Problem Description

You start with two integers num1 and num2.

You can perform an operation where you:

1. Choose any integer i in the range [0, 60]
2. Subtract 2^i + num2 from num1

The goal is to make num1 equal to 0 using the minimum number of such operations.

For example, if you choose i = 3, you would subtract 2^3 + num2 = 8 + num2 from num1.

You need to return the minimum number of operations required to reduce num1 to exactly 0. If it's impossible to achieve this, return -1.

The key insight is that after performing k operations with different values of i, you're essentially subtracting k * num2 plus some combination of powers of 2. The problem becomes determining whether num1 - k * num2 can be expressed as a sum of exactly k powers of 2 (where powers can be repeated).

For a solution to exist with k operations:

• The value x = num1 - k * num2 must be non-negative (since powers of 2 are positive)
• The number of 1s in the binary representation of x must be at most k (this is the minimum number of distinct powers of 2 needed)
• The value x itself must be at least k (since each power of 2 is at least 2^0 = 1)

Intuition

Let's think about what happens when we perform these operations. Each operation subtracts 2^i + num2 from num1. If we perform k operations total, we're subtracting:

• num2 exactly k times (once per operation)
• Various powers of 2 (one per operation)

So after k operations, we have: num1 - k * num2 - (sum of k powers of 2) = 0

Rearranging: num1 - k * num2 = sum of k powers of 2

Let's call x = num1 - k * num2. The question becomes: can we express x as the sum of exactly k powers of 2?

Here's the key insight: any positive integer can be represented as a sum of powers of 2 (that's just its binary representation). For example, 13 = 8 + 4 + 1 = 2^3 + 2^2 + 2^0, which uses 3 powers of 2.

But we need exactly k powers of 2, and we can use the same power multiple times. For instance, 6 can be written as:

• 4 + 2 (2 powers)
• 2 + 2 + 2 (3 powers)
• 2 + 1 + 1 + 1 + 1 (5 powers)
• 1 + 1 + 1 + 1 + 1 + 1 (6 powers)

The minimum number of powers needed is the number of 1s in the binary representation (bit count). The maximum is the value itself (using all 2^0 = 1s).

Therefore, x can be expressed as the sum of exactly k powers of 2 if and only if:

• x ≥ 0 (can't make negative numbers with positive powers)
• bit_count(x) ≤ k ≤ x (between minimum and maximum possible)

We try each value of k starting from 1, and return the first k that works. If x becomes negative, we know larger values of k won't work either (since we're subtracting more), so we can stop.

Solution Approach

The solution uses a simple enumeration strategy to find the minimum number of operations.

We iterate through possible values of k (number of operations) starting from 1:

1. Calculate the target value: For each k, compute x = num1 - k * num2. This represents the sum of powers of 2 we need to achieve.

2. Check feasibility conditions:

   • If x < 0, we break immediately. Since powers of 2 are positive, we can't form a negative sum. Moreover, as k increases, x will only become more negative, so there's no point checking larger values of k.

   • If x ≥ 0, we check if x can be expressed as exactly k powers of 2:

     • The minimum number of powers needed is x.bit_count() (the number of 1s in binary representation)
     • The maximum number of powers possible is x itself (using all 1s)
     • Therefore, we need x.bit_count() ≤ k ≤ x

3. Return the result:

   • If we find a valid k, return it immediately as it's the minimum (since we're checking in ascending order)
   • If the loop breaks without finding a solution, return -1

The implementation uses Python's count(1) from itertools to generate an infinite sequence starting from 1, and bit_count() method to efficiently count the number of 1s in the binary representation.

Example walkthrough: If num1 = 5 and num2 = -2:

• k = 1: x = 5 - 1*(-2) = 7. Binary of 7 is 111, so bit_count = 3. Check: 3 ≤ 1 ≤ 7? No (3 > 1).
• k = 2: x = 5 - 2*(-2) = 9. Binary of 9 is 1001, so bit_count = 2. Check: 2 ≤ 2 ≤ 9? Yes! Return 2.

The algorithm efficiently finds the minimum k with a time complexity that depends on the answer size, typically very small in practice.

Example Walkthrough

Let's walk through the solution with num1 = 3 and num2 = -4.

