Python O(n)✅ Solution with Explanation | Binomial Coefficients and Precompute 
IntuitionThe key insight is to recognize that each reduction step can be modeled using binomial coefficients modulo 10. Instead of simulating each step, we comp
# Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
The key insight is to recognize that each reduction step can be modeled using binomial coefficients modulo 10. Instead of simulating each step, we compute the final two digits directly using these coefficients. Here's the detailed approach:

1. Binomial Coefficients Modulo 10: The final two digits can be derived using binomial coefficients modulo 10. This involves:

    - Using Lucas's theorem to compute binomial coefficients modulo 2 and 5.
    - Combining these results using the Chinese Remainder Theorem (CRT) to get the result modulo 10.

2. Efficient Calculation: Precompute values for binomial coefficients modulo 5 using a lookup table and compute modulo 2 using bitwise operations. Combine these results using a precomputed CRT table for efficiency.

# Approach
<!-- Describe your approach to solving the problem. -->

Precompute Tables: The mod5_table precomputes binomial coefficients modulo 5 for small values. The crt_table maps combinations of modulo 2 and modulo 5 results to their equivalent modulo 10 values.

Combination Calculation: The comb_mod5 function uses Lucas's theorem to compute binomial coefficients modulo 5 efficiently by breaking down the numbers into their base-5 digits.

Coefficient Calculation: For each possible index, compute the binomial coefficient modulo 10 using the precomputed tables and bitwise operations for modulo 2.

Sum Calculation: Using the precomputed coefficients, compute the sums for the two final digits directly, then check if they are equal.


# Complexity
- Time complexity: $$O(n)$$
<!-- Add your time complexity here, e.g. $$O(n)$$ -->

- Space complexity: $$O(n)$$
<!-- Add your space complexity here, e.g. $$O(n)$$ -->

# Code
```python3 []
class Solution:
    def hasSameDigits(self, s: str) -> bool:
        # Precompute the mod5 combination table
        mod5_table = [
            [1, 0, 0, 0, 0],  # ni=0
            [1, 1, 0, 0, 0],  # ni=1
            [1, 2, 1, 0, 0],  # ni=2
            [1, 3, 3, 1, 0],  # ni=3
            [1, 4, 1, 4, 1],  # ni=4
        ]
        
        # CRT lookup table: (mod2, mod5) -> mod10
        crt_table = {
            (0, 0): 0,
            (0, 1): 6,
            (0, 2): 2,
            (0, 3): 8,
            (0, 4): 4,
            (1, 0): 5,
            (1, 1): 1,
            (1, 2): 7,
            (1, 3): 3,
            (1, 4): 9,
        }

        def comb_mod5(n, k):
            res = 1
            while n > 0 or k > 0:
                ni = n % 5
                ki = k % 5
                if ki > ni:
                    return 0
                res = (res * mod5_table[ni][ki]) % 5
                n = n // 5
                k = k // 5
            return res

        m = len(s)
        if m < 3:
            return False  # Though per problem constraints, this won't happen
        k = m - 2

        # Precompute coefficients for i in 0..k
        coeff = []
        for i in range(k + 1):
            # Compute mod2
            mod2 = 1 if (k & i) == i else 0
            # Compute mod5
            mod5 = comb_mod5(k, i)
            # Get mod10 from CRT table
            coeff.append(crt_table[(mod2, mod5)])

        # Compute sum1 and sum2
        sum1 = 0
        for i in range(k + 1):
            sum1 += int(s[i]) * coeff[i]
        sum1 %= 10

        sum2 = 0
        for i in range(k + 1):
            sum2 += int(s[i + 1]) * coeff[i]
        sum2 %= 10

        return sum1 == sum2
```
['Python3']