leetcode-problemset/leetcode/originData/maximum-sum-queries.json

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            "title": "Maximum Sum Queries",
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            "content": "<p>You are given two <strong>0-indexed</strong> integer arrays <code>nums1</code> and <code>nums2</code>, each of length <code>n</code>, and a <strong>1-indexed 2D array</strong> <code>queries</code> where <code>queries[i] = [x<sub>i</sub>, y<sub>i</sub>]</code>.</p>\n\n<p>For the <code>i<sup>th</sup></code> query, find the <strong>maximum value</strong> of <code>nums1[j] + nums2[j]</code> among all indices <code>j</code> <code>(0 &lt;= j &lt; n)</code>, where <code>nums1[j] &gt;= x<sub>i</sub></code> and <code>nums2[j] &gt;= y<sub>i</sub></code>, or <strong>-1</strong> if there is no <code>j</code> satisfying the constraints.</p>\n\n<p>Return <em>an array </em><code>answer</code><em> where </em><code>answer[i]</code><em> is the answer to the </em><code>i<sup>th</sup></code><em> query.</em></p>\n\n<p>&nbsp;</p>\n<p><strong class=\"example\">Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> nums1 = [4,3,1,2], nums2 = [2,4,9,5], queries = [[4,1],[1,3],[2,5]]\n<strong>Output:</strong> [6,10,7]\n<strong>Explanation:</strong> \nFor the 1st query <code node=\"[object Object]\">x<sub>i</sub> = 4</code>&nbsp;and&nbsp;<code node=\"[object Object]\">y<sub>i</sub> = 1</code>, we can select index&nbsp;<code node=\"[object Object]\">j = 0</code>&nbsp;since&nbsp;<code node=\"[object Object]\">nums1[j] &gt;= 4</code>&nbsp;and&nbsp;<code node=\"[object Object]\">nums2[j] &gt;= 1</code>. The sum&nbsp;<code node=\"[object Object]\">nums1[j] + nums2[j]</code>&nbsp;is 6, and we can show that 6 is the maximum we can obtain.\n\nFor the 2nd query <code node=\"[object Object]\">x<sub>i</sub> = 1</code>&nbsp;and&nbsp;<code node=\"[object Object]\">y<sub>i</sub> = 3</code>, we can select index&nbsp;<code node=\"[object Object]\">j = 2</code>&nbsp;since&nbsp;<code node=\"[object Object]\">nums1[j] &gt;= 1</code>&nbsp;and&nbsp;<code node=\"[object Object]\">nums2[j] &gt;= 3</code>. The sum&nbsp;<code node=\"[object Object]\">nums1[j] + nums2[j]</code>&nbsp;is 10, and we can show that 10 is the maximum we can obtain. \n\nFor the 3rd query <code node=\"[object Object]\">x<sub>i</sub> = 2</code>&nbsp;and&nbsp;<code node=\"[object Object]\">y<sub>i</sub> = 5</code>, we can select index&nbsp;<code node=\"[object Object]\">j = 3</code>&nbsp;since&nbsp;<code node=\"[object Object]\">nums1[j] &gt;= 2</code>&nbsp;and&nbsp;<code node=\"[object Object]\">nums2[j] &gt;= 5</code>. The sum&nbsp;<code node=\"[object Object]\">nums1[j] + nums2[j]</code>&nbsp;is 7, and we can show that 7 is the maximum we can obtain.\n\nTherefore, we return&nbsp;<code node=\"[object Object]\">[6,10,7]</code>.\n</pre>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> nums1 = [3,2,5], nums2 = [2,3,4], queries = [[4,4],[3,2],[1,1]]\n<strong>Output:</strong> [9,9,9]\n<strong>Explanation:</strong> For this example, we can use index&nbsp;<code node=\"[object Object]\">j = 2</code>&nbsp;for all the queries since it satisfies the constraints for each query.\n</pre>\n\n<p><strong class=\"example\">Example 3:</strong></p>\n\n<pre>\n<strong>Input:</strong> nums1 = [2,1], nums2 = [2,3], queries = [[3,3]]\n<strong>Output:</strong> [-1]\n<strong>Explanation:</strong> There is one query in this example with <code node=\"[object Object]\">x<sub>i</sub></code> = 3 and <code node=\"[object Object]\">y<sub>i</sub></code> = 3. For every index, j, either nums1[j] &lt; <code node=\"[object Object]\">x<sub>i</sub></code> or nums2[j] &lt; <code node=\"[object Object]\">y<sub>i</sub></code>. Hence, there is no solution. \n</pre>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>nums1.length == nums2.length</code>&nbsp;</li>\n\t<li><code>n ==&nbsp;nums1.length&nbsp;</code></li>\n\t<li><code>1 &lt;= n &lt;= 10<sup>5</sup></code></li>\n\t<li><code>1 &lt;= nums1[i], nums2[i] &lt;= 10<sup>9</sup>&nbsp;</code></li>\n\t<li><code>1 &lt;= queries.length &lt;= 10<sup>5</sup></code></li>\n\t<li><code>queries[i].length ==&nbsp;2</code></li>\n\t<li><code>x<sub>i</sub>&nbsp;== queries[i][1]</code></li>\n\t<li><code>y<sub>i</sub> == queries[i][2]</code></li>\n\t<li><code>1 &lt;= x<sub>i</sub>, y<sub>i</sub> &lt;= 10<sup>9</sup></code></li>\n</ul>\n",
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                    "code": "class Solution {\npublic:\n    vector<int> maximumSumQueries(vector<int>& nums1, vector<int>& nums2, vector<vector<int>>& queries) {\n        \n    }\n};",
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                    "code": "class Solution {\n    public int[] maximumSumQueries(int[] nums1, int[] nums2, int[][] queries) {\n        \n    }\n}",
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                    "lang": "Python",
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                    "code": "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        ",
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                    "code": "object Solution {\n    def maximumSumQueries(nums1: Array[Int], nums2: Array[Int], queries: Array[Array[Int]]): Array[Int] = {\n        \n    }\n}",
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                    "code": "class Solution {\n    fun maximumSumQueries(nums1: IntArray, nums2: IntArray, queries: Array<IntArray>): IntArray {\n        \n    }\n}",
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                    "code": "impl Solution {\n    pub fn maximum_sum_queries(nums1: Vec<i32>, nums2: Vec<i32>, queries: Vec<Vec<i32>>) -> Vec<i32> {\n        \n    }\n}",
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                "Sort (x, y) tuples and queries by x-coordinate descending. Don’t forget to index queries before sorting so that you can answer them in the correct order.",
                "Before answering a query (min_x, min_y), add all (x, y) pairs with x >= min_x to some data structure.",
                "Use a monotone descending map to store (y, x + y) pairs. A monotone map has ascending keys and descending values. When inserting a pair (y, x + y), remove all pairs (y', x' + y') with y' < y and x' + y' <= x + y.",
                "To find the insertion position use binary search (built-in in many languages).",
                "When querying for max (x + y) over y >= y', use binary search to find the first pair (y, x + y) with y >= y'. It will have the maximum value of x + y because the map has monotone descending values."
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