Skip to content
Related Articles

Related Articles

Improve Article
Number of ways to convert a character X to a string Y
  • Difficulty Level : Medium
  • Last Updated : 10 Jun, 2021

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:
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:
 

 

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



  • Initialize 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: 
 

  • Initialize 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 <bits/stdc++.h>
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 occurrences 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 occurrences 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 occurrences 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 occurrences 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

Javascript




<script>
 
    // JavaScript implementation of the approach
     
    let MOD = 1000000007;
       
    // Function to find the modular-inverse
    function modInv(a)
    {
        let p = MOD - 2;
        let 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 = parseInt(p / 2, 10);
        }
       
        // Return the final answer
        return (a * s) % MOD;
    }
       
    // Function to return the count of ways
    function findCnt(x, y)
    {
        // To store the final answer
        let ans = 0;
       
        // To store pre-computed factorials
        let fact = new Array(y.length + 1);
        fact.fill(0);
           
        for(let i = 0; i < y.length + 1; i++)
            fact[i] = 1;
       
        // Computing factorials
        for (let i = 1; i <= y.length; i++)
            fact[i] = (fact[i - 1] * i) % MOD;
       
        // Loop to find the occurrences of x
        // and update the ans
        for (let 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)]*0;
        ans = (ans+6)%MOD;
       
        // Return the final answer
        return ans;
    }
     
    let x = 'a';
    let y = "xxayy";
 
    document.write(findCnt(x, y));
     
</script>
Output: 
6

 

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with industry experts, please refer Geeks Classes Live 




My Personal Notes arrow_drop_up
Recommended Articles
Page :