483. Smallest Good Base

For an integer n, we call k>=2 a good base of n, if all digits of n base k are 1.

Now given a string representing n, you should return the smallest good base of n in string format.

Example 1:

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

Example 2:

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

Example 3:

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

Note:

1. The range of n is [3, 10^18].
2. The string representing n is always valid and will not have leading zeros.

 

Approach #1:

class Solution {
public:
    string smallestGoodBase(string n) {
        unsigned long long tn = (unsigned long long)stoll(n);
        unsigned long long x = 1;
        for (int i = 62; i >= 1; --i) {
            if ((x<<i) < tn) {
                unsigned long long temp = solve(tn, i);
                if (temp != 0) return to_string(temp);
            }
        }
        return to_string(tn-1);
    }
private:
    unsigned long long solve(unsigned long long num, int d) {
        double tn = (double) num;
        unsigned long long r = (unsigned long long)(pow(tn, 1.0/d)+1);
        unsigned long long l = 1;
        while (l <= r) {
            unsigned long long sum = 1;
            unsigned long long cur = 1;
            unsigned long long m = l + (r - l) / 2;
            for (int i = 1; i <= d; ++i) {
                cur *= m;
                sum += cur;
            }
            if (sum == num) return m;
            if (sum < num) l = m + 1;
            else r = m - 1;
        }
        return 0;
    }
};

Runtime: 4 ms, faster than 49.59% of C++ online submissions for Smallest Good Base.

 Analysis:

The input can be stored in a long long int, here I use unsigned long long int for a larger range. We need to find k, for 1+k^1+k^2+k^3+...+k^d=n. The smallest possible base is k=2, with has the longest possible representation, i.e., largest d. So, to find the smallest base means to find the longest possible representation "11111....1" based on k. As n<=10^18, so n<(1<<62). We iterate the length of the representation from 62 to 2 (2 can always be valid, with base=n-1), and check whether a given length can be valid.

For a given length d, we use binary search to check whether there is a base k which satisfies 1+k^1+k^2+...k^d=n. The left limit is 1, and the right limit is pow(n,1/d)+1, i.e., the d th square root of n. The code is shown below.

come from: https://leetcode.com/problems/smallest-good-base/discuss/96590/3ms-AC-C%2B%2B-long-long-int-%2B-binary-search