2749. Minimum Operations to Make the Integer Zero

This is a “Medium-Hard” problem that combines Mathematical Logic with Bit Manipulation. At first glance, it looks like a Breadth-First Search (BFS) or Dynamic Programming problem because it asks for the “minimum operations.” However, due to the large constraints, the only viable solution is a mathematical observation.

This problem evaluates your ability to recognize that an integer can be represented as a sum of k powers of 2 if and only if k falls within a specific range of bit counts.

Problem

You are given two integers, num1 and num2.

In one operation, you can choose any integer i and subtract (2^i + num2) from num1.

Your task: Return the minimum number of operations to make num1 equal to 0. If it is impossible, return -1.

Rules and constraints:

• 1 <= num1 <= 10^9
• -10^9 <= num2 <= 10^9

Solution

Let k be the number of operations. If we perform k operations, we subtract (2^i1 + num2), (2^i2 + num2), ..., (2^ik + num2) from num1.

Setting the result to zero gives the equation: num1 - (2^i1 + 2^i2 + ... + 2^ik + k * num2) = 0

Rearranging this, we get: num1 - k * num2 = 2^i1 + 2^i2 + ... + 2^ik

Let target = num1 - k * num2. We need to find the smallest k such that target can be represented as a sum of exactly k powers of 2.

Key Idea: The Range of Bit Representation A positive integer V can be represented as a sum of k powers of 2 if and only if:

1. Minimum bits: k is greater than or equal to the number of set bits (1s) in the binary form of V.
2. Maximum bits: k is less than or equal to V itself (since the smallest power of 2 is 2^0 = 1).

Algorithm Outline

1. Iterate through possible values of k starting from 1 up to 60.
2. For each k:
   • Calculate target = num1 - k * num2.
   • If target < k, it is impossible for this k (because even using the smallest power 2^0, the sum of k powers would be at least k).
   • Count the number of set bits (1s) in target.
   • Check if set_bits <= k <= target.
   • If this condition is met, return k.
3. If the loop completes without finding a k, return -1.

Sample Implementation (Python)

Complexity

• Time: \( O(log(num1)) \). We check a small number of possible k values (up to 60), and bit counting is very fast.
• Space: \( O(1) \). We only store a few integer variables.