Introduction

In this blog post, we will explore how to solve the LeetCode problem "3007" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

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

Explanation

Here's the solution to the problem:

• Binary Search: Perform a binary search on the range of possible numbers (1 to a sufficiently large upper bound, e.g., 2**60) to find the greatest cheap number.
• Calculate Accumulated Price: For each number considered during the binary search, calculate its accumulated price.

• Check Cheapness: Compare the accumulated price with k to determine if the number is cheap.

• Runtime Complexity: O(log(2**60) * N), where N is the highest number (close to 2**60). Since x is small, the calculation of the price for one number can be considered O(1). Thus, the complexity simplifies to O(log(2**60) * log(2**60)), as the complexity of accumulated price calculation will be O(log(2**60)) in the worst case due to repeated bitwise calculation. Storage Complexity: O(1)

Code

    def solve():
    k, x = map(int, input().split())

    def calculate_price(num, x):
        price = 0
        binary_representation = bin(num)[2:]
        for i in range(x - 1, len(binary_representation), x):
            if binary_representation[len(binary_representation) - 1 - i] == '1':
                price += 1
        return price

    def calculate_accumulated_price(num, x):
        accumulated_price = 0
        for i in range(1, num + 1):
            accumulated_price += calculate_price(i, x)
        return accumulated_price

    def is_cheap(num, k, x):
        return calculate_accumulated_price(num, x) <= k

    low = 1
    high = 2**60  # A sufficiently large upper bound
    ans = 0

    while low <= high:
        mid = (low + high) // 2
        if is_cheap(mid, k, x):
            ans = mid
            low = mid + 1
        else:
            high = mid - 1

    print(ans)

solve()