Introduction

In this blog post, we will explore how to solve the LeetCode problem "2571" 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 a positive integer n, you can do the following operation any number of times: Add or subtract a power of 2 from n. Return the minimum number of operations to make n equal to 0. A number x is power of 2 if x == 2i where i >= 0. Example 1: Input: n = 39 Output: 3 Explanation: We can do the following operations: - Add 20 = 1 to n, so now n = 40. - Subtract 23 = 8 from n, so now n = 32. - Subtract 25 = 32 from n, so now n = 0. It can be shown that 3 is the minimum number of operations we need to make n equal to 0. Example 2: Input: n = 54 Output: 3 Explanation: We can do the following operations: - Add 21 = 2 to n, so now n = 56. - Add 23 = 8 to n, so now n = 64. - Subtract 26 = 64 from n, so now n = 0. So the minimum number of operations is 3. Constraints: 1 <= n <= 105

Explanation

Here's the approach to solve this problem, followed by the code:

• Key Idea: The optimal strategy is to repeatedly bring n closer to the nearest power of 2 (either greater or smaller). This is because powers of 2 are the only operations allowed.
• Greedy Approach: At each step, determine which power of 2 is closest to the current value of n. Add or subtract that power to move closer to zero.

• Bit Manipulation: We can efficiently find the nearest powers of 2 using bit manipulation.

• Runtime Complexity: O(log n). Storage Complexity: O(1).

Code

    def min_operations(n: int) -> int:
    """
    Calculates the minimum number of operations to make n equal to 0 by adding or subtracting powers of 2.

    Args:
        n: A positive integer.

    Returns:
        The minimum number of operations.
    """
    operations = 0
    while n > 0:
        msb = n.bit_length()
        lower_power_of_2 = 1 << (msb - 1)
        upper_power_of_2 = 1 << msb

        if n == lower_power_of_2:
            n -= lower_power_of_2
        else:
            diff_lower = n - lower_power_of_2
            diff_upper = upper_power_of_2 - n

            if diff_lower <= diff_upper:
                n -= lower_power_of_2
            else:
                n += lower_power_of_2
        operations += 1
    return operations