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Mathematical Proof for "Check If Digits Are Equal in String After Operations II"Problem StatementGiven a string s consisting of digits, perform the following op
# Mathematical Proof for "Check If Digits Are Equal in String After Operations II"

## Problem Statement

Given a string $s$ consisting of digits, perform the following operation repeatedly until the string has exactly two digits:

* For each pair of consecutive digits in $s$, calculate a new digit as the sum of the two digits modulo 10.
* Replace $s$ with the sequence of newly calculated digits, maintaining the order.

Return `true` if the final two digits in $s$ are the same; otherwise, return `false`.

## Key Insight

The solution uses a combinatorial approach based on the fact that after successive operations, each position in the resulting string can be expressed as a linear combination of the original digits, with coefficients being binomial coefficients.

## Mathematical Derivation

Let $s^{(k)}[j]$ represent the $j$-th digit after $k$ operations. By induction, we can prove:

$$s^{(k)}[j] = \left(\sum_{i=0}^{k} \binom{k}{i} \cdot s[j+i]\right) \bmod 10$$

After $n-2$ operations (where $n$ is the string length), we have:

$$s^{(n-2)}[0] = \left(\sum_{i=0}^{n-2} \binom{n-2}{i} \cdot s[i]\right) \bmod 10$$

$$s^{(n-2)}[1] = \left(\sum_{i=0}^{n-2} \binom{n-2}{i} \cdot s[i+1]\right) \bmod 10$$

For these digits to be equal, their difference must be 0 (mod 10):

$$\left(\sum_{i=0}^{n-2} \binom{n-2}{i} \cdot (s[i] - s[i+1])\right) \bmod 10 = 0$$

## Implementation Analysis

### Number Representation

The solution uses tuples $(y, a, b)$ where a number is represented as $y \cdot 2^a \cdot 5^b$ with $\gcd(y,10)=1$. This factorization efficiently handles modular arithmetic with powers of 10.

### Binomial Coefficient Calculation

The `compute_comb` function calculates binomial coefficients using the recurrence relation:

$$\binom{N}{k} = \binom{N-1}{k-1} \cdot \frac{N-k+1}{k}$$

### Modular Arithmetic Optimizations

The solution only needs coefficient values modulo 10, handled efficiently by the `toInt` function which leverages these properties:

* If both $a \geq 1$ and $b \geq 1$, then $2^a \cdot 5^b \equiv 0 \pmod{10}$
* If $b \geq 1$, then $y \cdot 5^b \equiv 5 \pmod{10}$ when $y$ is coprime to 10
* Powers of 2 modulo 10 follow the cycle $[6, 2, 4, 8]$

### Final Calculation

The `hasSameDigits` function computes:

$$\sum_{i=0}^{n-2} \left(\text{toInt}(C[i]) \cdot (s[i] - s[i+1])\right) \bmod 10$$

where $C[i]$ represents $\binom{n-2}{i}$ in our factorized representation.

## Time Complexity Analysis

The solution achieves $O(n)$ time complexity versus $O(n^2)$ for a simulation-based approach, as it directly computes the final result without simulating each intermediate step.

## Correctness

The algorithm correctly determines if the final two digits will be equal by computing their mathematical relationship to the original digits, rather than performing the operations explicitly.

## Implementation Details

The code leverages several optimizations:

1. The `factor()` function decomposes integers into the $(y, a, b)$ representation
2. The `inverse()` function efficiently computes modular multiplicative inverses
3. The `toInt()` function handles special cases for efficient modular evaluation
4. The combinatorial identity allows us to compute the result in linear time

By using these mathematical properties, the solution avoids the expensive operation of actually simulating the digit transformations, resulting in a much more efficient algorithm.

``` C+++
using int3 = array<int, 3>; // (y, a, b) where x = y * 2^a * 5^b and gcd(y,10) = 1

int3 C[100000];
int inv10[10]={0, 1, 0, 7, 0, 0, 0, 3, 0, 9};// inv10[x]*x=1 (mod 10) for x coprime to 10

class Solution {
public:
    int3 factor(int x) {
        int a = countr_zero((unsigned)x);
        x >>= a;
        int b = 0;
        while (x % 5 == 0) {
            x /= 5;
            b++;
        }
        return {x % 10, a, b}; // Only keep mod 10 for y
    }

    int3 inverse(const int3& x) {
        int x0 = x[0], a = x[1], b = x[2];
        int x0_inv = inv10[x0];
        return {x0_inv, -a, -b};
    }

    int3 mult(const int3& x, const int3& y) {
        return {x[0]*y[0]%10, x[1]+y[1], x[2]+y[2]};
    }

    int toInt(const int3& x) { // mod 10
        if (x[1] >= 1 && x[2] >= 1) return 0;
        if (x[2] >= 1) return 5;
        int B[4] = {6, 2, 4, 8}; // 2^x(mod 10) cycle
        if (x[1] == 0) return x[0];
        int B2 = B[x[1] % 4]; // 
        return x[0] * B2 % 10;
    }

    void compute_comb(int N) {
        C[0] = {1, 0, 0}; 
        if (N == 0) return; // Prevent C[N] access if N=0
        C[N] = {1, 0, 0}; 

        for (int k = 1; k <= N/2; k++) {
            int3 P = factor(N-k + 1);
            int3 Q = inverse(factor(k));
            C[k]=C[N-k] = mult(mult(C[k-1], P), Q);
        }
    }

    bool hasSameDigits(string& s) {
        int n = s.size();

        fill(C, C+(n-2), int3{0, 0, 0}); // Prevent overflow
        compute_comb(n-2);

        int sum=0;
        for (int i=0; i <= n-2; i++) {
            int C_val= (toInt(C[i])*(s[i]-s[i+1]))%10;
            sum=(sum+C_val+10) % 10; // Ensure non-negative
        }

     //   cout << sum << endl;
        return sum == 0;
    }
};
```
['Math', 'Combinatorics', 'Number Theory', 'C++']