We need to find the minimum number of operations k such that num1 - k * num2 can be expressed as a sum of exactly k powers of 2.

k = 1:

• Calculate x = num1 - k * num2 = 3 - 1*(-4) = 3 + 4 = 7
• Binary representation of 7: 111 (which is 4 + 2 + 1)
• Number of 1s in binary (bit_count): 3
• Check if bit_count(7) ≤ k ≤ x: Is 3 ≤ 1 ≤ 7?
• No, because 3 > 1. We need at least 3 powers of 2, but k only allows 1.

k = 2:

• Calculate x = num1 - k * num2 = 3 - 2*(-4) = 3 + 8 = 11
• Binary representation of 11: 1011 (which is 8 + 2 + 1)
• Number of 1s in binary (bit_count): 3
• Check if bit_count(11) ≤ k ≤ x: Is 3 ≤ 2 ≤ 11?
• No, because 3 > 2. We need at least 3 powers of 2, but k only allows 2.

k = 3:

• Calculate x = num1 - k * num2 = 3 - 3*(-4) = 3 + 12 = 15
• Binary representation of 15: 1111 (which is 8 + 4 + 2 + 1)
• Number of 1s in binary (bit_count): 4
• Check if bit_count(15) ≤ k ≤ x: Is 4 ≤ 3 ≤ 15?
• No, because 4 > 3. We need at least 4 powers of 2, but k only allows 3.

k = 4:

• Calculate x = num1 - k * num2 = 3 - 4*(-4) = 3 + 16 = 19
• Binary representation of 19: 10011 (which is 16 + 2 + 1)
• Number of 1s in binary (bit_count): 3
• Check if bit_count(19) ≤ k ≤ x: Is 3 ≤ 4 ≤ 19?
• Yes! All conditions are satisfied.
• We can express 19 as a sum of 4 powers of 2, for example: 16 + 1 + 1 + 1 or 8 + 8 + 2 + 1

Answer: 4 operations

This means we perform 4 operations, each subtracting 2^i + num2 from num1. The specific choices of i values would sum to 19 (like choosing powers that give us 16 + 1 + 1 + 1), and combined with subtracting 4 * num2 = -16, we get 3 - (-16) - 19 = 3 + 16 - 19 = 0.

Solution Implementation

Time and Space Complexity

The time complexity is O(log num1) where num1 is the input parameter. The loop iterates at most O(log num1) times because:

• Each iteration checks if x = num1 - k * num2 satisfies certain conditions
• The loop breaks when x < 0, which happens when k > num1/num2 (assuming num2 > 0)
• The valid range for k is bounded by the constraint x.bit_count() <= k <= x
• Since x = num1 - k * num2 and we need k <= x, we get k <= num1 - k * num2, which gives us k <= num1/(1 + num2)
• The maximum value of k that needs to be checked is approximately O(log num1) because the bit count of x is at most O(log x), and we need x.bit_count() <= k

The space complexity is O(1) as the algorithm only uses a constant amount of extra space for variables k and x, regardless of the input size.

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Not Understanding the Upper Bound Condition (k <= target)

A common mistake is forgetting or misunderstanding why we need k <= target. Students often only check target.bit_count() <= k without realizing that we also need an upper bound.

Why this happens: The condition k <= target ensures we can actually form target using exactly k powers of 2. Even if we use the smallest power (2^0 = 1) repeatedly, we need at least target number of 1s to sum up to target.

Example of the pitfall:

# Incorrect implementation - missing upper bound check
if target.bit_count() <= k:  # Wrong! Missing k <= target
    return k

Solution: Always include both bounds in the feasibility check:

if target.bit_count() <= k <= target:
    return k

2. Continuing the Loop Too Long

Another pitfall is not recognizing when to stop the iteration early. Some implementations might use an arbitrary upper limit like 60 or 100 iterations.

Why this happens: The problem mentions choosing i in range [0, 60], which might mislead you into thinking the loop should run up to 60 times.

Example of the pitfall:

# Inefficient - using arbitrary upper bound
for k in range(1, 61):  # Arbitrary limit
    target = num1 - k * num2
    if target < 0:
        continue  # Wrong! Should break here
    if target.bit_count() <= k <= target:
        return k
return -1

Solution: Break immediately when target becomes negative, as it will only get more negative with increasing k:

if target < 0:
    break  # Correct - no point checking further

3. Integer Overflow Concerns (Language-Specific)

In languages like C++ or Java, calculating num1 - k * num2 might cause integer overflow, especially when num2 is negative and k is large.

Example scenario: If num2 = -10^9 and we're checking large values of k, the multiplication k * num2 could overflow.

Solution for other languages:

• Use long long in C++ or long in Java
• Add overflow checks or use safe arithmetic operations
• Consider breaking early if target exceeds reasonable bounds (e.g., if target > 60 since we can't use more than 60 different powers of 2 given the constraint on i)

// C++ example with overflow consideration
long long target = (long long)num1 - (long long)k * num2;
if (target < 0 || target > 60) break;  // Early termination