LeetCode 2749 - Minimum Operations to Make the Integer Zero

LeetCode Problem 2749

Difficulty: 🟡 Medium
Topics: Bit Manipulation, Brainteaser, Enumeration

Solution

Problem Understanding

The problem gives us two integers, num1 and num2. In one operation, we may choose any integer i between 0 and 60, inclusive, and subtract the value 2^i + num2 from num1.

The goal is to determine the minimum number of operations required to make num1 become exactly 0. If it is impossible, we return -1.

The important detail is that every operation has two components:

1. A fixed contribution from num2
2. A power of two, 2^i

Suppose we perform exactly k operations. Then the total amount subtracted is:

$$k \cdot num2 + \sum 2^{i_j}$$

We want this total subtraction to equal num1, so:

$$num1 = k \cdot num2 + \sum 2^{i_j}$$

Rearranging:

$$num1 - k \cdot num2 = \sum 2^{i_j}$$

This transformed equation is the key to solving the problem efficiently.

The right side is a sum of exactly k powers of two. Since powers of two correspond directly to binary representation, the number of set bits becomes extremely important.

The constraints are:

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

These constraints are large enough that brute force simulation over all operation sequences is impossible. We need a mathematical observation rather than explicit search.

There are several important edge cases:

• num2 may be negative, meaning operations can actually increase the value being subtracted in unusual ways.
• The intermediate value can become negative during the process.
• The required number of operations may be quite small even for large inputs.
• Some values are impossible because the binary structure cannot be represented with the chosen number of operations.

For example, if after k operations we compute:

$$target = num1 - k \cdot num2$$

then:

• target must be positive
• target must be representable as the sum of exactly k powers of two

This representation condition drives the entire solution.

Approaches

Brute Force Approach

A naive approach would attempt to simulate every possible sequence of operations.

At each operation, we may choose any i from 0 to 60, giving 61 choices per step. We could recursively try all possibilities and see whether we eventually reach zero.

This approach is correct because it explores every valid sequence of operations. If a solution exists, brute force will eventually discover it.

However, this is computationally infeasible. Even with only 10 operations, the number of possibilities becomes:

$$61^{10}$$

which is astronomically large.

Additionally, the value of num1 can fluctuate because num2 may be negative, making pruning difficult.

We therefore need a mathematical characterization of valid solutions.

Optimal Approach

The key insight is to stop thinking about the individual operations and instead think about the total effect after exactly k operations.

If we perform k operations:

$$num1 - k \cdot num2 = \sum 2^{i_j}$$

The right side is a sum of exactly k powers of two.

Now consider the binary representation of a number:

• The minimum number of powers of two needed to form a number equals its popcount, the number of set bits.
• We can always split powers of two further. For example:

$$8 = 4 + 4 = 2 + 2 + 4$$

So a number can be represented using exactly k powers of two if:

$$\text{popcount}(target) \le k \le target$$

where:

$$target = num1 - k \cdot num2$$

This transforms the problem into checking small candidate values of k.

Since num1 is at most 10^9, trying values of k up to around 60 is sufficient.

Approach Comparison

Approach	Time Complexity	Space Complexity	Notes
Brute Force	O(61^k)	O(k)	Explores every possible operation sequence
Optimal	O(60)	O(1)	Uses binary representation and enumeration

Algorithm Walkthrough

1. Iterate over possible numbers of operations k.

We try k from 1 upward because we want the minimum number of operations. 2. Compute the remaining target value.

For a fixed k:

$$target = num1 - k \cdot num2$$

This is the total amount that must be formed using powers of two. 3. Check whether target is positive.

If target <= 0, then it cannot be represented as a positive sum of powers of two in the required way. 4. Compute the number of set bits in target.

The popcount tells us the minimum number of powers of two required to form target. 5. Verify representability conditions.

We need:

$$\text{popcount}(target) \le k$$

because we need at least one power of two for each set bit.

We also need:

$$k \le target$$

because the smallest power of two is 1, so the minimum possible sum using k terms is k. 6. Return the first valid k.

Since we enumerate k in increasing order, the first valid value is the minimum answer. 7. If no valid k exists, return -1.

Why it works

The algorithm relies on a fundamental property of binary numbers.

Any positive integer can be expressed as a sum of powers of two. The minimum number of terms required equals the number of set bits in its binary representation.

We may also split powers of two into smaller powers repeatedly, allowing us to increase the number of terms while preserving the total sum.

Therefore, a number target can be written as the sum of exactly k powers of two if and only if:

$$\text{popcount}(target) \le k \le target$$

Since every sequence of operations corresponds exactly to such a representation, the algorithm checks all valid possibilities and returns the smallest feasible k.

Python Solution

class Solution:
    def makeTheIntegerZero(self, num1: int, num2: int) -> int:
        for operations in range(1, 61):
            target = num1 - operations * num2

            if target <= 0:
                continue

            set_bits = target.bit_count()

            if set_bits <= operations <= target:
                return operations

        return -1

The implementation directly follows the mathematical derivation.

We iterate through possible operation counts from 1 to 60. For each candidate count, we compute the remaining value target after subtracting the contribution from num2.

