2568

2568. Minimum Impossible OR (Medium)

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

Related Topics:
Array, Bit Manipulation, Brainteaser

Solution 1.

We can find that powers of 2 are hard to get using OR.

If we have 1,2,4, we can get all numbers with 3 bits. The next impossible number is 8.

If we have 1,3,4, we can't get 2.

So, the minimum impossible number we can't get is the smallest power of 2 that doesn't exist in A.

// OJ: https://leetcode.com/problems/minimum-impossible-or
// Author: github.com/lzl124631x
// Time: O(NlogN)
// Space: O(1)
class Solution {
public:
    int minImpossibleOR(vector<int>& A) {
        int seen = 0;
        for (int i : A) {
            if ((i & (i - 1)) == 0) seen |= i;
        }
        for (int i = 0; i < 32; ++i) {
            if ((seen >> i & 1) == 0) return 1 << i;
        }
        return -1;
    }
};