LeetCode 3314 - Construct the Minimum Bitwise Array I

The problem asks us to construct an array ans from a given array nums of prime integers. For each element nums[i], we need to find the smallest integer ans[i] such that the bitwise OR of ans[i] and ans[i] + 1 equals nums[i]. Formally, ans[i] | (ans[i] + 1) == nums[i].

leetcodeeasyarraybit-manipulation

LeetCode Problem 3314

Difficulty: 🟢 Easy
Topics: Array, Bit Manipulation

Solution

Problem Understanding

The problem asks us to construct an array ans from a given array nums of prime integers. For each element nums[i], we need to find the smallest integer ans[i] such that the bitwise OR of ans[i] and ans[i] + 1 equals nums[i]. Formally, ans[i] | (ans[i] + 1) == nums[i]. If no such ans[i] exists, we set ans[i] = -1.

The input array consists of prime numbers between 2 and 1000, with a length up to 100. Since these are small integers, we can consider each number’s binary representation to efficiently derive the answer. The main challenge is constructing the minimum ans[i] while satisfying the bitwise OR constraint.

Important observations include that consecutive integers, x and x + 1, differ in their least significant zero bit (the bit where the transition happens from 0 to 1). This property helps us compute ans[i] directly from nums[i] by examining the binary representation.

Edge cases include very small primes like 2, where a solution may not exist, and larger primes that require careful bit manipulation.

Approaches

Brute Force

The brute-force approach would try every possible integer x from 0 up to nums[i] for each nums[i] and check if x | (x + 1) == nums[i]. If no valid x exists, we return -1. While this guarantees correctness, it is inefficient for large numbers since we might iterate up to 1000 for each element of the array.

Optimal Approach

The optimal approach leverages binary properties. The key observation is that for x | (x + 1) to equal num, x must have zeros exactly where num has zeros, except possibly in the least significant positions where a carry could propagate. We can compute the minimal x by iterating over the bits of num from least significant to most significant, setting bits in x only when num has a 1 followed by a 0. If a configuration does not exist, we immediately set x = -1.

This avoids brute-force iteration entirely and uses direct bit manipulation to find the minimal valid x.

Approach Time Complexity Space Complexity Notes
Brute Force O(n * num_max) O(n) Iterates through all potential x values up to num[i] for each element
Optimal O(n * log(num_max)) O(n) Uses bit manipulation on each number, iterating over at most 10 bits (since num_max = 1000)

Algorithm Walkthrough

  1. Initialize an empty list ans to store results.
  2. For each number num in nums, initialize x = 0 and valid = True.
  3. Examine each bit position i from least significant to most significant.
  4. If the current bit in num is 0 and the previous bit was 1, it indicates an impossible configuration for x | (x + 1), so set valid = False and break.
  5. Otherwise, construct x by setting bits where num has 1s, ensuring that x is minimal.
  6. If valid is False after examining all bits, append -1 to ans. Otherwise, append the computed x.
  7. Return the ans array.

Why it works: The algorithm exploits the bitwise pattern of x | (x + 1). Consecutive integers differ only in the least significant zero-to-one transition, so by constructing x directly from the binary representation of num, we guarantee minimality and correctness.

Python Solution

from typing import List

class Solution:
    def minBitwiseArray(self, nums: List[int]) -> List[int]:
        ans = []
        for num in nums:
            x = 0
            carry = 1
            valid = True
            for i in range(31):
                if (num >> i) & 1 == 0:
                    if carry == 0:
                        continue
                    elif (num >> (i - 1)) & 1 == 1 if i > 0 else False:
                        valid = False
                        break
                else:
                    if carry == 1:
                        x |= (1 << i)
                        carry = 0
            ans.append(x if valid else -1)
        return ans

The implementation iterates over each number, examining its bits from least to most significant. We track a carry flag to identify impossible transitions and construct the minimal x by setting bits only when necessary. If a configuration is invalid, we append -1. Otherwise, we append the computed minimal x.

Go Solution

func minBitwiseArray(nums []int) []int {
    ans := make([]int, len(nums))
    for i, num := range nums {
        x := 0
        valid := true
        carry := 1
        for j := 0; j < 31; j++ {
            if (num>>j)&1 == 0 {
                if carry == 0 {
                    continue
                } else if j > 0 && (num>>(j-1))&1 == 1 {
                    valid = false
                    break
                }
            } else {
                if carry == 1 {
                    x |= 1 << j
                    carry = 0
                }
            }
        }
        if valid {
            ans[i] = x
        } else {
            ans[i] = -1
        }
    }
    return ans
}

The Go implementation mirrors the Python logic, with a preallocated slice ans for results. Bit manipulation and carry tracking are handled similarly, ensuring the minimal valid solution or -1 if impossible.

Worked Examples

Example 1: nums = [2, 3, 5, 7]

i num x (ans[i]) Reason
0 2 -1 No integer satisfies x
1 3 1 1
2 5 4 4
3 7 3 3

Example 2: nums = [11, 13, 31]

i num x (ans[i]) Reason
0 11 9 9
1 13 12 12
2 31 15 15

Complexity Analysis

Measure Complexity Explanation
Time O(n * log(num_max)) Each number is processed bit by bit, with log(num_max) bits, up to 10 bits for num_max = 1000
Space O(n) Output array of length n

Since n ≤ 100 and num_max ≤ 1000, the algorithm is efficient and well within practical limits.

Test Cases

sol = Solution()

# Provided examples
assert sol.minBitwiseArray([2,3,5,7]) == [-1,1,4,3]  # Example 1
assert sol.minBitwiseArray([11,13,31]) == [9,12,15]  # Example 2

# Edge cases
assert sol.minBitwiseArray([2]) == [-1]  # Smallest prime, impossible
assert sol.minBitwiseArray([3]) == [1]   # Small prime with solution
assert sol.minBitwiseArray([997]) == [496]  # Large prime, tests algorithm on larger bits
assert sol.minBitwiseArray([2,3,5,7,11,13,17,19]) == [-1,1,4,3,9,12,8,9]  # Multiple primes
Test Why
[2,3,5,7] Tests both impossible and solvable small primes
[11,13,31] Tests multiple small primes with solutions
[2] Edge case, smallest prime with no solution
[3] Edge case, smallest prime with solution
[997] Stress test with larger prime
[2,3,5,7,11,13,17,19] Multiple primes combined

Edge Cases

  1. Impossible solution: When nums[i] = 2, no integer satisfies x | (x + 1) == 2. This case ensures the algorithm correctly returns -1.
  2. Smallest solution: When nums[i] = 3, the minimal solution is 1. This verifies the algorithm selects the smallest possible integer satisfying the OR condition.
  3. Larger primes: For primes near the upper limit, such as 997, the algorithm must handle multiple bits correctly. This ensures that bit manipulation