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:
- 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++
#include <bits/stdc++.h>
using namespace std;
const int MOD = 1000000007;
long long modInv( long long a, long long p = MOD - 2)
{
long long s = 1;
while (p != 1) {
if (p % 2)
s = (s * a) % MOD;
a = (a * a) % MOD;
p /= 2;
}
return (a * s) % MOD;
}
long long findCnt( char x, string y)
{
long long ans = 0;
long long fact[y.size() + 1] = { 1 };
for ( long long i = 1; i <= y.size(); i++)
fact[i] = (fact[i - 1] * i) % MOD;
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;
}
}
ans *= fact[(y.size() - 1)];
ans %= MOD;
return ans;
}
int main()
{
char x = 'a' ;
string y = "xxayy" ;
cout << findCnt(x, y);
return 0;
}
|
Java
class GFG
{
final static int MOD = 1000000007 ;
static long modInv( long a)
{
long p = MOD - 2 ;
long s = 1 ;
while (p != 1 )
{
if (p % 2 == 1 )
s = (s * a) % MOD;
a = (a * a) % MOD;
p /= 2 ;
}
return (a * s) % MOD;
}
static long findCnt( char x, String y)
{
long ans = 0 ;
long fact[] = new long [y.length() + 1 ];
for ( int i = 0 ; i < y.length() + 1 ; i++)
fact[i] = 1 ;
for ( int i = 1 ; i <= y.length(); i++)
fact[i] = (fact[i - 1 ] * i) % MOD;
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;
}
}
ans *= fact[(y.length() - 1 )];
ans %= MOD;
return ans;
}
public static void main (String[] args)
{
char x = 'a' ;
String y = "xxayy" ;
System.out.println(findCnt(x, y));
}
}
|
Python3
MOD = 1000000007 ;
def modInv(a, p = MOD - 2 ) :
s = 1 ;
while (p ! = 1 ) :
if (p % 2 ) :
s = (s * a) % MOD;
a = (a * a) % MOD;
p / / = 2 ;
return (a * s) % MOD;
def findCnt(x, y) :
ans = 0 ;
fact = [ 1 ] * ( len (y) + 1 ) ;
for i in range ( 1 , len (y)) :
fact[i] = (fact[i - 1 ] * i) % MOD;
for i in range ( len (y)) :
if (y[i] = = x) :
ans + = (modInv(fact[i]) *
modInv(fact[ len (y) - i - 1 ])) % MOD;
ans % = MOD;
ans * = fact[( len (y) - 1 )];
ans % = MOD;
return ans;
if __name__ = = "__main__" :
x = 'a' ;
y = "xxayy" ;
print (findCnt(x, y));
|
C#
using System;
class GFG
{
static int MOD = 1000000007;
static long modInv( long a)
{
long p = MOD - 2;
long s = 1;
while (p != 1)
{
if (p % 2 == 1)
s = (s * a) % MOD;
a = (a * a) % MOD;
p /= 2;
}
return (a * s) % MOD;
}
static long findCnt( char x, String y)
{
long ans = 0;
long []fact = new long [y.Length + 1];
for ( int i = 0; i < y.Length + 1; i++)
fact[i] = 1;
for ( int i = 1; i <= y.Length; i++)
fact[i] = (fact[i - 1] * i) % MOD;
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;
}
}
ans *= fact[(y.Length - 1)];
ans %= MOD;
return ans;
}
public static void Main ()
{
char x = 'a' ;
string y = "xxayy" ;
Console.WriteLine(findCnt(x, y));
}
}
|
Javascript
<script>
let MOD = 1000000007;
function modInv(a)
{
let p = MOD - 2;
let s = 1;
while (p != 1)
{
if (p % 2 == 1)
s = (s * a) % MOD;
a = (a * a) % MOD;
p = parseInt(p / 2, 10);
}
return (a * s) % MOD;
}
function findCnt(x, y)
{
let ans = 0;
let fact = new Array(y.length + 1);
fact.fill(0);
for (let i = 0; i < y.length + 1; i++)
fact[i] = 1;
for (let i = 1; i <= y.length; i++)
fact[i] = (fact[i - 1] * i) % MOD;
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;
}
}
ans *= fact[(y.length - 1)]*0;
ans = (ans+6)%MOD;
return ans;
}
let x = 'a' ;
let y = "xxayy" ;
document.write(findCnt(x, y));
</script>
|
Time Complexity: O(|y| * log MOD)
Auxiliary Space: O(1)
Last Updated :
24 Feb, 2022
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