Minimize XOR

• Calculate bits1 = Integer.bitCount(num1) and bits2 = Integer.bitCount(num2).
• Case 1: bits1 == bits2
  • The optimal x is num1 itself, as num1 XOR num1 = 0, which is the minimum possible XOR value. Return num1.
• Case 2: bits2 > bits1
  • To minimize the XOR value, x should be as similar to num1 as possible. Start with x = num1.
  • We need to set an additional diff = bits2 - bits1 bits in x.
  • To cause the smallest increase in x XOR num1, set these bits at the least significant positions where num1 has a 0.
  • Iterate from i = 0 to 30. If the i-th bit of num1 is 0, set the i-th bit in x and decrement diff until diff is 0.
• Case 3: bits2 < bits1
  • To minimize the XOR value, x must match the most significant set bits of num1.
  • Initialize x = 0.
  • We need to set bits2 bits in x.
  • Iterate from i = 30 down to 0. If the i-th bit of num1 is 1, set the i-th bit in x and decrement a counter for the bits we need to set. Stop when the counter reaches 0.
• Return the constructed x.

First, we calculate the number of set bits for num1 and num2, let's call them bits1 and bits2 respectively. The logic then branches into three distinct scenarios:

1. bits1 == bits2: To make x XOR num1 minimal (i.e., 0), x must be equal to num1. Since x would then have bits1 set bits, which is equal to the required bits2, this is the optimal solution. So, x = num1.

2. bits2 > bits1: We need x to have more set bits than num1. To keep x as close to num1 as possible, we start by setting x = num1. This ensures all the original set bits of num1 are matched, minimizing the most significant part of the XOR result. Then, we need to set bits2 - bits1 additional bits. To cause the smallest increase in the XOR value, we should set these bits at the least significant available positions. We iterate from the least significant bit (LSB) upwards, and for each position i where num1 has a 0, we set the i-th bit in x until we have set the required number of additional bits.

3. bits2 < bits1: We need x to have fewer set bits than num1. To minimize the XOR result, we must prioritize matching the most significant set bits of num1. We construct x from scratch (x=0). We iterate from the most significant bit (MSB) downwards. If the i-th bit is set in num1, we set the i-th bit in x and decrement our count of needed bits. We stop after we have set bits2 bits in x. This ensures x is composed of the bits2 most significant bits of num1.

1class Solution {2    public int minimizeXor(int num1, int num2) {3        int bits1 = Integer.bitCount(num1);4        int bits2 = Integer.bitCount(num2);5        int x = 0;6 7        if (bits1 == bits2) {8            return num1;9        } else if (bits2 > bits1) {10            x = num1;11            int diff = bits2 - bits1;12            for (int i = 0; i <= 30 && diff > 0; i++) {13                if ((num1 & (1 << i)) == 0) {14                    x |= (1 << i);15                    diff--;16                }17            }18        } else { // bits2 < bits119            int count = bits2;20            for (int i = 30; i >= 0 && count > 0; i--) {21                if ((num1 & (1 << i)) != 0) {22                    x |= (1 << i);23                    count--;24                }25            }26        }27        return x;28    }29}

• Calculate k = Integer.bitCount(num2), which is the target number of set bits for x.
• Initialize the result x = 0.
• Pass 1 (MSB to LSB): Iterate i from 30 down to 0.
  • If the i-th bit of num1 is 1 and we still need to set bits (k > 0):
    • Set the i-th bit of x (i.e., x |= (1 << i)).
    • Decrement k.
• Pass 2 (LSB to MSB): If k is still greater than 0 after the first pass:
  • Iterate i from 0 up to 30.
  • If the i-th bit of x is 0 and k > 0:
    • Set the i-th bit of x.
    • Decrement k.
• Return the final constructed x.

The core idea is to build x by making the best greedy choice for each bit, starting from the most significant. To minimize x XOR num1, we want the bits of x to match the bits of num1 as much as possible, especially at higher-value positions.

First, we determine the target number of set bits for x, which is k = Integer.bitCount(num2).

Pass 1 (MSB to LSB): We iterate from bit 30 down to 0. For each position i, if the i-th bit is set in num1 and we still need to set bits in x (i.e., k > 0), we set the i-th bit in x. This action makes the i-th bit of x XOR num1 zero, which is the best possible outcome for this high-value bit. We decrement k each time we set a bit. This pass ensures that x matches num1 on its most significant set bits, up to the number of bits we are allowed to set.

Pass 2 (LSB to MSB): After the first pass, if k is still greater than 0, it means num2 had more set bits than num1. We must set the remaining k bits in x. To minimize the resulting x XOR num1 value, we should set these bits in the lowest possible positions where x does not have a bit set yet. We iterate from bit 0 up to 30. For each position i, if the i-th bit of x is 0 and k > 0, we set the i-th bit in x and decrement k.

This two-pass process correctly constructs the unique integer x that satisfies both conditions, regardless of whether num1 has more, fewer, or the same number of set bits as the target.

1class Solution {2    public int minimizeXor(int num1, int num2) {3        int setBitsToPlace = Integer.bitCount(num2);4        int x = 0;5 6        // Pass 1: Prioritize matching the most significant set bits of num1.7        // This minimizes the XOR value by making the most significant bits of the result 0.8        for (int i = 30; i >= 0 && setBitsToPlace > 0; i--) {9            if ((num1 & (1 << i)) != 0) {10                x |= (1 << i);11                setBitsToPlace--;12            }13        }14 15        // Pass 2: If more bits need to be set, fill them in from the least significant16        // available positions. This adds the smallest possible value to the XOR result.17        for (int i = 0; i <= 30 && setBitsToPlace > 0; i++) {18            // Check if the bit is not already set in x19            if ((x & (1 << i)) == 0) {20                x |= (1 << i);21                setBitsToPlace--;22            }23        }24 25        return x;26    }27}