{ "data": { "question": { "questionId": "3878", "questionFrontendId": "3569", "categoryTitle": "Algorithms", "boundTopicId": 3688677, "title": "Maximize Count of Distinct Primes After Split", "titleSlug": "maximize-count-of-distinct-primes-after-split", "content": "
You are given an integer array nums having length n and a 2D integer array queries where queries[i] = [idx, val].
For each query:
\n\nnums[idx] = val.k with 1 <= k < n to split the array into the non-empty prefix nums[0..k-1] and suffix nums[k..n-1] such that the sum of the counts of distinct prime values in each part is maximum.Note: The changes made to the array in one query persist into the next query.
\n\nReturn an array containing the result for each query, in the order they are given.
\n\n\n
Example 1:
\n\nInput: nums = [2,1,3,1,2], queries = [[1,2],[3,3]]
\n\nOutput: [3,4]
\n\nExplanation:
\n\nnums = [2, 1, 3, 1, 2].nums = [2, 2, 3, 1, 2]. Split nums into [2] and [2, 3, 1, 2]. [2] consists of 1 distinct prime and [2, 3, 1, 2] consists of 2 distinct primes. Hence, the answer for this query is 1 + 2 = 3.nums = [2, 2, 3, 3, 2]. Split nums into [2, 2, 3] and [3, 2] with an answer of 2 + 2 = 4.[3, 4].Example 2:
\n\nInput: nums = [2,1,4], queries = [[0,1]]
\n\nOutput: [0]
\n\nExplanation:
\n\nnums = [2, 1, 4].nums = [1, 1, 4]. There are no prime numbers in nums, hence the answer for this query is 0.[0].\n
Constraints:
\n\n2 <= n == nums.length <= 5 * 1041 <= queries.length <= 5 * 1041 <= nums[i] <= 1050 <= queries[i][0] < nums.length1 <= queries[i][1] <= 105给你一个长度为 'n' 的整数数组 nums,以及一个二维整数数组 queries,其中 queries[i] = [idx, val]。
对于每个查询:
\n\nnums[idx] = val。1 <= k < n 的整数 k ,将数组分为非空前缀 nums[0..k-1] 和后缀 nums[k..n-1],使得每部分中 不同 质数的数量之和 最大 。注意:每次查询对数组的更改将持续到后续的查询中。
\n\n返回一个数组,包含每个查询的结果,按给定的顺序排列。
\n\n质数是大于 1 的自然数,只有 1 和它本身两个因数。
\n\n\n\n
示例 1:
\n\n输入: nums = [2,1,3,1,2], queries = [[1,2],[3,3]]
\n\n输出: [3,4]
\n\n解释:
\n\nnums = [2, 1, 3, 1, 2]。nums = [2, 2, 3, 1, 2]。将 nums 分为 [2] 和 [2, 3, 1, 2]。[2] 包含 1 个不同的质数,[2, 3, 1, 2] 包含 2 个不同的质数。所以此查询的答案是 1 + 2 = 3。nums = [2, 2, 3, 3, 2]。将 nums 分为 [2, 2, 3] 和 [3, 2],其答案为 2 + 2 = 4。[3, 4]。示例 2:
\n\n输入: nums = [2,1,4], queries = [[0,1]]
\n\n输出: [0]
\n\n解释:
\n\nnums = [2, 1, 4]。nums = [1, 1, 4]。此时数组中没有质数,因此此查询的答案为 0。[0]。\n\n
提示:
\n\n2 <= n == nums.length <= 5 * 1041 <= queries.length <= 5 * 1041 <= nums[i] <= 1050 <= queries[i][0] < nums.length1 <= queries[i][1] <= 105max(nums) with a sieve to enable O(1) primality checks.",
"For each prime p, record its occurrence indices; if it appears at least twice, treat [first, last] as a segment, and note that the split position k with the most overlapping segments equals the number of primes counted on both sides.",
"Use a segment tree with lazy propagation over all possible k to maintain, update, and query the sum of distinct-prime counts in the prefix and suffix, adjusting for overlaps."
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