LeetCode in Kotlin
2044. Count Number of Maximum Bitwise-OR Subsets
Medium
Given an integer array nums, find the maximum possible bitwise OR of a subset of nums and return the number of different non-empty subsets with the maximum bitwise OR.
An array a is a subset of an array b if a can be obtained from b by deleting some (possibly zero) elements of b. Two subsets are considered different if the indices of the elements chosen are different.
The bitwise OR of an array a is equal to a[0] **OR** a[1] **OR** ... **OR** a[a.length - 1] (0-indexed).
Example 1:
Input: nums = [3,1]
Output: 2
Explanation: The maximum possible bitwise OR of a subset is 3. There are 2 subsets with a bitwise OR of 3:
-
[3]
-
[3,1]
Example 2:
Input: nums = [2,2,2]
Output: 7
Explanation: All non-empty subsets of [2,2,2] have a bitwise OR of 2. There are 23 - 1 = 7 total subsets.
Example 3:
Input: nums = [3,2,1,5]
Output: 6
Explanation: The maximum possible bitwise OR of a subset is 7. There are 6 subsets with a bitwise OR of 7:
-
[3,5]
-
[3,1,5]
-
[3,2,5]
-
[3,2,1,5]
-
[2,5]
-
[2,1,5]
Constraints:
1 <= nums.length <= 161 <= nums[i] <= 105
Solution
class Solution {
private var count = 0
fun countMaxOrSubsets(nums: IntArray): Int {
var lookfor = 0
for (i in nums) {
lookfor = lookfor or i
}
countsub(nums, 0, lookfor, 0)
return count
}
private fun countsub(nums: IntArray, index: Int, lookfor: Int, sofar: Int) {
if (lookfor == sofar) {
count++
}
if (index >= nums.size) {
return
}
for (start in index until nums.size) {
countsub(nums, start + 1, lookfor, sofar or nums[start])
}
}
}