- Difficulty: Medium
- Tags: LeetCode, Medium, Bit Manipulation, Binary Search, Dynamic Programming, leetcode-3007, Math, O(max(logk, x) * log((logk) / x)), O((logk) / x), Combinatorics
Problem
You are given an integer k
and an integer x
. The price of a number num
is calculated by the count of set bits at positions x
, 2x
, 3x
, etc., in its binary representation, starting from the least significant bit. The following table contains examples of how price is calculated.
x | num | Binary Representation | Price |
---|---|---|---|
1 | 13 | 000001101 | 3 |
2 | 13 | 000001101 | 1 |
2 | 233 | 011101001 | 3 |
3 | 13 | 000001101 | 1 |
3 | 362 | 101101010 | 2 |
The accumulated price of num
is the total price of numbers from 1
to num
. num
is considered cheap if its accumulated price is less than or equal to k
.
Return the greatest cheap number.
Example 1:
Input: k = 9, x = 1
Output: 6
Explanation:
As shown in the table below, 6
is the greatest cheap number.
x | num | Binary Representation | Price | Accumulated Price |
---|---|---|---|---|
1 | 1 | 001 | 1 | 1 |
1 | 2 | 010 | 1 | 2 |
1 | 3 | 011 | 2 | 4 |
1 | 4 | 100 | 1 | 5 |
1 | 5 | 101 | 2 | 7 |
1 | 6 | 110 | 2 | 9 |
1 | 7 | 111 | 3 | 12 |
Example 2:
Input: k = 7, x = 2
Output: 9
Explanation:
As shown in the table below, 9
is the greatest cheap number.
x | num | Binary Representation | Price | Accumulated Price |
---|---|---|---|---|
2 | 1 | 0001 | 0 | 0 |
2 | 2 | 0010 | 1 | 1 |
2 | 3 | 0011 | 1 | 2 |
2 | 4 | 0100 | 0 | 2 |
2 | 5 | 0101 | 0 | 2 |
2 | 6 | 0110 | 1 | 3 |
2 | 7 | 0111 | 1 | 4 |
2 | 8 | 1000 | 1 | 5 |
2 | 9 | 1001 | 1 | 6 |
2 | 10 | 1010 | 2 | 8 |
Constraints:
1 <= k <= 1015
1 <= x <= 8