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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.

• Java
• C++
• Python
• Go
• TypeScript

• 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;
          }
      }
  }
  
  

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