#3314

Easy

LeetCode Problem Workspace

Construct the Minimum Bitwise Array I

Learn how to construct an array minimizing each element while matching given primes using bitwise operations efficiently.

array bit manipulation

Problem Statement

Given an array nums containing n prime integers, construct an array ans of length n. Each ans[i] must satisfy ans[i] OR (ans[i] + 1) == nums[i], ensuring the bitwise relationship is correct for all elements.

Your goal is to minimize each ans[i] while respecting the bitwise OR constraint. The array length is between 1 and 100, and each nums[i] is a prime between 2 and 1000, so simple iteration over candidates is feasible.

Examples

Example 1

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

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

Example 2

Input: nums = [11,13,31]

Output: [9,12,15]

Constraints

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

Solution Approach

Iterate Over Candidates

For each nums[i], test all possible integers x starting from 0 until x OR (x + 1) equals nums[i]. Choose the first valid x, which guarantees the minimum value for ans[i].

Apply Bit Manipulation Directly

Use the property of OR to efficiently check each candidate. Since nums[i] is prime, the binary pattern allows early termination when the OR exceeds nums[i], reducing unnecessary checks.

Construct the Result Array

Once the minimal x is found for each index, append it to ans. After processing all indices, return ans as the final solution, which ensures each element is minimized while satisfying ans[i] OR (ans[i] + 1) == nums[i].

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

Time complexity is O(n * m) where n is the array length and m is the maximum value to check for each nums[i]; space complexity is O(n) for the result array.

What Interviewers Usually Probe

Check small constraints to hint at brute-force feasibility.
Focus on minimizing each element instead of any valid OR solution.
Expect you to leverage bit properties of primes for efficiency.

Common Pitfalls or Variants

Common pitfalls

Assuming any x that satisfies OR is minimal without testing smaller candidates.
Overlooking the +1 in ans[i] OR (ans[i] + 1), causing incorrect results.
Ignoring that nums[i] are prime, missing opportunities for early termination in iteration.

Follow-up variants

Construct the Maximum Bitwise Array using the same OR rule but maximizing values.
Use AND instead of OR to define a similar minimal array problem.
Solve with non-prime numbers, requiring careful checks for multiple valid x values.

FAQ

What is the main pattern in Construct the Minimum Bitwise Array I?

The main pattern combines array construction with bit manipulation, focusing on minimizing each ans[i] while satisfying ans[i] OR (ans[i] + 1) == nums[i].

Can we solve this problem without iterating over possible values?

Due to small constraints, direct iteration is simple and reliable; bit manipulation shortcuts may exist but are unnecessary for correctness.

Why must nums[i] be prime?

Primes simplify the binary pattern, ensuring a unique minimal ans[i] exists and allowing early termination in candidate checks.

What happens if multiple candidates satisfy the OR condition?

Always select the smallest x to ensure ans[i] is minimized, which is the core requirement of the problem.

How does GhostInterview prevent common mistakes here?

It highlights edge cases with the +1 in OR, ensures minimal values are chosen, and avoids skipping candidates that satisfy the bitwise condition.

terminal

Solution

Solution 1: Bit Manipulation

For an integer $a$, the result of $a \lor (a + 1)$ is always odd. Therefore, if $\text{nums[i]}$ is even, then $\text{ans}[i]$ does not exist, and we directly return $-1$. In this problem, $\textit{nums}[i]$ is a prime number, so to check if it is even, we only need to check if it equals $2$.

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class Solution:
    def minBitwiseArray(self, nums: List[int]) -> List[int]:
        ans = []
        for x in nums:
            if x == 2:
                ans.append(-1)
            else:
                for i in range(1, 32):
                    if x >> i & 1 ^ 1:
                        ans.append(x ^ 1 << (i - 1))
                        break
        return ans

Continue Practicing

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