LeetCode in Kotlin
3287. Find the Maximum Sequence Value of Array
Hard
You are given an integer array nums and a positive integer k.
The value of a sequence seq of size 2 * x is defined as:
(seq[0] OR seq[1] OR ... OR seq[x - 1]) XOR (seq[x] OR seq[x + 1] OR ... OR seq[2 * x - 1]).
Return the maximum value of any subsequence of nums having size 2 * k.
Example 1:
Input: nums = [2,6,7], k = 1
Output: 5
Explanation:
The subsequence [2, 7] has the maximum value of 2 XOR 7 = 5.
Example 2:
Input: nums = [4,2,5,6,7], k = 2
Output: 2
Explanation:
The subsequence [4, 5, 6, 7] has the maximum value of (4 OR 5) XOR (6 OR 7) = 2.
Constraints:
2 <= nums.length <= 4001 <= nums[i] < 271 <= k <= nums.length / 2
Solution
import kotlin.math.max
class Solution {
fun maxValue(nums: IntArray, k: Int): Int {
val n = nums.size
val left: Array<Array<MutableSet<Int>>> =
Array<Array<MutableSet<Int>>>(n) { Array<MutableSet<Int>>(k + 1) { mutableSetOf() } }
val right: Array<Array<MutableSet<Int>>> =
Array<Array<MutableSet<Int>>>(n) { Array<MutableSet<Int>>(k + 1) { mutableSetOf() } }
left[0][0].add(0)
left[0][1].add(nums[0])
for (i in 1 until n - k) {
left[i][0].add(0)
for (j in 1..k) {
left[i][j].addAll(left[i - 1][j])
for (v in left[i - 1][j - 1]) {
left[i][j].add(v or nums[i])
}
}
}
right[n - 1][0].add(0)
right[n - 1][1].add(nums[n - 1])
var result = 0
if (k == 1) {
for (l in left[n - 2][k]) {
result = max(result, (l xor nums[n - 1]))
}
}
for (i in n - 2 downTo k) {
right[i][0].add(0)
for (j in 1..k) {
right[i][j].addAll(right[i + 1][j])
for (v in right[i + 1][j - 1]) {
right[i][j].add(v or nums[i])
}
}
for (l in left[i - 1][k]) {
for (r in right[i][k]) {
result = max(result, (l xor r))
}
}
}
return result
}
}