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Java-based LeetCode algorithm problem solutions, regularly updated

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package g2401_2500.s2429_minimize_xor;

// #Medium #Greedy #Bit_Manipulation #2022_12_07_Time_0_ms_(100.00%)_Space_39.6_MB_(87.47%)

/**
 * 2429 - Minimize XOR\.
 *
 * Medium
 *
 * Given two positive integers `num1` and `num2`, find the positive integer `x` such that:
 *
 * *   `x` has the same number of set bits as `num2`, and
 * *   The value `x XOR num1` is **minimal**.
 *
 * Note that `XOR` is the bitwise XOR operation.
 *
 * Return _the integer_ `x`. The test cases are generated such that `x` is **uniquely determined**.
 *
 * The number of **set bits** of an integer is the number of `1`'s in its binary representation.
 *
 * **Example 1:**
 *
 * **Input:** num1 = 3, num2 = 5
 *
 * **Output:** 3
 *
 * **Explanation:** The binary representations of num1 and num2 are 0011 and 0101, respectively. The integer **3** has the same number of set bits as num2, and the value `3 XOR 3 = 0` is minimal.
 *
 * **Example 2:**
 *
 * **Input:** num1 = 1, num2 = 12
 *
 * **Output:** 3
 *
 * **Explanation:** The binary representations of num1 and num2 are 0001 and 1100, respectively. The integer **3** has the same number of set bits as num2, and the value `3 XOR 1 = 2` is minimal.
 *
 * **Constraints:**
 *
 * *   1 <= num1, num2 <= 10⁹
**/
public class Solution {
    public int minimizeXor(int num1, int num2) {
        int bits = Integer.bitCount(num2);
        int result = 0;
        for (int i = 30; i >= 0 && bits > 0; --i) {
            if ((1 << i & num1) != 0) {
                --bits;
                result = result ^ (1 << i);
            }
        }
        for (int i = 0; i <= 30 && bits > 0; ++i) {
            if ((1 << i & num1) == 0) {
                --bits;
                result = result ^ (1 << i);
            }
        }
        return result;
    }
}