binomial coefficient reduction for final digit equivalence
Intuitionthe idea is that after repeatedly summing adjacent digits mod 10, the final two digits can be expressed as a weighted sum of the original digits. the w
# Intuition
the idea is that after repeatedly summing adjacent digits mod 10, the final two digits can be expressed as a weighted sum of the original digits. the weights come from the binomial coefficients in the n-2 row of pascal's triangle, taken mod 10.

# Approach
- convert the input string to a list of digits.
- note that the final digits are given by:
  - f0 = sum(binom(n-2, i) * digit[i]) mod 10
  - f1 = sum(binom(n-2, i) * digit[i+1]) mod 10
- compute the binomial coefficients mod 10 using lucas' theorem by calculating them mod 2 and mod 5, then combining the results.
-- compare the two final digits and return true if they are the same; otherwise, return false.

# Complexity
- Time complexity:
o(n)
- Space complexity:
o(n)

# Code
```python3 []
class Solution:
    def hasSameDigits(self, s: str) -> bool:
        n = len(s)
        if n == 2:
            return s[0] == s[1]
        
        digits = list(map(int, s))
        N = n - 2  
        mod = 10
        F0, F1 = 0, 0
        
        for i in range(n - 1):
            c2 = self.binom_mod2(N, i)      
            c5 = self.binom_mod5(N, i, 5)    
            r = (5 * c2 + 6 * c5) % mod
            
            F0 = (F0 + r * digits[i]) % mod
            F1 = (F1 + r * digits[i+1]) % mod
        
        return F0 == F1

    def binom_mod2(self, n: int, k: int) -> int:
        if k > n:
            return 0
        return 1 if (n & k) == k else 0

    def binom_mod5(self, n: int, k: int, p: int) -> int:
        if k > n:
            return 0
        res = 1
        while n or k:
            n_i = n % p
            k_i = k % p
            if k_i > n_i:
                return 0
            res = (res * self.small_binom(n_i, k_i, p)) % p
            n //= p
            k //= p
        return res

    def small_binom(self, n: int, k: int, p: int) -> int:
        if k > n:
            return 0
        fact = [1] * p
        for i in range(1, p):
            fact[i] = (fact[i-1] * i) % p
        inv_fact = [1] * p
        inv_fact[p-1] = pow(fact[p-1], p-2, p)
        for i in range(p-2, -1, -1):
            inv_fact[i] = (inv_fact[i+1] * (i+1)) % p
        return (fact[n] * inv_fact[k] * inv_fact[n-k]) % p
```
['Python3']