Welcome to Subscribe On Youtube
Formatted question description: https://leetcode.ca/all/2568.html
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) { 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; } } } };
-
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) { s.add(x); } for (let i = 0; ; ++i) { if (!s.has(1 << i)) { return 1 << i; } } }