2571. Minimum Operations to Reduce an Integer to 0

Problem Description

You are given a positive integer n. The task is to transform n to 0 by repeatedly performing the following operation: add or subtract a power of 2 to/from n. The available powers of 2 range from 2^0 up to any higher power (as 2^1, 2^2, and so on). The goal is to find the minimum number of such operations required to achieve n = 0.

In simpler words, you're essentially allowed to repeatedly increase or decrease the integer n by any value that is a power of 2 (like 1, 2, 4, 8, and so on), and you need to find the least number of times you have to do this to get to 0.

Intuition

The solution involves a strategic approach rather than trying out different powers of 2 at random. We look at the binary representation of the given integer n. Since binary representation is based on powers of 2, incrementing or decrementing by powers of 2 can be visualized as flipping specific bits in the binary form.

So, we scan through the binary digits (bits) of n from least significant to most significant (from the right-hand side to the left).

• If we encounter a 1 bit, this represents a power of 2 that we need to cancel out. We keep a running total of consecutive 1 bits because a string of 1s may be dealt with more efficiently together rather than individually.

• When we encounter a 0 bit, this provides an opportunity to address the consecutive 1 bits encountered before it. If there's only one 1 bit, we can perform a single subtract operation to remove it. If there are multiple 1s, we can turn all but one of them into 0 by adding a power of 2 that's just higher than the string of 1s. This adds 1 to the count of subtract operations needed.

In the end, we might be left with a string of 1s at the most significant end. We'll need two operations if there are multiple 1s (add and then subtract the next higher power of 2), or just one if there's a single 1.

The algorithm hence devised is cautious; it chooses the most effective operation at each step due to its bitwise considerations, ensuring fewer total operations.

Learn more about Greedy and Dynamic Programming patterns.

Solution Approach

The Python code provided implements the algorithm through a mix of bitwise operations and a greedy approach.

Let's walk through the implementation:

1. Initialize ans and cnt. The variable ans keeps track of the total number of operations required, and cnt counts the consecutive 1 bits in the binary representation of n.

2. Process the binary digits of n from least significant to most significant. This is done by examining n bit by bit in a loop where the iteration involves a right shift operation n >>= 1, effectively moving to the next bit towards the left.

3. When the current bit is 1 (checked by if n & 1), increment the cnt by 1. This is because each 1 represents a power of 2 that must be addressed.

4. If the current bit is 0 and cnt is not zero, this indicates we have just finished processing a sequence of 1 bits. Now, determine how to handle them:

   • If cnt equals 1, increment ans by 1, as we need one operation to subtract this power of 2.
   • If cnt is more than 1, increment ans by 1 and set cnt to 1. The rationale here is that we've added a power of 2 that turns all but one of the consecutive 1 bits into 0. We're left with one last subtraction operation to perform later.

5. After processing all bits, there may be a leftover cnt:

   • If cnt is 1, we only need one more operation, so add 1 to ans.
   • If cnt is greater than 1, we need a two-step operation where we first add and then subtract the next power of 2, hence add 2 to ans.

6. Finally, return the accumulated result ans, which gives us the minimum number of operations needed to transform n into 0.

By leveraging bitwise operations, the solution is both efficient and straightforward. There's no need for iteration or recursion that directly considers powers of 2; instead, the solution operates implicitly on powers of 2 by manipulating binary digits, which are intrinsically linked to powers of 2.

Example Walkthrough

Let's illustrate the solution approach using the example of n = 23. The binary representation of 23 is 10111. We will apply the solution approach step by step:

1. Initialize ans = 0 and cnt = 0. No operations have been done yet.
2. As n is 10111, we process from the rightmost bit to the left:
   • Bit 1 (rightmost): n & 1 is true so increment cnt to 1.
   • Bit 2: n & 1 is true so increment cnt to 2.
   • Bit 3: n & 1 is true so increment cnt to 3.
   • Bit 4: n & 1 is false, and cnt is 3. Therefore, increment ans to 1 and reset cnt to 1 (because we perform an add operation to turn the sequence 111 into 1000, which leaves one operation for later).
   • Last shift will happen, and now n becomes 2 (10 in binary), and we've processed all bits.
3. Leftover cnt is 1 because our binary representation now starts with 1. We increment ans by 1 because a final subtract operation is needed.
4. Finally, ans is 2 because we performed 2 operations in total: converting 10111 to 11000 (+1 operation) and then to 0 (-1 operation).

The minimum number of operations required to transform n = 23 into 0 using the solution approach is 2.

Solution Implementation

Time and Space Complexity

The time complexity of the provided code is O(log n) because the primary operation within the while loop is a bitwise right shift on n, which effectively divides n by 2. The number of iterations is directly proportional to the number of bits in the binary representation of n, which is log n base 2.

The space complexity of the code is O(1) since the extra space used is constant, irrespective of the size of n. The variables ans and cnt use a fixed amount of space during the execution of the algorithm.

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