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
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)
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;
}
}
}
}
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;
}
}
}
};
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) {
s.add(x);
}
for (let i = 0; ; ++i) {
if (!s.has(1 << i)) {
return 1 << i;
}
}
}