Minimize XOR

class Solution {
public:
    int minimizeXOR(int num1, int num2) {
        int xorResult = 0;

        int countBitsInNum2 = __builtin_popcount(num2);
        int bitsToSet = 0;
        int currentPosition = 31;  // Checking from the MSB.

        // Set bits in results based on num1
        while (bitsToSet < countBitsInNum2) {
            // Ensure the current bit is set, or fill remaining required bits
            if (bitIsSet(num1, currentPosition) ||
                (countBitsInNum2 - bitsToSet > currentPosition)) {
                applyBit(xorResult, currentPosition);
                bitsToSet++;
            }
            currentPosition--;  // Proceeding to lower bit position.
        }

        return xorResult;
    }

private:
    // Check if specific bit is set in the integer x at "bit" position
    bool bitIsSet(int x, int bit) { return x & (1 << bit); }

    // Set a specific bit in integer x at "bit" position
    void applyBit(int &x, int bit) { x |= (1 << bit); }
};

In this solution, the task concentrates on minimizing the XOR value between two integers, num1 and num2, in C++. The implementation in the minimizeXOR function involves bit manipulation strategies. Below, you'll find how the solution approaches the problem:

• Calculate the number of bits that are set (1 values) in num2 using the built-in function __builtin_popcount.
• Initialize xorResult to store the result and bitsToSet to count how many bits to set during processing.
• Start from the most significant bit (position 31) and calculate downwards.
• For each bit position:
  • Check if that bit is set in num1 or if the number of remaining bits to set in xorResult is higher than the current position; this is done using conditions in the if statement involving two functions: bitIsSet, which checks whether a specific bit is set, and applyBit, which sets a specific bit in the resuling number.
  • The decision to set a bit is based on achieving a minimal XOR while maximizing similarity with the bit pattern of num1.
• The loop continues to check bit positions decrementally until the number of required bits (bitsToSet) matches those set (countBitsInNum2).

Two helper functions facilitate checking and setting bits:

• bitIsSet: Determines if a particular bit is 1 in an integer by masking and shifting.
• applyBit: Sets a bit at a specified position in xorResult to 1.

By managing which bits to select from num1 and maintaining the count, the function strives to minimize the resultant XOR value, effectively employing bit-wise manipulation for precision and optimal results in a bit-level context.

public class Solution {

    public int optimizeXor(int value1, int value2) {
        int calcResult = 0;

        int bitsToSet = Integer.bitCount(value2);
        int currentSetBits = 0;
        int processingBit = 31; // Start from the highest bit.

        // Loop while zeroes-to-ones transitions are less than target amount
        while (currentSetBits < bitsToSet) {
            if (
                isBitSet(value1, processingBit) ||
                (bitsToSet - currentSetBits > processingBit)
            ) {
                calcResult = toggleBitOn(calcResult, processingBit);
                currentSetBits++;
            }
            processingBit--; // Decrement to check the next lower bit.
        }

        return calcResult;
    }

    private boolean isBitSet(int num, int position) {
        return (num & (1 << position)) != 0;
    }

    private int toggleBitOn(int num, int position) {
        return num | (1 << position);
    }
}

This Java solution involves minimizing the XOR value between two integers, value1 and value2, by setting the binary bits of an initially zero integer (calcResult) in such a way that it matches the number of set bits in value2 while considering the bit positions in value1.

• The function optimizeXor initializes calcResult to zero which will hold the result value after bitwise operations.
• The number of bits that needs to be set (bitsToSet) is determined using Integer.bitCount(value2), which returns the count of one-bits in the binary representation of value2.
• A loop runs until the number of currently set bits in calcResult (tracked by currentSetBits) equals bitsToSet.
• In each iteration, the bit at the position processingBit (starting from 31, the highest bit in a traditional integer representation) is considered:
  • If the corresponding bit in value1 is set or if setting this bit would not exceed the total required bitsToSet, the bit at processingBlIT is set on calcResult using the helper function toggleBitOn.
  • After potentially setting the bit, currentSetBits is incremented and processingBit is decremented to move to the next lower bit.
• Helper functions isBitSet and toggleBitOn are used to check if a specific bit is set and to set a specific bit on an integer, respectively.

The solution uses bitwise operators efficiently to build the resultant integer with a focused bit manipulation approach, ensuring the total number of bits set is aligned with value2 while considering the state of bits in value1.

class Solution:
    def reduceXor(self, value1: int, value2: int) -> int:
        optimized_result = 0

        necessary_set_bits = bin(value2).count("1")
        current_set_bits = 0
        bit_pointer = 31  # Start from the highest bit.

        # Ensure optimized_result matches the set bits count of value2
        while current_set_bits < necessary_set_bits:
            # Set bit if it's set in value1 or need to set remaining bits in optimized_result
            if self.bit_is_set(value1, bit_pointer) or (
                necessary_set_bits - current_set_bits > bit_pointer
            ):
                optimized_result = self.set_specific_bit(optimized_result, bit_pointer)
                current_set_bits += 1
            bit_pointer -= 1  # Move to the lower significant bit.

        return optimized_result

    # Check if specific bit in number is set
    def bit_is_set(self, number: int, bit_position: int) -> bool:
        return (number & (1 << bit_position)) != 0

    # Set specific bit in number
    def set_specific_bit(self, number: int, bit_position: int):
        return number | (1 << bit_position)

The provided solution aims to minimize the XOR value by aligning the number of set bits in two given integers, value1 and value2. The core objective here is to modify value1 such that its binary representation has the same number of set bits as value2, and the positions of these set bits should strategically minimize the XOR result with value2.

Here's a breakdown of how the solution is implemented:

1. Define a class Solution with the method reduceXor that takes two integers, value1 and value2, and returns an integer.

2. Calculate the number of necessary set bits in value2 using the binary representation and counting the "1"s.

3. Initialize variables: optimized_result to store the result, current_set_bits to track the number of set bits added, and bit_pointer starting from the highest significant bit (31, assuming 32-bit integer).

4. Use a loop to set bits in optimized_result based on the bits set in value1 and the bits still required to match the count of set bits in value2. The loop runs until current_set_bits matches the number of set bits in value2.

5. For each bit position from highest to lowest, check conditions:

   • If the current bit in value1 is set OR if the difference in necessary set bits and current set bits needs to be filled up without exact matches in positions.

6. Use helper methods bit_is_set to check if a specific bit is set in the given number, and set_specific_bit to set a specific bit at a position in the number.

7. Return the optimized_result which should now have the same number of set bits as value2 and is expected to yield a minimized XOR value when computed against value2.

The thoughtful bit manipulation and calculation ensures that the resultant optimized_result has the optimized set bits configuration to align with value2 for XOR minimization. This approach effectively handles bit-level operations to tailor value1's representation for the desired outcome.