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class Solution:
    def makeTheIntegerZero(self, num1: int, num2: int) -> int:
        for k in range(61):
            target = num1 - k * num2
            if target >= 0 and target.bit_count() <= k <= target:
                return k
        return -1

This solution uses an important observation: a number num1 can be decomposed into at most num1 operations, and the number of 1 bits in the binary representation of num1 is the minimum number of operations.

Here’s how this solution works:

  1. The code iterates through k from 0 to 60. k represents the number of operations being considered.

  2. In each iteration, it computes target = num1 - k * num2. This target is what num1 would be if k operations were performed on it. If the operation can be performed, target will be non-negative (since we cannot perform an operation that would make num1 negative).

  3. If target is non-negative and k is greater or equal than the number of 1s in the binary representation of target (target.bit_count()) and k is less than or equal to target, then k is a valid number of operations.

    • The check k <= target is based on the observation that num1 can be decomposed into at most num1 operations.

    • The check target.bit_count() <= k is based on the observation that the number of 1 bits in the binary representation of num1 is the minimum number of operations.

  4. If k is a valid number of operations, the function returns k as the minimum number of operations needed to make num1 equal to 0.

  5. If no valid k is found after checking all possibilities, the function returns -1, indicating that it is impossible to make num1 equal to 0 with the given operation.

Problem Classification

This problem belongs to the domain of Algorithm Design and Analysis, particularly Greedy Algorithms. It can also be considered under Mathematics given the numerical operations involved.

Components:

  • What: The primary components of the problem are the two given integers, num1 and num2.
  • What: The objective is to transform num1 to 0 through a specific operation involving num2 and a choice of i from [0,60].
  • What: The operation consists of subtracting 2^i + num2 from num1.
  • What: The goal is to determine the minimum number of operations required to achieve this transformation. If impossible, the problem asks for a return value of -1.

This problem could be categorized as a Transformation Problem where the task is to transform one state (in this case num1) to another (0) using a specific operation. It also has elements of Optimization as we are required to minimize the number of operations.

The problem could also be seen as a Decision Problem where we need to decide if a transformation is possible, and if so, how it can be achieved with the least number of steps.

The challenge and complexity in this problem come from the fact that the impact of the operation varies depending on the value of i chosen, and there is a large range of i to choose from. This necessitates a methodical approach to determining the optimal sequence of operations.

Constraints

Here are some useful characteristics or conditions we can exploit:

  1. Number Range: The range of num1 and num2 is quite large (from -10^9 to 10^9), which suggests that an efficient algorithm is necessary to avoid time complexity issues.

  2. Operations Limit: The value of i in the operation 2^i + num2 can range from 0 to 60. This limitation sets a ceiling on the range of the operation, which can be used in designing an efficient solution.

  3. Power of 2: The operation involves the use of 2^i. Since i is an integer, 2^i will always be a power of 2. We know that powers of 2 have specific properties in binary representation - specifically, in binary form, they will have only one bit set (i.e., be 1 at a single position and 0 at all others). This can be used to our advantage if we use bitwise operations or think in terms of binary numbers.

  4. Subtraction Operation: The operation involves subtracting 2^i + num2 from num1. This means that we should always aim to make 2^i + num2 as large as possible without exceeding num1 to reduce num1 to 0 in the minimum number of steps. This suggests a possible greedy approach.

  5. Negative num2: The num2 can be negative, which can increase the value to be subtracted. This can be taken into consideration while finding the optimal i value.

By understanding these aspects of the problem’s constraints, we can approach the solution in a more directed and efficient manner.

Thought Process

The key cues in the problem statement are:

  1. The operations involve subtraction of 2^i + num2 from num1 where i is in the range of [0, 60].
  2. We aim to minimize the number of operations needed to make num1 equal to 0.
  3. If it’s impossible to reach 0, we return -1.

From these points, it suggests that a greedy approach could be used to solve this problem. The idea is to always subtract the largest possible amount from num1 that doesn’t exceed num1 itself, in order to minimize the number of operations. We could start from the largest possible value of i (which is 60), and work our way down to i=0, subtracting 2^i + num2 from num1 at each step if it does not exceed num1.

One crucial insight is to consider the effects of num2. If num2 is negative, it can help to make the subtraction larger; but if num2 is positive, it will make the subtraction smaller.

This function uses a for loop to check all possible i from 60 to 0. For each i, it calculates the operation value (2 ** i) + num2 and checks if it can be subtracted from num1. If yes, it performs the operation and increments the operation count. If num1 becomes 0, it immediately returns the operation count. If the loop finishes without num1 reaching 0, it returns -1 if num1 is still positive, or the operation count if num1 has become negative.

Solution Approach and Analysis

Here’s a possible approach for solving this problem:

  1. Identify Approach: This problem seems to lend itself to a greedy approach, where we aim to subtract the largest possible amount from num1 at each step. Given the operation is 2^i + num2, we want to choose the largest i such that 2^i + num2 does not exceed num1.

