LeetCode-in-Java

3314. Construct the Minimum Bitwise Array I

Easy

You are given an array nums consisting of n prime integers.

You need to construct an array ans of length n, such that, for each index i, the bitwise OR of ans[i] and ans[i] + 1 is equal to nums[i], i.e. ans[i] OR (ans[i] + 1) == nums[i].

Additionally, you must minimize each value of ans[i] in the resulting array.

If it is not possible to find such a value for ans[i] that satisfies the condition, then set ans[i] = -1.

Example 1:

Input: nums = [2,3,5,7]

Output: [-1,1,4,3]

Explanation:

For i = 0, as there is no value for ans[0] that satisfies ans[0] OR (ans[0] + 1) = 2, so ans[0] = -1.
For i = 1, the smallest ans[1] that satisfies ans[1] OR (ans[1] + 1) = 3 is 1, because 1 OR (1 + 1) = 3.
For i = 2, the smallest ans[2] that satisfies ans[2] OR (ans[2] + 1) = 5 is 4, because 4 OR (4 + 1) = 5.
For i = 3, the smallest ans[3] that satisfies ans[3] OR (ans[3] + 1) = 7 is 3, because 3 OR (3 + 1) = 7.

Example 2:

Input: nums = [11,13,31]

Output: [9,12,15]

Explanation:

For i = 0, the smallest ans[0] that satisfies ans[0] OR (ans[0] + 1) = 11 is 9, because 9 OR (9 + 1) = 11.
For i = 1, the smallest ans[1] that satisfies ans[1] OR (ans[1] + 1) = 13 is 12, because 12 OR (12 + 1) = 13.
For i = 2, the smallest ans[2] that satisfies ans[2] OR (ans[2] + 1) = 31 is 15, because 15 OR (15 + 1) = 31.

Constraints:

1 <= nums.length <= 100
2 <= nums[i] <= 1000
nums[i] is a prime number.

Solution

import java.util.List;

public class Solution {
    public int[] minBitwiseArray(List<Integer> nums) {
        int l = nums.size();
        int[] r = new int[l];
        for (int i = 0; i < l; i++) {
            r[i] = check(nums.get(i));
        }
        return r;
    }

    private int check(int v) {
        if (v % 2 == 0) {
            return -1;
        }
        for (int j = 1; j < v; j++) {
            if ((j | (j + 1)) == v) {
                return j;
            }
        }
        return -1;
    }
}

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