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:
- Left append ‘x’ (“xa”), left append ‘x’ (“xxa”), right append y(“xxay”).
- Left append ‘x’ (“xa”), right append y(“xay”), left append ‘x’ (“xxay”).
- Right append y(“ay”), left append ‘x’ (“xay”), left append ‘x’ (“xxay”).
Input: X = ‘a’, Y = “cd”
Output: 0
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-inverselong 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 wayslong 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 codeint 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
Time Complexity: O(|y| * log MOD)
Auxiliary Space: O(1)
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