2917. Find the K-or of an Array
Description
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 <= 50
0 <= nums[i] < 231
1 <= k <= nums.length
Solutions
Solution 1: Enumeration
We can enumerate each bit $i$ in the range $[0, 32)$, and count the number of numbers in the array $nums$ whose $i$-th bit is $1$, denoted as $cnt$. If $cnt \ge k$, we add $2^i$ to the answer.
After the enumeration, we return the answer.
The time complexity is $O(n \times \log M)$, where $n$ and $M$ are the length of the array $nums$ and the maximum value in $nums$, respectively. The space complexity is $O(1)$.
1 2 3 4 5 6 7 8 |
|
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 |
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1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 |
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1 2 3 4 5 6 7 8 9 10 11 12 |
|
1 2 3 4 5 6 7 8 9 10 11 12 13 |
|