Problem Description

The problem asks us to find a positive integer x that adheres to two conditions. First, x must have the same number of set bits (bits which are 1) as in another given positive integer num2. Second, when x is XORed with a given positive integer num1, the result should be as small as possible. This problem guarantees a unique solution for the given integers.

To give an example, if we have num1 = 2 (binary 10) and num2 = 3 (binary 11), an x with the same number of set bits as num2 would be 3 (binary 11), and x XOR num1 would result in a minimal value, which is 1.

Set bits are important in computing because they often represent boolean flags or other binary indicators within a number's binary representation.

Intuition

To solve this problem, we need to balance between matching the number of set bits in x and num2, and minimizing the XOR with num1. Creatively manipulating the bits is essential here.

Beginning with the count of set bits (bit_count) in both num1 and num2 provides us with the targets we need to hit. If num1 has more set bits than num2, we need to turn off some bits in num1. We do this by performing the operation num1 &= num1 - 1 which turns off the rightmost set bit in num1.

Conversely, if num1 has fewer set bits than num2, we need to turn on some bits in num1. The most efficient way to do this is to turn on the rightmost unset bit, which can be done using num1 |= num1 + 1.

Use these operations to adjust the number of set bits in num1 to match num2. This process relies on the bitwise nature of numbers and the properties of XOR to ensure that the value remains minimal. No sorting or array transformation is necessary, making this algorithm efficient and elegant.

We are certain that this solution is unique because of the constraints set by the problem and the deterministic nature of the operations used to adjust the set bits. By focusing on changing the rightmost bits first, we affect the value of num1 by the smallest possible amounts, guaranteeing the minimal possible result for x XOR num1.

Learn more about Greedy patterns.

Solution Approach

The implementation of the solution to the given problem follows a simple yet elegant approach, which uses standard bitwise operations to directly manipulate the bits of the input numbers, thereby maintaining a very efficient time complexity.

Here are the steps involved in the implementation using Python:

1. Counting Set Bits: We begin by counting the number of set bits (1s) in the binary representation of num1 and num2. Python's bit_count() function gives us this value directly. This is critical because we need to make the number of set bits in x match num2.

   cnt1 = num1.bit_count()
   cnt2 = num2.bit_count()

2. Reducing Excess Set Bits: If num1 has more set bits than num2 (cnt1 > cnt2), it means we need to reduce the number of set bits in num1. We do this by turning off set bits from the least significant end. The operation num1 &= num1 - 1 will unset the rightmost set bit in num1 each time it is executed. We continue this in a loop until the number of set bits in num1 (cnt1) is the same as in num2 (cnt2).

   while cnt1 > cnt2:
       num1 &= num1 - 1
       cnt1 -= 1

3. Increasing Insufficient Set Bits: If num1 has fewer set bits than num2 (cnt1 < cnt2), we need to increase the number of set bits in num1. We target the least significant bits to have minimal impact on the resulting value. The operation num1 |= num1 + 1 effectively turns on the first off bit (from the right), increasing the count by one. We repeat this until the count matches that of num2.

   while cnt1 < cnt2:
       num1 |= num1 + 1
       cnt1 += 1

4. Returning the Result: After ensuring that num1 has the same number of set bits as num2, and thus transforming num1 into x, we return x which will satisfy the condition that x XOR num1 is minimized.

   return num1

No additional data structures are needed. The pattern lies in bitwise manipulation, specifically toggling bits based on their state and position, all the while prioritizing changes to bits of lower significance to achieve the minimal XOR value. By repeatedly applying basic bitwise operations, we incrementally mold num1 to into the desired x that meets all the problem's conditions.

Example Walkthrough

Let's illustrate the solution approach with a small example. Suppose we have num1 = 8 (binary 1000) and num2 = 5 (binary 101). We want to find x, a positive integer that matches the number of set bits in num2 and minimizes x XOR num1.

Step 1: Counting Set Bits

• For num1 (1000), the number of set bits (cnt1) is 1.
• For num2 (101), the number of set bits (cnt2) is 2.

Step 2 & 3: Adjusting Set Bits

• cnt1 < cnt2 (1 < 2), so we need to increase the number of set bits in num1.
• We perform the operation num1 |= num1 + 1. The binary representation of num1 + 1 is 1001.
• After using the |= operation, num1 turns into 1001 (which is 9 in decimal), and now cnt1 = 2, matching cnt2.

Step 4: Returning the Result

• No further actions needed because cnt1 now equals cnt2.
• The x we've found is 9, which maintains the set bit requirement.

The result is minimal because we've only turned on the least significant bit that was originally off in num1, which has the smallest impact on the value when XORed with num1.

Finally, x XOR num1 equals 9 XOR 8 which is 1 in binary (0001), and indeed this is the smallest possible result for x XOR num1. Thus, x = 9 is the solution.

Solution Implementation

Time and Space Complexity

Time Complexity

The given code has two main operations that affect the time complexity:

1. Reducing the number of 1 bits in num1 to match the number of 1 bits in num2 when num1 has more 1 bits than num2. This operation involves a loop that repeats as long as cnt1 is greater than cnt2. Inside the loop, it performs num1 &= num1 - 1, which removes a 1 bit from num1 each time. Since this operation depends on the number of 1 bits to be removed, its time complexity is O(K), where K is the difference in the number of 1 bits between num1 and num2.

2. Increasing the number of 1 bits in num1 to match the number of 1 bits in num2 when num1 has fewer 1 bits than num2. This operation involves another loop that repeats while cnt1 is less than cnt2, performing num1 |= num1 + 1 to add 1 bits to num1. Since num1 + 1 includes at least one 1 bit at the lowest-order position (binary form), there's a guarantee that a new 1 bit is added to num1 each time this operation completes. Thus, its time complexity is also O(K).

Hence, the overall time complexity of the code is O(K), where K is the absolute difference in the number of 1 bits between num1 and num2.

Space Complexity

The space complexity of the given code is O(1). Apart from the input numbers num1 and num2, the code only uses a fixed number of integer variables (cnt1 and cnt2), and no additional structures or recursive calls that consume memory proportional to the input size.

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