              fact[i] = 1;                 // Computing factorials          for (int i = 1; i <= y.Length; i++)              fact[i] = (fact[i - 1] * i) % MOD;                 // Loop to find the occurences of x          // and update the ans          for (int i = 0; i < y.Length; i++)          {              if (y[i] == x)              {                  ans += (modInv(fact[i])                      * modInv(fact[y.Length - i - 1])) % MOD;                                     ans %= MOD;              }          }                 // Multiplying the answer by (n - 1)!          ans *= fact[(y.Length - 1)];          ans %= MOD;                 // Return the final answer          return ans;      }             // Driver code      public static void Main ()      {          char x = 'a';          string y = "xxayy";                 Console.WriteLine(findCnt(x, y));      }  }     // This code is contributed by AnkitRai01

Output:

6

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