LeetCode in Kotlin
3755. Find Maximum Balanced XOR Subarray Length
Medium
Given an integer array nums, return the length of the longest non-empty subarrays that has a bitwise XOR of zero and contains an equal number of even and odd numbers. If no such subarray exists, return 0.
Example 1:
Input: nums = [3,1,3,2,0]
Output: 4
Explanation:
The subarray [1, 3, 2, 0] has bitwise XOR 1 XOR 3 XOR 2 XOR 0 = 0 and contains 2 even and 2 odd numbers.
Example 2:
Input: nums = [3,2,8,5,4,14,9,15]
Output: 8
Explanation:
The whole array has bitwise XOR 0 and contains 4 even and 4 odd numbers.
Example 3:
Input: nums = [0]
Output: 0
Explanation:
No non-empty subarray satisfies both conditions.
Constraints:
1 <= nums.length <= 1050 <= nums[i] <= 109
Solution
import kotlin.math.max
class Solution {
fun maxBalancedSubarray(nums: IntArray): Int {
val n = nums.size
var ans = 0
var xor = 0
var diff = n
val pos: MutableMap<Long?, Int?> = HashMap<Long?, Int?>(n + 1, 1f)
pos[xor.toLong() shl 20 or diff.toLong()] = -1
for (i in 0..<n) {
xor = xor xor nums[i]
diff += if (nums[i] % 2 != 0) 1 else -1
val key = xor.toLong() shl 20 or diff.toLong()
val j = pos[key]
if (j != null) {
ans = max(ans, i - j)
} else {
pos[key] = i
}
}
return ans
}
}