Given a positive integer k, you need to find the length of the smallest positive integer n such that n is divisible by k, and n only contains the digit 1. Return the length of n. If there is no such n, return -1. Note: n may not fit in a 64-bit sig

smallestRepunitDivByK.js

/**
 * @param {number} k
 * @return {number}
 */
var smallestRepunitDivByK = function(k) {
    // Step 1: Handle impossible cases
    // Any number made only of '1's is odd and not divisible by 2 or 5.
    // So if k has a factor of 2 or 5, return -1 immediately.
    if (k % 2 === 0 || k % 5 === 0) {
        return -1;
    }

    // Step 2: Initialize variables
    // remainder will track (current number % k) without building the full number.
    // length will track how many '1's we've added so far.
    let remainder = 0;
    let length = 0;

    // Step 3: Iterate until we find a remainder of 0
    // At each step, we "append" another '1' to the number.
    // new_remainder = (old_remainder * 10 + 1) % k
    while (true) {
        remainder = (remainder * 10 + 1) % k; // simulate appending '1'
        length++; // increase the length of the repunit

        // If remainder becomes 0, we found a number divisible by k
        if (remainder === 0) {
            return length;
        }

        // Safety check: If length exceeds k, we're cycling through remainders
        // Without hitting 0, so no solution exists.
        if (length > k) {
            return -1;
        }
    }
};