LeetCode 483: Smallest Good Base Solution
Master LeetCode problem 483 (Smallest Good Base), a hard challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.
483. Smallest Good Base
Problem Explanation
Explanation
To find the smallest good base for a given number n, we can iterate through all possible values of the base from 2 to n-1. For each base k, we can calculate the geometric progression sum using the formula (k^m - 1) / (k - 1), where m is the number of digits in the base k representation of n.
We can use binary search to find the optimal value for m, as the number of digits in the base k representation of n will be in the range [2, log2(n)]. By iterating through possible values of m, we can find the smallest good base for n.
Solution Code
class Solution {
public String smallestGoodBase(String n) {
long num = Long.parseLong(n);
for (int m = (int)(Math.log(num + 1) / Math.log(2)); m > 2; m--) {
long l = 2, r = (long)Math.pow(num, 1.0 / (m - 1)) + 1;
while (l < r) {
long k = l + (r - l) / 2;
long sum = 0;
for (int i = 0; i < m; i++) {
sum = sum * k + 1;
}
if (sum == num) {
return String.valueOf(k);
} else if (sum < num) {
l = k + 1;
} else {
r = k;
}
}
}
return String.valueOf(num - 1);
}
}Try It Yourself
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