LeetCode in Kotlin

3845. Maximum Subarray XOR with Bounded Range

Hard

You are given a non-negative integer array nums and an integer k.

You must select a non-empty subarrays of nums such that the difference between its maximum and minimum elements is at most k. The value of this subarray is the bitwise XOR of all elements in the subarray.

Return an integer denoting the maximum possible value of the selected subarray.

Example 1:

Input: nums = [5,4,5,6], k = 2

Output: 7

Explanation:

• Select the subarray [5, **4, 5, 6**].
• The difference between its maximum and minimum elements is 6 - 4 = 2 <= k.
• The value is 4 XOR 5 XOR 6 = 7.

Example 2:

Input: nums = [5,4,5,6], k = 1

Output: 6

Explanation:

• Select the subarray [5, 4, 5, **6**].
• The difference between its maximum and minimum elements is 6 - 6 = 0 <= k.
• The value is 6.

Constraints:

• 1 <= nums.length <= 4 * 104
• 0 <= nums[i] < 215
• 0 <= k < 215

Solution

import java.util.ArrayDeque
import java.util.Deque
import kotlin.math.max

class Solution {
    companion object {
        private const val BITS = 15
    }

    class TrieNode {
        val next = arrayOfNulls<TrieNode>(2)
        var count = 0
    }

    private fun insert(root: TrieNode, num: Int) {
        var node = root
        for (i in BITS - 1 downTo 0) {
            val bit = (num ushr i) and 1
            if (node.next[bit] == null) {
                node.next[bit] = TrieNode()
            }
            node = node.next[bit]!!
            node.count++
        }
    }

    private fun remove(root: TrieNode, num: Int) {
        var node = root
        for (i in BITS - 1 downTo 0) {
            val bit = (num ushr i) and 1
            val child = node.next[bit]!!
            child.count--
            if (child.count == 0) {
                node.next[bit] = null
            }
            node = child
        }
    }

    private fun query(root: TrieNode, num: Int): Int {
        var node = root
        var res = 0
        for (i in BITS - 1 downTo 0) {
            val bit = (num ushr i) and 1
            val targetBit = bit xor 1
            if (node.next[targetBit] != null) {
                res = res or (1 shl i)
                node = node.next[targetBit]!!
            } else {
                node = node.next[bit]!!
            }
        }
        return res
    }

    fun maxXor(nums: IntArray, k: Int): Int {
        val n = nums.size
        val prefix = IntArray(n + 1)
        for (i in 0 until n) {
            prefix[i + 1] = prefix[i] xor nums[i]
        }
        val maxDeque: Deque<Int> = ArrayDeque()
        val minDeque: Deque<Int> = ArrayDeque()
        val root = TrieNode()
        var maxXor = 0
        var l = 0
        insert(root, 0)
        for (r in 0 until n) {
            // maintain max deque
            while (maxDeque.isNotEmpty() && nums[r] > nums[maxDeque.peekLast()]) {
                maxDeque.pollLast()
            }
            maxDeque.offerLast(r)
            // maintain min deque
            while (minDeque.isNotEmpty() && nums[r] < nums[minDeque.peekLast()]) {
                minDeque.pollLast()
            }
            minDeque.offerLast(r)
            // shrink window if max - min > k
            while (maxDeque.isNotEmpty() &&
                minDeque.isNotEmpty() &&
                nums[maxDeque.peekFirst()] - nums[minDeque.peekFirst()] > k
            ) {
                remove(root, prefix[l])
                l++
                if (maxDeque.peekFirst() < l) {
                    maxDeque.pollFirst()
                }
                if (minDeque.peekFirst() < l) {
                    minDeque.pollFirst()
                }
            }
            // query and insert prefix[r + 1]
            maxXor = max(maxXor, query(root, prefix[r + 1]))
            insert(root, prefix[r + 1])
        }
        return maxXor
    }
}

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