LeetCode-in-Java

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

public class Solution {
    private void update(NavigableMap<Integer, Integer> map, int num, int sum) {
        Map.Entry<Integer, Integer> 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<Integer, Integer> map, int num) {
        Map.Entry<Integer, Integer> 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<int[]> 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<Integer> ind = new ArrayList<>();
        for (int i = 0; i < m; i++) {
            ind.add(i);
        }
        ind.sort((a, b) -> queries[b][0] - queries[a][0]);
        TreeMap<Integer, Integer> 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;
    }
}

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