  2. Implementation Details: We can implement the greedy approach by starting from i=60 (the maximum possible value for i) and work our way down to i=0. For each i, we check if 2^i + num2 is less than or equal to num1. If it is, we subtract 2^i + num2 from num1 and increment a counter to keep track of the number of operations.

  3. Termination and Edge Cases: The process continues until num1 is zero, or until we reach i=0 and num1 is not zero yet. If num1 is zero, the counter gives the minimum number of operations. If num1 is not zero even after i=0, it means it’s impossible to make num1 equal to 0, and we return -1.

Now let’s apply this approach to an example:

Example: num1 = 13, num2 = -2

Starting from i=60, we look for the largest i such that 2^i + num2 ≤ num1. We find that i=3 fits because 2^3 + (-2) = 6 is less than 13. We subtract 6 from 13 to get 7, and increment the counter.

Now, num1 = 7. We repeat the process, starting from i=3 this time. We find i=2 fits because 2^2 + (-2) = 2 is less than 7. We subtract 2 from 7 to get 5, and increment the counter.

Now, num1 = 5. We repeat the process, starting from i=2 this time. We find i=2 fits because 2^2 + (-2) = 2 is less than 5. We subtract 2 from 5 to get 3, and increment the counter.

Now, num1 = 3. We repeat the process, starting from i=2 this time. We find i=1 fits because 2^1 + (-2) = 0 is less than 3. We subtract 0 from 3 to get 3, and increment the counter.

Finally, num1 is still 3. We repeat the process, starting from i=1 this time. We find i=1 fits because 2^1 + (-2) = 0 is less than 3. We subtract 0 from 3 to get 3, and increment the counter.

We continue this process, but num1 never decreases below 3. Thus, it’s impossible to make num1 equal to 0 with the given operation, and we return -1.

Please note that this approach assumes that num2 is always subtracted. If num2 can be added or subtracted, then the problem becomes more complex as we would need to explore both possibilities at each step, leading to a combinatorial explosion.

Language Agnostic Coding Drills

  1. Dissecting the code into distinct concepts:

    • Iteration: The code makes use of a for loop to iterate from 0 to 60. This concept is about understanding how loops work in programming languages. It’s a fundamental skill and important to understand to solve this problem.

    • Arithmetic operations: The code performs arithmetic operations, such as subtraction and multiplication. This concept is about understanding how to use arithmetic operators in programming languages.

    • Conditional statements: The code makes use of if conditions to check for certain conditions. This concept is about understanding how to use conditional statements in programming languages.

    • Binary Representation and Bit Counting: The bit_count() method is used to get the count of 1 bits in the binary representation of a number. This concept is about understanding binary representation and how to count bits in a programming language.

  2. List of concepts in order of increasing difficulty:

    • Iteration: This is a basic concept and foundation to most programming languages.

    • Arithmetic operations: This is also a basic concept, although it might require a bit more understanding of how different operations work.

    • Conditional statements: This is slightly more advanced as it requires logical thinking and understanding of the conditions to be checked.

    • Binary Representation and Bit Counting: This is the most advanced concept in this problem, as it requires an understanding of binary numbers and bit manipulation, which can be complex for beginners.

  3. Problem-solving approach:

    • Understand the problem: The first step in the problem-solving approach is to understand the problem. The problem here is to reduce a number num1 to zero, by subtracting 2^i + num2 in each operation, and find the minimum number of such operations.

    • Iterate over possible operations: The code then iterates from 0 to 60, each iteration representing a potential operation.

    • Calculate potential target after operation: For each operation, it calculates what the number would be after the operation, using the arithmetic operation num1 - k * num2.

    • Check if operation is valid: The code then checks whether this operation would be valid. It does this by checking whether the target is non-negative, and whether k is both greater or equal to the bit count of the target and less than or equal to the target.

    • Return result: If such an operation is found, it is returned as the result. If no such operation is found after iterating over all possibilities, the function returns -1 to indicate that it is impossible to reduce num1 to zero with the given operation.

Targeted Drills in Python

Sure, let’s implement the four concepts identified previously:

  1. Iteration: Python code for a simple loop from 0 to 10. This is an essential concept as it allows us to repeat a piece of code several times.

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    for i in range(11):
    print(i)

  2. Arithmetic operations: Python code performing basic arithmetic operations. In the problem, these operations are used to calculate the potential target after each operation.

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    a = 10
b = 2
print("Addition:", a + b)
print("Subtraction:", a - b)
print("Multiplication:", a * b)

  3. Conditional statements: Python code demonstrating conditional statements. In our problem, these are used to check whether a potential operation is valid or not.

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    a = 10
if a > 5:
    print("a is greater than 5")
elif a == 5:
    print("a is equal to 5")
else:
    print("a is less than 5")

  4. Binary Representation and Bit Counting: Python code for calculating the number of bits in the binary representation of a number. This is used in the problem to check whether a potential operation is valid or not.

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    a = 10
print("Binary representation of a:", bin(a))
print("Number of bits in a:", bin(a).count("1"))

Now, how would these drills come together to form the solution for our problem?

