✅✅✅Easy To Understand C++ Code Using Binomial Coefficients
IntuitionThe problem requires reducing a string of digits by repeatedly summing pairs of consecutive digits modulo 10 until only two digits remain. We need to c
# Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
The problem requires reducing a string of digits by repeatedly summing pairs of consecutive digits modulo 10 until only two digits remain. We need to check if these final two digits are identical.

The key observation is that the operations performed resemble a binomial expansion, where each digit contributes to the final result based on the number of ways it can be formed. Therefore, the problem can be solved efficiently using properties of binomial coefficients modulo 10.

# Approach
<!-- Describe your approach to solving the problem. -->
1. Binomial Coefficients Modulo 10:

- The contribution of each digit to the final two digits depends on the number of ways it can appear, which can be calculated using binomial coefficients.
- Since we're interested in the last two digits, we only need the coefficients modulo 10.

2. Efficient Computation Using Lucas' Theorem:

- To calculate binomial coefficients modulo 10, we split the problem into calculations modulo 2 and 5 (since 10 = 2 × 5).
- Lucas' Theorem is used to efficiently compute binomial coefficients modulo prime powers.
   - binmod2(n, k): Computes binomial coefficient modulo 2.
   - binmod5(n, k): Computes binomial coefficient modulo 5 using   factorial properties and modular inverses.
   
- Combine the results using the Chinese Remainder Theorem to get the result modulo 10 (binmod10(n, k)).

3. Calculate Final Two Digits:
- Use the precomputed binomial coefficients to accumulate the contribution of each digit in the original string.
- Keep track of the contributions to the last two digits (s1 and s2) using modular arithmetic.
4. Check and Return Result:

- Return true if the final two digits are the same (s1 == s2), otherwise return false.

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

- Space complexity:
<!-- Add your space complexity here, e.g. $$O(n)$$ -->
# Code
```cpp []
class Solution {
public:
    int binmod10(int n, int k) {
        int mod2 = binmod2(n, k);
        int mod5 = binmod5(n, k);
        for (int i = 0; i < 10; i++) {
            if (i % 2 == mod2 && i % 5 == mod5)
                return i;
        }
        return 0;
    }

    int binmod2(int n, int k) {
        while (k > 0) {
            if ((k & 1) > (n & 1))
                return 0;
            n >>= 1;
            k >>= 1;
        }
        return 1;
    }

    int binmod5(int n, int k) {
        int res = 1;
        while (n > 0 || k > 0) {
            int nd = n % 5;
            int kd = k % 5;
            if (kd > nd)
                return 0;
            res = (res * binsmall(nd, kd)) % 5;
            n /= 5;
            k /= 5;
        }
        return res;
    }

    int binsmall(int n, int k) {
        if (k > n)
            return 0;
        int fact[5] = {1, 1, 2, 1, 4};
        int numerator = fact[n];
        int denominator = (fact[k] * fact[n - k]) % 5;
        int deninv = 0;
        for (int i = 0; i < 5; i++) {
            if ((denominator * i) % 5 == 1) {
                deninv = i;
                break;
            }
        }
        return (numerator * deninv) % 5;
    }

    bool hasSameDigits(string s) {
        int L = s.size();
        int m = L - 2;
        int s1 = 0, s2 = 0;
        for (int i = 0; i <= m; i++) {
            int val = binmod10(m, i);
            int d1 = s[i] - '0';
            int d2 = s[i + 1] - '0';
            s1 = (s1 + val * d1) % 10;
            s2 = (s2 + val * d2) % 10;
        }
        return s1 == s2;
    }
};

```
['C++']