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}