483. Smallest Good Base

Problem

Given an integer n represented as a string, return **the smallest *good base* of** n.

We call k >= 2 a good base of n, if all digits of n base k are 1's.

  Example 1:

Input: n = "13"
Output: "3"
Explanation: 13 base 3 is 111.

Example 2:

Input: n = "4681"
Output: "8"
Explanation: 4681 base 8 is 11111.

Example 3:

Input: n = "1000000000000000000"
Output: "999999999999999999"
Explanation: 1000000000000000000 base 999999999999999999 is 11.

  Constraints:

• n is an integer in the range [3, 1018].

• n does not contain any leading zeros.

Solution

/**
 * @param {string} n
 * @return {string}
 */
var smallestGoodBase = function(n) {
   const N = BigInt(n), bigint2 = BigInt(2), bigint1 = BigInt(1), bigint0 = BigInt(0)
   let maxLen = countLength(N, bigint2) // result at most maxLen 1s
   const findInHalf = (length, smaller = bigint2, bigger = N) => {
      if (smaller > bigger){
         return [false];
      };
      if (smaller == bigger) {
         return [valueOf1s(smaller, length) == N, smaller]
      };
      let mid = (smaller + bigger) / bigint2;
      let val = valueOf1s(mid, length);
      if(val == N){
         return [true, mid];
      };
      if (val > N){
         return findInHalf(length, smaller, mid - bigint1);
      };
      return findInHalf(length, mid + bigint1, bigger);
   };
   for (let length = maxLen; length > 0; length--) {
      let [found, base] = findInHalf(length);
      if(found){
         return '' + base;
      }
   };
   return '' + (N - 1);
   function valueOf1s(base, lengthOf1s) {
      let t = bigint1
      for (let i = 1; i < lengthOf1s; i++) {
         t *= base
         t += bigint1
      }
      return t
   }
   function countLength(N, base) {
      let t = N, len = 0
      while (t > bigint0) {
         t /= base
         len++
      }
      return len
   }
};

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).