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3314. Construct the Minimum Bitwise Array I

Description

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.

Solutions

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 \(2\).

If \(\text{nums[i]}\) is odd, suppose \(\text{nums[i]} = \text{0b1101101}\). Since \(a \lor (a + 1) = \text{nums[i]}\), this is equivalent to changing the last \(0\) bit of \(a\) to \(1\). To solve for \(a\), we need to change the bit after the last \(0\) in \(\text{nums[i]}\) to \(0\). We start traversing from the least significant bit (index \(1\)) and find the first \(0\) bit. If it is at position \(i\), we change the \((i - 1)\)-th bit of \(\text{nums[i]}\) to \(1\), i.e., \(\text{ans}[i] = \text{nums[i]} \oplus 2^{i - 1}\).

By traversing all elements in \(\text{nums}\), we can obtain the answer.

The time complexity is \(O(n \times \log M)\), where \(n\) and \(M\) are the length of the array \(\text{nums}\) and the maximum value in the array, respectively. Ignoring the space consumption of the answer array, the space complexity is \(O(1)\).

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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

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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 x = nums.get(i);
            if (x == 2) {
                ans[i] = -1;
            } else {
                for (int j = 1; j < 32; ++j) {
                    if ((x >> j & 1) == 0) {
                        ans[i] = x ^ 1 << (j - 1);
                        break;
                    }
                }
            }
        }
        return ans;
    }
}

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class Solution {
public:
    vector<int> minBitwiseArray(vector<int>& nums) {
        vector<int> ans;
        for (int x : nums) {
            if (x == 2) {
                ans.push_back(-1);
            } else {
                for (int i = 1; i < 32; ++i) {
                    if (x >> i & 1 ^ 1) {
                        ans.push_back(x ^ 1 << (i - 1));
                        break;
                    }
                }
            }
        }
        return ans;
    }
};

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func minBitwiseArray(nums []int) (ans []int) {
    for _, x := range nums {
        if x == 2 {
            ans = append(ans, -1)
        } else {
            for i := 1; i < 32; i++ {
                if x>>i&1 == 0 {
                    ans = append(ans, x^1<<(i-1))
                    break
                }
            }
        }
    }
    return
}

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function minBitwiseArray(nums: number[]): number[] {
    const ans: number[] = [];
    for (const x of nums) {
        if (x === 2) {
            ans.push(-1);
        } else {
            for (let i = 1; i < 32; ++i) {
                if (((x >> i) & 1) ^ 1) {
                    ans.push(x ^ (1 << (i - 1)));
                    break;
                }
            }
        }
    }
    return ans;
}

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