3314. Construct the Minimum Bitwise Array I

Description

You are given an array nums consisting of n 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.

 

Solutions

Solution: Bit Manipulation

• Time complexity: O(n)
• Space complexity: O(n)

 

JavaScript

/**
 * @param {number[]} nums
 * @return {number[]}
 */
const minBitwiseArray = function (nums) {
  return nums.map(num => {
    let leadingOne = 1;

    while (num & leadingOne) {
      leadingOne <<= 1;
    }

    leadingOne >>= 1;

    if (!leadingOne) return -1;

    return num - leadingOne;
  });
};