Has Same Digits Javascript Solution
Intuition
This problem requires checking whether alternating sums of binomial coefficients modulo 10 are equal.
The key observation is that binomial coefficient
# Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
* This problem requires checking whether alternating sums of binomial coefficients modulo 10 are equal.
 * The key observation is that binomial coefficients have properties that can be used for efficient computation.
# Approach
<!-- Describe your approach to solving the problem. -->
 * 1. Compute alternating sums using binomial coefficients modulo 10.
 * 2. Use modular arithmetic properties of binomial coefficients with respect to 2 and 5.
 * 3. Compare the computed sums.
# Complexity
- Time complexity: O(n), where n is the length of the string (each binomial coefficient calculation is efficient).
<!-- Add your time complexity here, e.g. $$O(n)$$ -->

- Space complexity: O(1), since only a few variables are used.
<!-- Add your space complexity here, e.g. $$O(n)$$ -->

# Code
```javascript []
var hasSameDigits = function (s) {
    // Step 1: Initialize Variables
    let size = s.length;
    let X = size - 2;
    let x = 0, y = 0;

    // Step 2: Compute Alternating Sum Using Binomial Coefficients
    for (let j = 0; j <= X; j++) {
        let coeff = binomialMod10(X, j);
        x = (x + coeff * (s.charCodeAt(j) - '0'.charCodeAt(0))) % 10;
        y = (y + coeff * (s.charCodeAt(j + 1) - '0'.charCodeAt(0))) % 10;
    }

    // Step 3: Compare Alternating Sums
    return x === y;
};

// Step 4: Compute Binomial Coefficient Modulo 10
var binomialMod10 = function (n, k) {
    let i = binomialMod2(n, k);
    let j = binomialMod5(n, k);

    for (let x = 0; x < 10; x++) {
        if (x % 2 === i && x % 5 === j) {
            return x;
        }
    }
    return 0;
};

// Step 5: Compute Binomial Coefficient Modulo 2
var binomialMod2 = function (n, k) {
    return ((n & k) === k) ? 1 : 0;
};

// Step 6: Compute Binomial Coefficient Modulo 5 Using Lookup Table
var binomialMod5 = function (n, k) {
    const tuples = [
        [1],
        [1, 1],
        [1, 2, 1],
        [1, 3, 3, 1],
        [1, 4, 1, 4, 1]
    ];
    let result = 1;

    while (n > 0 || k > 0) {
        let nthd = n % 5;
        let kthd = k % 5;

        if (kthd > nthd) return 0;

        result = (result * tuples[nthd][kthd]) % 5;
        n = Math.floor(n / 5);
        k = Math.floor(k / 5);
    }
    return result;
};

```
['JavaScript']