The condition target <= 0 is skipped immediately because a nonpositive value cannot be represented as a sum of positive powers of two.

The method bit_count() efficiently computes the number of set bits in the binary representation of target.

Finally, we check the two required inequalities:

• set_bits <= operations
• operations <= target

If both hold, then target can be decomposed into exactly operations powers of two, meaning the operations are feasible.

Because we iterate in increasing order, the first valid answer is minimal.

Go Solution

func makeTheIntegerZero(num1 int, num2 int) int {
	for operations := 1; operations <= 60; operations++ {
		target := int64(num1) - int64(operations)*int64(num2)

		if target <= 0 {
			continue
		}

		setBits := 0
		value := target

		for value > 0 {
			setBits += int(value & 1)
			value >>= 1
		}

		if setBits <= operations && int64(operations) <= target {
			return operations
		}
	}

	return -1
}

The Go implementation is almost identical to the Python version conceptually.

One important difference is integer handling. Since multiplication involving num2 may exceed 32 bit integer limits, we explicitly use int64 during arithmetic.

Go does not provide a built in bit_count() method for plain integers in the same way Python does, so we manually count set bits using bit operations.

The rest of the logic remains unchanged.

Worked Examples

Example 1

Input: num1 = 3, num2 = -2

We test increasing values of k.

k	target = num1 - k*num2	Binary	Popcount	Valid?
1	3 - 1*(-2) = 5	101	2	No
2	3 - 2*(-2) = 7	111	3	No
3	3 - 3*(-2) = 9	1001	2	Yes

For k = 3:

• popcount(9) = 2
• 2 <= 3
• 3 <= 9

So 9 can be decomposed into exactly 3 powers of two.

One valid decomposition:

$$9 = 4 + 4 + 1$$

Thus the answer is 3.

Example 2

Input: num1 = 5, num2 = 7

Try several values:

k	target = num1 - k*num2
1	-2
2	-9
3	-16

The target immediately becomes negative and only decreases further.

No valid representation exists, so the answer is -1.

Complexity Analysis

Measure	Complexity	Explanation
Time	O(60)	We test at most 60 candidate operation counts
Space	O(1)	Only a few integer variables are used

The algorithm is effectively constant time because the loop bound is fixed.

Each iteration performs only arithmetic operations and a popcount computation on a 64 bit integer, both of which are extremely efficient.

Test Cases

solution = Solution()

# Provided examples
assert solution.makeTheIntegerZero(3, -2) == 3  # example 1
assert solution.makeTheIntegerZero(5, 7) == -1  # example 2

# Simple direct case
assert solution.makeTheIntegerZero(1, 0) == 1  # subtract 1 once

# Already representable with one operation
assert solution.makeTheIntegerZero(8, 0) == 1  # subtract 2^3

# Multiple operations required
assert solution.makeTheIntegerZero(10, 1) == 2  # achievable in two operations

# Negative num2 helps
assert solution.makeTheIntegerZero(4, -1) == 2  # can expand target

# Large num1
assert solution.makeTheIntegerZero(10**9, 0) > 0  # large power decomposition

# Impossible due to fast negative growth
assert solution.makeTheIntegerZero(2, 100) == -1  # impossible immediately

# Exact popcount match
assert solution.makeTheIntegerZero(7, 0) == 3  # 111 requires three powers

# Requires splitting powers
assert solution.makeTheIntegerZero(15, 1) == 2  # decomposition possible

# Boundary style case
assert solution.makeTheIntegerZero(999999999, -1000000000) > 0  # large negative num2

Test Summary

Test	Why
(3, -2)	Validates provided example with negative num2
(5, 7)	Validates impossible scenario
(1, 0)	Smallest basic solvable input
(8, 0)	Single power of two
(10, 1)	Requires multiple operations
(4, -1)	Tests negative num2 behavior
(10^9, 0)	Large input stress test
(2, 100)	Quickly impossible due to large positive num2
(7, 0)	Popcount exactly equals operations
(15, 1)	Requires splitting powers of two
(999999999, -1000000000)	Large magnitude negative num2

Edge Cases

One important edge case occurs when target <= 0.

After choosing a number of operations k, we compute:

$$target = num1 - k \cdot num2$$

If this value becomes zero or negative, it cannot be represented as a positive sum of powers of two. A buggy implementation might accidentally accept zero because its popcount is zero. The solution explicitly skips all nonpositive targets.

Another critical edge case involves numbers whose popcount exceeds the number of operations.

For example, suppose:

$$target = 7$$

Its binary form is 111, which requires at least three powers of two. If we only allow two operations, representation is impossible. The implementation correctly checks:

$$\text{popcount}(target) \le k$$

before accepting a solution.

A third important edge case happens when num2 is negative.

Negative num2 increases the effective target value as k grows:

$$target = num1 - k \cdot num2$$

A naive implementation might assume the target always decreases. Instead, the algorithm handles both positive and negative num2 uniformly using the same mathematical conditions.

Finally, large integer arithmetic can become problematic in languages with fixed width integers. In Go, multiplying operations * num2 may overflow a 32 bit integer, so the implementation explicitly uses int64 for all intermediate computations.