3315. Construct the Minimum Bitwise Array II

/**
 * Problem: LeetCode 3315 - Construct the Minimum Bitwise Array II
 * Difficulty: Medium
 * Language: Java
 * Author: CodeByTushu
 */
import java.util.List;

class Solution {
    public int[] minBitwiseArray(List<Integer> nums) {
        int n = nums.size();
        int[] ans = new int[n];

        for (int i = 0; i < n; i++) {
            int p = nums.get(i);
            // Even prime (2) cannot be formed by x | (x + 1) because x | (x + 1) is always odd
            if (p == 2) {
                ans[i] = -1;
            } else {
                // (p + 1) & -(p + 1) isolates the lowest set bit of (p + 1), which equals 2^k
                // where k is the number of trailing ones in p.
                // Shifting right by 1 gives the highest set bit in the trailing ones block of p.
                ans[i] = p - (((p + 1) & -(p + 1)) >> 1);
            }
        }

        return ans;
    }
}

"""
Problem: LeetCode 3315 - Construct the Minimum Bitwise Array II
Difficulty: Medium
Language: Python
Author: CodeByTushu
"""
from typing import List

class Solution:
    def minBitwiseArray(self, nums: List[int]) -> List[int]:
        ans = []
        for p in nums:
            # Even prime (2) cannot be formed by x | (x + 1) because x | (x + 1) is always odd
            if p == 2:
                ans.append(-1)
            else:
                # Isolate the lowest set bit of p + 1 and divide by 2 to get
                # the highest set bit in the contiguous trailing 1s of p.
                lowbit = (p + 1) & -(p + 1)
                ans.append(p - (lowbit >> 1))
        return ans

/**
 * Problem: LeetCode 3315 - Construct the Minimum Bitwise Array II
 * Difficulty: Medium
 * Language: C++
 * Author: CodeByTushu
 */
#include <vector>

using namespace std;

class Solution {
public:
    vector<int> minBitwiseArray(vector<int>& nums) {
        int n = nums.size();
        vector<int> ans(n);

        for (int i = 0; i < n; i++) {
            int p = nums[i];
            // Even prime (2) cannot be formed by x | (x + 1) because x | (x + 1) is always odd
            if (p == 2) {
                ans[i] = -1;
            } else {
                // Find lowest set bit of (p + 1) and shift right by 1
                ans[i] = p - (((p + 1) & -(p + 1)) >> 1);
            }
        }

        return ans;
    }
};

/**
 * Problem: LeetCode 3315 - Construct the Minimum Bitwise Array II
 * Difficulty: Medium
 * Language: JavaScript
 * Author: CodeByTushu
 */
/**
 * @param {number[]} nums
 * @return {number[]}
 */
var minBitwiseArray = function(nums) {
    const n = nums.length;
    const ans = new Array(n);

    for (let i = 0; i < n; i++) {
        const p = nums[i];
        // Even prime (2) cannot be formed by x | (x + 1) because x | (x + 1) is always odd
        if (p === 2) {
            ans[i] = -1;
        } else {
            // Find lowest set bit of (p + 1), then right shift by 1 to get highest bit in trailing 1s of p
            ans[i] = p - (((p + 1) & -(p + 1)) >> 1);
        }
    }

    return ans;
};