Introduction

In this blog post, we will explore how to solve the LeetCode problem "2749" 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 two integers num1 and num2. In one operation, you can choose integer i in the range [0, 60] and subtract 2i + num2 from num1. Return the integer denoting the minimum number of operations needed to make num1 equal to 0. If it is impossible to make num1 equal to 0, return -1. Example 1: Input: num1 = 3, num2 = -2 Output: 3 Explanation: We can make 3 equal to 0 with the following operations: - We choose i = 2 and subtract 22 + (-2) from 3, 3 - (4 + (-2)) = 1. - We choose i = 2 and subtract 22 + (-2) from 1, 1 - (4 + (-2)) = -1. - We choose i = 0 and subtract 20 + (-2) from -1, (-1) - (1 + (-2)) = 0. It can be proven, that 3 is the minimum number of operations that we need to perform. Example 2: Input: num1 = 5, num2 = 7 Output: -1 Explanation: It can be proven, that it is impossible to make 5 equal to 0 with the given operation. Constraints: 1 <= num1 <= 109 -109 <= num2 <= 109

Explanation

Here's the solution approach, complexity analysis, and Python code:

• Core Idea: The problem can be reframed as finding the minimum number of terms of the form 2^i + num2 that sum to num1. Since we want to minimize the number of operations, we can iterate through possible number of operations (starting from 1) and for each number of operations, check if it is possible to achieve num1 using that many operations.
• Iterative Check: For each possible number of operations k, we try all combinations of i values such that num1 can be represented as sum of k terms 2^i + num2.

• Early Exit: If after a certain number of operations no solution has been found, and the possible values of terms 2^i + num2 will only increase, it can be concluded that no solution is possible, returning -1.

• Complexity: O(log(num1)^2) time complexity and O(1) space complexity.

Code

    def minOperations(num1: int, num2: int) -> int:
    """
    Calculates the minimum number of operations needed to make num1 equal to 0.

    Args:
        num1: The initial value of num1.
        num2: The value to be added to 2^i in each operation.

    Returns:
        The minimum number of operations needed, or -1 if impossible.
    """

    for k in range(1, 32):  # Iterate through possible number of operations
        for i in range(k + 1):
            # Check if num1 can be expressed as a sum of k terms of the form (2^j + num2)
            # where we have i terms with different j values and k - i is number of terms with j =0, when num2 can be negative
            remaining = num1 - (k * num2)
            if remaining < 0:
                continue

            if remaining == 0 and k > 0:
                if k == 1 and num2 == num1:
                    return 1
                if num2 == 0:
                    return -1 # num1 must be a sum of powers of 2.

                if num2 != 0:
                   if k == 1:
                        return -1 if (num1-num2) & (num1-num2 -1) !=0 else 1
                   elif k > 1:
                       return k


            # If remaining is positive, check if it can be represented as a sum of k powers of 2

            if remaining > 0:
                if bin(remaining).count('1') <= k:
                    return k

    return -1