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++ 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 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 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# 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 |
<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> |
6
Time Complexity: O(|y| * log MOD)
Auxiliary Space: O(1)