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2736. Maximum Sum Queries

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Description

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

Solutions

Python3

Java

class Solution {
    public int[] maximumSumQueries(int[] nums1, int[] nums2, int[][] q) {
        int n = nums1.length, m = q.length;
        int[][] a = new int[n][2];
        for (int i = 0; i < n; i++) {
            a[i][0] = nums1[i];
            a[i][1] = nums2[i];
        }
        int[][] b = new int[m][3];
        for (int i = 0; i < m; i++) {
            b[i][0] = q[i][0];
            b[i][1] = q[i][1];
            b[i][2] = i;
        }
        Arrays.sort(a, (o1, o2) -> o1[0] - o2[0]);
        Arrays.sort(b, (o1, o2) -> o1[0] - o2[0]);
        TreeMap<Integer, Integer> map = new TreeMap<>();
        int[] res = new int[m];
        int max = -1;
        for (int i = m - 1, j = n - 1; i >= 0; i--) {
            int x = b[i][0], y = b[i][1], idx = b[i][2];
            while (j >= 0 && a[j][0] >= x) {
                if (max < a[j][1]) {
                    max = a[j][1];
                    Integer key = map.floorKey(a[j][1]);
                    while (key != null && map.get(key) <= a[j][0] + a[j][1]) {
                        map.remove(key);
                        key = map.floorKey(key);
                    }
                    map.put(max, a[j][0] + a[j][1]);
                }
                j--;
            }
            Integer key = map.ceilingKey(y);
            if (key == null)
                res[idx] = -1;
            else
                res[idx] = map.get(key);
        }
        return res;
    }
}

C++

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