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Java-based LeetCode algorithm problem solutions, regularly updated
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