LeetCode in Kotlin

1707. Maximum XOR With an Element From Array

Hard

You are given an array nums consisting of non-negative integers. You are also given a queries array, where queries[i] = [xi, mi].

The answer to the ith query is the maximum bitwise XOR value of xi and any element of nums that does not exceed mi. In other words, the answer is max(nums[j] XOR xi) for all j such that nums[j] <= mi. If all elements in nums are larger than mi, then the answer is -1.

Return an integer array answer where answer.length == queries.length and answer[i] is the answer to the ith query.

Example 1:

Input: nums = [0,1,2,3,4], queries = [[3,1],[1,3],[5,6]]

Output: [3,3,7]

Explanation:

1) 0 and 1 are the only two integers not greater than 1. 0 XOR 3 = 3 and 1 XOR 3 = 2. The larger of the two is 3.

2) 1 XOR 2 = 3.

3) 5 XOR 2 = 7.

Example 2:

Input: nums = [5,2,4,6,6,3], queries = [[12,4],[8,1],[6,3]]

Output: [15,-1,5]

Constraints:

• 1 <= nums.length, queries.length <= 105
• queries[i].length == 2
• 0 <= nums[j], xi, mi <= 109

Solution

@Suppress("NAME_SHADOWING")
class Solution {
    internal class QueryComparator : Comparator<IntArray> {
        override fun compare(a: IntArray, b: IntArray): Int {
            return a[1].compareTo(b[1])
        }
    }

    internal class Node {
        var zero: Node? = null
        var one: Node? = null
    }

    fun maximizeXor(nums: IntArray, queries: Array<IntArray>): IntArray {
        nums.sort()
        val len = queries.size
        val queryWithIndex = Array(len) { IntArray(3) }
        for (i in 0 until len) {
            queryWithIndex[i][0] = queries[i][0]
            queryWithIndex[i][1] = queries[i][1]
            queryWithIndex[i][2] = i
        }
        queryWithIndex.sortWith(QueryComparator())
        var numId = 0
        val ans = IntArray(len)
        val root = Node()
        for (i in 0 until len) {
            while (numId < nums.size && nums[numId] <= queryWithIndex[i][1]) {
                addNumToTree(nums[numId], root)
                numId++
            }
            ans[queryWithIndex[i][2]] = maxXOR(queryWithIndex[i][0], root)
        }
        return ans
    }

    private fun addNumToTree(num: Int, node: Node) {
        var node: Node? = node
        for (i in 31 downTo 0) {
            val digit = num shr i and 1
            if (digit == 1) {
                if (node!!.one == null) {
                    node.one = Node()
                }
                node = node.one
            } else {
                if (node!!.zero == null) {
                    node.zero = Node()
                }
                node = node.zero
            }
        }
    }

    private fun maxXOR(num: Int, node: Node): Int {
        var node: Node? = node
        if (node!!.one == null && node.zero == null) {
            return -1
        }
        var ans = 0
        var i = 31
        while (i >= 0 && node != null) {
            val digit = num shr i and 1
            if (digit == 1) {
                if (node.zero != null) {
                    ans += 1 shl i
                    node = node.zero
                } else {
                    node = node.one
                }
            } else {
                if (node.one != null) {
                    ans += 1 shl i
                    node = node.one
                } else {
                    node = node.zero
                }
            }
            i--
        }
        return ans
    }
}

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