  1. We would start with a loop (Iteration concept) from 0 to 60.
  2. For each iteration, we perform arithmetic operations (Arithmetic operations concept) to calculate the potential target after the operation.
  3. We then use a conditional statement (Conditional statements concept) to check whether the operation is valid.
  4. Part of this condition involves checking the number of bits in the binary representation of the potential target (Binary Representation and Bit Counting concept).
  5. If the operation is valid, we immediately return the number of operations (which is the current iteration of our loop). If no valid operation is found after the loop finishes, we return -1.

Q&A

Why is the answer bounded by 60?

The operation value subtracted from num1 in each step is (2**i) + num2, where i is in the range [0, 60]. The reason for this range is that the maximum possible value of num1 as per the problem constraints is 10^9.

If you consider the binary representation of num1, it can be seen that the binary length of num1 can go up to 30 bits (since 2^30 is roughly equal to 10^9).

In the worst case, where num2 is a large negative number (say -10^9), the value (2**i) + num2 can become very small or even negative. This could potentially require us to subtract numbers of the form 2^i twice, once for a positive value and once for a corresponding negative value, hence the range up to 60.

It’s important to note that this range is more than sufficient for the given problem constraints, and the range could potentially be smaller depending on the actual values of num1 and num2. This range essentially guarantees that we can always find an operation that can subtract from num1 and eventually make it zero, if it’s possible.

Is this not supposed to be a bfs with pruning type of question?

In a way, you could think of it like that. However, it’s not a traditional Breadth-First Search (BFS) problem because there isn’t a clear tree or graph structure that we’re traversing.

The “BFS” part comes in when you consider each possible operation (i from 0 to 60) as a “neighbor” that you could visit from the current state. And the “pruning” part comes in when you notice that if a certain operation leads to a negative num1 value or a previously visited num1 value, then you don’t need to explore that operation further.

But note that a direct BFS approach can be inefficient due to a large number of operations, especially when num2 is a large negative number. The problem can be solved more efficiently by directly considering the binary representation of num1, as it gives a direct mapping to the operations needed to make num1 zero. This is because each bit in the binary representation of num1 corresponds to an operation subtracting a power of 2, and we just need to account for the addition of num2 in the operation.

What’s the exit condition? or not put condition. I was stuck there. Tried multiple variations but neither did work

The exit condition of the loop in this solution is when num1 becomes zero. This is because the goal of the problem is to make num1 equal to zero. If num1 is already zero, the loop doesn’t need to start, and the function should return the count of operations, which is zero in this case.

However, in some situations, num1 may not be able to become zero (for example, when num2 is a positive number greater than num1), it might run indefinitely. To avoid this, the function should also keep track of whether num1 has become negative or not. If num1 becomes negative, the function should immediately return -1 because it is impossible to make num1 equal to zero in this case. This can serve as an additional exit condition.

Therefore, the exit conditions for this problem are when num1 becomes zero or when num1 becomes negative.

Cant this be done using recursion or…bfs precisely? The answer must come..because at most its gonna take 60 steps to reach answer.

Yes, theoretically, it could be solved using recursion or Breadth-First Search (BFS) due to the limit on the number of operations. However, it’s important to note that these methods would typically require significant resources in terms of time and space, and could be less efficient than the given solution.

Here’s how you might approach this problem with recursion or BFS:

  • Recursion: You would start with num1 and in each recursive call, you would subtract 2^i + num2 from num1 for all i in the range [0, 60]. You would then recursively call the function for the new value of num1. If num1 reaches 0, you would return the number of steps taken. If num1 becomes negative, you would stop the recursion and return -1. The minimum number of steps from all recursive calls would be the final answer.

  • BFS: You would use a queue to store each num1 value and its associated step count. Starting with the initial num1, you would dequeue each value, subtract 2^i + num2 for all i in the range [0, 60] from num1, and enqueue the new value with its step count. If num1 reaches 0, you would return the number of steps. If num1 becomes negative, you would continue with the next value in the queue. The first time num1 becomes 0, you have found the minimum number of steps, because BFS always finds the shortest path first.

However, the primary issue with these methods is that they can generate a very large number of recursive calls or queue entries due to the range of i, even though the answer is bounded by 60. This could potentially cause stack overflow in recursion or consume a lot of memory in BFS, and it would also increase the time complexity. The given solution avoids these problems by directly calculating the maximum i that can be used in each step, making it more efficient.

Is this problem related to nearest power of 2

Yes, the problem has a relationship with powers of two due to the operation 2^i which is being applied at each step. Specifically, we are interested in subtracting the largest power of two that we can in each step to minimize the number of steps needed.

So, we could say the problem is asking us to represent num1 as a sum of distinct values of the form 2^i + num2 for i in the range [0, 60], with the additional constraint that 2^i + num2 must not exceed num1.

This is similar to expressing a number in binary (which involves powers of two), but with the additional operation of adding num2 at each step. Therefore, finding the nearest (or largest) power of 2 that is less than or equal to num1 minus num2 at each step can indeed help us solve the problem more efficiently.