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Java-based LeetCode algorithm problem solutions, regularly updated
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package g2701_2800.s2736_maximum_sum_queries;

// #Hard #Array #Sorting #Binary_Search #Stack #Monotonic_Stack #Segment_Tree #Binary_Indexed_Tree
// #2023_09_23_Time_66_ms_(78.43%)_Space_84.1_MB_(94.12%)

import java.util.ArrayList;
import java.util.Comparator;
import java.util.List;
import java.util.Map;
import java.util.NavigableMap;
import java.util.TreeMap;

/**
 * 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] = [x_i, y_i].
 *
 * For the i^th query, find the **maximum value** of `nums1[j] + nums2[j]` among all indices `j` `(0 <= j < n)`, where nums1[j] >= x_i and nums2[j] >= y_i, or **\-1** if there is no `j` satisfying the constraints.
 *
 * Return _an array_ `answer` _where_ `answer[i]` _is the answer to the_ i^th _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 x_i = 4 and y_i = 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 x_i = 1 and y_i = 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 x_i = 2 and y_i = 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 x_i = 3 and y_i = 3. For every index, j, either nums1[j] < x_i or nums2[j] < y_i. 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`
 * *   x_i == queries[i][1]
 * *   y_i == queries[i][2]
 * *   1 <= x_i, y_i <= 10⁹
**/
public class Solution {
    private void update(NavigableMap map, int num, int sum) {
        Map.Entry entry = map.floorEntry(num);
        while (entry != null && entry.getValue() <= sum) {
            map.remove(entry.getKey());
            int x = entry.getKey();
            entry = map.floorEntry(x);
        }
        entry = map.ceilingEntry(num);
        if (entry == null || entry.getValue() < sum) {
            map.put(num, sum);
        }
    }

    private int queryVal(NavigableMap map, int num) {
        Map.Entry entry = map.ceilingEntry(num);
        if (entry == null) {
            return -1;
        }
        return entry.getValue();
    }

    public int[] maximumSumQueries(int[] nums1, int[] nums2, int[][] queries) {
        int n = nums1.length;
        int m = queries.length;
        List v = new ArrayList<>();
        for (int i = 0; i < n; i++) {
            v.add(new int[] {nums1[i], nums2[i]});
        }
        v.sort(Comparator.comparingInt(a -> a[0]));
        List ind = new ArrayList<>();
        for (int i = 0; i < m; i++) {
            ind.add(i);
        }
        ind.sort((a, b) -> queries[b][0] - queries[a][0]);
        TreeMap values = new TreeMap<>();
        int j = n - 1;
        int[] ans = new int[m];
        for (int i : ind) {
            int a = queries[i][0];
            int b = queries[i][1];
            for (; j >= 0 && v.get(j)[0] >= a; j--) {
                update(values, v.get(j)[1], v.get(j)[0] + v.get(j)[1]);
            }
            ans[i] = queryVal(values, b);
        }
        return ans;
    }
}