# 2568. Minimum Impossible OR

## 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.

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

• 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;
}
}
}
};

• 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)


• 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
}
}
}

• function minImpossibleOR(nums: number[]): number {
const s: Set<number> = new Set();
for (const x of nums) {