LiveCodeBench / 2839 / maximum-sum-queries

contest date

2023-06-11T00:00:00

contest id

weekly-contest-349

difficulty

hard

platform

leetcode

question content

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 <= 10^5
1 <= nums1[i], nums2[i] <= 10^9 
1 <= queries.length <= 10^5
queries[i].length == 2
x_i == queries[i][1]
y_i == queries[i][2]
1 <= x_i, y_i <= 10^9

question title

maximum-sum-queries

starter code

Code

class Solution:
    def maximumSumQueries(self, nums1: List[int], nums2: List[int], queries: List[List[int]]) -> List[int]: