Easy Java Solution
IntuitionWe reinterpret the repeated adjacent-sum operation as a convolution with the vector [1, 1]. Repeating this operation (n–2) times is equivalent to convo
# Intuition
We reinterpret the repeated adjacent-sum operation as a convolution with the vector [1, 1]. Repeating this operation (n–2) times is equivalent to convolving the original digit sequence with the binomial coefficients from the expansion of (1+x)^(n–2) modulo 10. Thus, the two final digits can be expressed as linear combinations of the original digits with these binomial coefficients. The "rare CP concept" here is to compute the required binomial coefficients modulo 10 using Lucas’s theorem (applied separately modulo 2 and 5) and then combine them via the Chinese Remainder Theorem.

# Approach
1. Represent the final two digits as:
   - f0 = sum_{j=0}^{n–2} (binom(n–2, j) * digit_j) mod 10
   - f1 = sum_{j=0}^{n–2} (binom(n–2, j) * digit_{j+1}) mod 10
2. Instead of computing huge binomials directly, use Lucas’s theorem to compute binom(n–2, j) modulo 2 and modulo 5. Then combine these two residues using the Chinese Remainder Theorem to get the coefficient modulo 10.
3. Sum up the contributions for f0 and f1 and check if they are equal modulo 10.

# Complexity
- Time complexity: O(n * (log(n) base p)) where p ∈ {2,5}; effectively O(n)
- Space complexity: O(1)

# Code
```java []
class Solution {
    int[][] binom2;
    int[][] binom5;
    
    public Solution() {
        binom2 = new int[2][2];
        binom2[0][0] = 1;
        binom2[1][0] = 1;
        binom2[1][1] = 1;
        binom5 = new int[5][5];
        for (int i = 0; i < 5; i++) {
            for (int j = 0; j <= i; j++) {
                if (j == 0 || j == i) 
                    binom5[i][j] = 1;
                else 
                    binom5[i][j] = (binom5[i - 1][j - 1] + binom5[i - 1][j]) % 5;
            }
        }
    }
    
    int lucas(int n, int k, int p) {
        int res = 1;
        while (n > 0 || k > 0) {
            int ni = n % p;
            int ki = k % p;
            if (ki > ni) return 0;
            if (p == 2) 
                res = (res * binom2[ni][ki]) % 2;
            else if (p == 5) 
                res = (res * binom5[ni][ki]) % 5;
            n /= p;
            k /= p;
        }
        return res;
    }
    
    int binomMod10(int n, int k) {
        int mod2 = lucas(n, k, 2);
        int mod5 = lucas(n, k, 5);
        for (int x = mod5; x < 10; x += 5) {
            if (x % 2 == mod2) return x;
        }
        return 0;
    }
    
    public boolean hasSameDigits(String s) {
        int n = s.length();
        int m = n - 2;
        int f0 = 0, f1 = 0;
        for (int j = 0; j <= m; j++) {
            int c = binomMod10(m, j);
            int d = s.charAt(j) - '0';
            f0 = (f0 + c * d) % 10;
        }
        for (int j = 0; j <= m; j++) {
            int c = binomMod10(m, j);
            int d = s.charAt(j + 1) - '0';
            f1 = (f1 + c * d) % 10;
        }
        return f0 == f1;
    }
}
```
['Java']