2568. Minimum Impossible OR

中文文档

Description

You are given a 0-indexed integer array nums.

We say that an integer x is expressible from nums if there exist some integers 0 <= index1 < index2 < ... < indexk < nums.length for which nums[index1] | nums[index2] | ... | nums[indexk] = x. In other words, an integer is expressible if it can be written as the bitwise OR of some subsequence of nums.

Return the minimum positive non-zero integer that is not expressible from nums.

 

Example 1:

Input: nums = [2,1]
Output: 4
Explanation: 1 and 2 are already present in the array. We know that 3 is expressible, since nums[0] | nums[1] = 2 | 1 = 3. Since 4 is not expressible, we return 4.

Example 2:

Input: nums = [5,3,2]
Output: 1
Explanation: We can show that 1 is the smallest number that is not expressible.

 

Constraints:

• 1 <= nums.length <= 105
• 1 <= nums[i] <= 109

Solutions

Solution 1: Enumerate Powers of 2

We start from the integer $1. If \1 is expressible, it must appear in the array `nums`. If \2 is expressible, it must also appear in the array `nums`. If both \1 and \2 are expressible, then their bitwise OR operation \3$ is also expressible, and so on.

Therefore, we can enumerate the powers of $2. If the currently enumerated \2^i is not in the array `nums`, then \2^i$ is the smallest unexpressible integer.

The time complexity is $O(n + \log M)$, and the space complexity is $O(n)$. Here, $n$ and $M$ are the length of the array nums and the maximum value in the array nums, respectively.

Python3

class Solution:
    def minImpossibleOR(self, nums: List[int]) -> int:
        s = set(nums)
        return next(1 << i for i in range(32) if 1 << i not in s)

Java

class Solution {
    public int minImpossibleOR(int[] nums) {
        Set<Integer> s = new HashSet<>();
        for (int x : nums) {
            s.add(x);
        }
        for (int i = 0;; ++i) {
            if (!s.contains(1 << i)) {
                return 1 << i;
            }
        }
    }
}

C++

class Solution {
public:
    int minImpossibleOR(vector<int>& nums) {
        unordered_set<int> s(nums.begin(), nums.end());
        for (int i = 0;; ++i) {
            if (!s.count(1 << i)) {
                return 1 << i;
            }
        }
    }
};

Go

func minImpossibleOR(nums []int) int {
	s := map[int]bool{}
	for _, x := range nums {
		s[x] = true
	}
	for i := 0; ; i++ {
		if !s[1<<i] {
			return 1 << i
		}
	}
}

TypeScript

function minImpossibleOR(nums: number[]): number {
    const s: Set<number> = new Set();
    for (const x of nums) {
        s.add(x);
    }
    for (let i = 0; ; ++i) {
        if (!s.has(1 << i)) {
            return 1 << i;
        }
    }
}