You are given an integer array nums, and an integer k. Let's introduce K-or operation by extending the standard bitwise OR. In K-or, a bit position in the result is set to 1 if at least k numbers in nums have a 1 in that position.
Return the K-or of nums.
Example 1:
Input: nums = [7,12,9,8,9,15], k = 4
Output: 9
Explanation:
Represent numbers in binary:
| Number | Bit 3 | Bit 2 | Bit 1 | Bit 0 |
|---|---|---|---|---|
| 7 | 0 | 1 | 1 | 1 |
| 12 | 1 | 1 | 0 | 0 |
| 9 | 1 | 0 | 0 | 1 |
| 8 | 1 | 0 | 0 | 0 |
| 9 | 1 | 0 | 0 | 1 |
| 15 | 1 | 1 | 1 | 1 |
| Result = 9 | 1 | 0 | 0 | 1 |
Bit 0 is set in 7, 9, 9, and 15. Bit 3 is set in 12, 9, 8, 9, and 15.
Only bits 0 and 3 qualify. The result is (1001)2 = 9.
Example 2:
Input: nums = [2,12,1,11,4,5], k = 6
Output: 0
Explanation: No bit appears as 1 in all six array numbers, as required for K-or with k = 6. Thus, the result is 0.
Example 3:
Input: nums = [10,8,5,9,11,6,8], k = 1
Output: 15
Explanation: Since k == 1, the 1-or of the array is equal to the bitwise OR of all its elements. Hence, the answer is 10 OR 8 OR 5 OR 9 OR 11 OR 6 OR 8 = 15.
Constraints:
1 <= nums.length <= 500 <= nums[i] < 2311 <= k <= nums.length[7,12,9,8,9,15] 4