FPEvalRepoPublic/LeetCodeMetaData

1{2  "id": 2839,3  "name": "maximum-sum-queries",4  "difficulty": "Hard",5  "link": "https://leetcode.com/problems/maximum-sum-queries/",6  "date": "2023-06-04",7  "task_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`",8  "test_case": [9    {10      "label": "Example 1",11      "input": "nums1 = [4,3,1,2], nums2 = [2,4,9,5], queries = [[4,1],[1,3],[2,5]]",12      "output": "[6,10,7] "13    },14    {15      "label": "Example 2",16      "input": "nums1 = [3,2,5], nums2 = [2,3,4], queries = [[4,4],[3,2],[1,1]]",17      "output": "[9,9,9] "18    },19    {20      "label": "Example 3",21      "input": "nums1 = [2,1], nums2 = [2,3], queries = [[3,3]]",22      "output": "[-1] "23    }24  ],25  "constraints": [26    "nums1.length == nums2.length",27    "n == nums1.length",28    "1 <= n <= 105",29    "1 <= nums1[i], nums2[i] <= 109",30    "1 <= queries.length <= 105",31    "queries[i].length == 2",32    "xi == queries[i][1]",33    "yi == queries[i][2]",34    "1 <= xi, yi <= 109"35  ],36  "python_template": "class Solution(object):\n    def maximumSumQueries(self, nums1, nums2, queries):\n        \"\"\"\n        :type nums1: List[int]\n        :type nums2: List[int]\n        :type queries: List[List[int]]\n        :rtype: List[int]\n        \"\"\"\n        ",37  "java_template": "class Solution {\n    public int[] maximumSumQueries(int[] nums1, int[] nums2, int[][] queries) {\n        \n    }\n}",38  "metadata": {39    "func_name": "maximumSumQueries"40  }41}