LeetCode in Kotlin

2736. Maximum Sum Queries

Hard

You are given two 0-indexed integer arrays nums1 and nums2, each of length n, and a 1-indexed 2D array queries where queries[i] = [xi, yi].

For the ith query, find the maximum value of nums1[j] + nums2[j] among all indices j (0 <= j < n), where nums1[j] >= xi and nums2[j] >= yi, or -1 if there is no j satisfying the constraints.

Return an array answer where answer[i] is the answer to the ith query.

Example 1:

Input: nums1 = [4,3,1,2], nums2 = [2,4,9,5], queries = [[4,1],[1,3],[2,5]]

Output: [6,10,7]

Explanation:

For the 1st query xi = 4 and yi = 1, we can select index j = 0 since nums1[j] >= 4 and nums2[j] >= 1. The sum nums1[j] + nums2[j] is 6, and we can show that 6 is the maximum we can obtain.

For the 2nd query xi = 1 and yi = 3, we can select index j = 2 since nums1[j] >= 1 and nums2[j] >= 3. The sum nums1[j] + nums2[j] is 10, and we can show that 10 is the maximum we can obtain.

For the 3rd query xi = 2 and yi = 5, we can select index j = 3 since nums1[j] >= 2 and nums2[j] >= 5. The sum nums1[j] + nums2[j] is 7, and we can show that 7 is the maximum we can obtain.

Therefore, we return [6,10,7].

Example 2:

Input: nums1 = [3,2,5], nums2 = [2,3,4], queries = [[4,4],[3,2],[1,1]]

Output: [9,9,9]

Explanation: For this example, we can use index j = 2 for all the queries since it satisfies the constraints for each query.

Example 3:

Input: nums1 = [2,1], nums2 = [2,3], queries = [[3,3]]

Output: [-1]

Explanation: There is one query in this example with xi = 3 and yi = 3. For every index, j, either nums1[j] < xi or nums2[j] < yi. Hence, there is no solution.

Constraints:

• nums1.length == nums2.length
• n == nums1.length
• 1 <= n <= 105
• 1 <= nums1[i], nums2[i] <= 109
• 1 <= queries.length <= 105
• queries[i].length == 2
• xi == queries[i][1]
• yi == queries[i][2]
• 1 <= xi, yi <= 109

Solution

import java.util.TreeMap

class Solution {
    private fun update(map: TreeMap<Int, Int>, num: Int, sum: Int) {
        var entry = map.floorEntry(num)
        while (entry != null && entry.value <= sum) {
            map.remove(entry.key)
            val x = entry.key
            entry = map.floorEntry(x)
        }
        entry = map.ceilingEntry(num)
        if (entry == null || entry.value < sum) map.put(num, sum)
    }

    private fun queryVal(map: TreeMap<Int, Int>, num: Int): Int {
        val (_, value) = map.ceilingEntry(num) ?: return -1
        return value
    }

    fun maximumSumQueries(nums1: IntArray, nums2: IntArray, queries: Array<IntArray>): IntArray {
        val n = nums1.size
        val m = queries.size
        val v: MutableList<IntArray> = ArrayList()
        for (i in 0 until n) {
            v.add(intArrayOf(nums1[i], nums2[i]))
        }
        v.sortWith(
            Comparator { a: IntArray, b: IntArray ->
                a[0] - b[0]
            },
        )
        val ind: MutableList<Int> = ArrayList()
        for (i in 0 until m) ind.add(i)
        ind.sortWith(Comparator { a: Int?, b: Int? -> queries[b!!][0] - queries[a!!][0] })
        val values = TreeMap<Int, Int>()
        var j = n - 1
        val ans = IntArray(m)
        for (i in ind) {
            val a = queries[i][0]
            val b = queries[i][1]
            while (j >= 0 && v[j][0] >= a) {
                update(values, v[j][1], v[j][0] + v[j][1])
                j--
            }
            ans[i] = queryVal(values, b)
        }
        return ans
    }
}

This site is open source. Improve this page.