You are given a 0-indexed array nums of size n consisting of positive integers.
You are also given a 2D array queries of size m where queries[i] = [indexi, ki].
Initially all elements of the array are unmarked.
\n\nYou need to apply m queries on the array in order, where on the ith query you do the following:
indexi if it is not already marked.ki unmarked elements in the array with the smallest values. If multiple such elements exist, mark the ones with the smallest indices. And if less than ki unmarked elements exist, then mark all of them.Return an array answer of size m where answer[i] is the sum of unmarked elements in the array after the ith query.
\n
Example 1:
\n\nInput: nums = [1,2,2,1,2,3,1], queries = [[1,2],[3,3],[4,2]]
\n\nOutput: [8,3,0]
\n\nExplanation:
\n\nWe do the following queries on the array:
\n\n1, and 2 of the smallest unmarked elements with the smallest indices if they exist, the marked elements now are nums = [1,2,2,1,2,3,1]. The sum of unmarked elements is 2 + 2 + 3 + 1 = 8.3, since it is already marked we skip it. Then we mark 3 of the smallest unmarked elements with the smallest indices, the marked elements now are nums = [1,2,2,1,2,3,1]. The sum of unmarked elements is 3.4, since it is already marked we skip it. Then we mark 2 of the smallest unmarked elements with the smallest indices if they exist, the marked elements now are nums = [1,2,2,1,2,3,1]. The sum of unmarked elements is 0.Example 2:
\n\nInput: nums = [1,4,2,3], queries = [[0,1]]
\n\nOutput: [7]
\n\nExplanation: We do one query which is mark the element at index 0 and mark the smallest element among unmarked elements. The marked elements will be nums = [1,4,2,3], and the sum of unmarked elements is 4 + 3 = 7.
\n
Constraints:
\n\nn == nums.lengthm == queries.length1 <= m <= n <= 1051 <= nums[i] <= 105queries[i].length == 20 <= indexi, ki <= n - 1给你一个长度为 n 下标从 0 开始的正整数数组 nums 。
同时给你一个长度为 m 的二维操作数组 queries ，其中 queries[i] = [indexi, ki] 。
一开始，数组中的所有元素都 未标记 。
\n\n你需要依次对数组执行 m 次操作，第 i 次操作中，你需要执行：
indexi 对应的元素还没标记，那么标记这个元素。ki 个数组中还没有标记的 最小 元素。如果有元素的值相等，那么优先标记它们中下标较小的。如果少于 ki 个未标记元素存在，那么将它们全部标记。请你返回一个长度为 m 的数组 answer ，其中 answer[i]是第 i 次操作后数组中还没标记元素的 和 。
\n\n
示例 1：
\n\n输入：nums = [1,2,2,1,2,3,1], queries = [[1,2],[3,3],[4,2]]
\n\n输出：[8,3,0]
\n\n解释：
\n\n我们依次对数组做以下操作：
\n\n1 的元素，同时标记 2 个未标记的最小元素。标记完后数组为 nums = [1,2,2,1,2,3,1] 。未标记元素的和为 2 + 2 + 3 + 1 = 8 。3 的元素，由于它已经被标记过了，所以我们忽略这次标记，同时标记最靠前的 3 个未标记的最小元素。标记完后数组为 nums = [1,2,2,1,2,3,1] 。未标记元素的和为 3 。4 的元素，由于它已经被标记过了，所以我们忽略这次标记，同时标记最靠前的 2 个未标记的最小元素。标记完后数组为 nums = [1,2,2,1,2,3,1] 。未标记元素的和为 0 。示例 2：
\n\n输入：nums = [1,4,2,3], queries = [[0,1]]
\n\n输出：[7]
\n\n解释：我们执行一次操作，将下标为 0 处的元素标记，并且标记最靠前的 1 个未标记的最小元素。标记完后数组为 nums = [1,4,2,3] 。未标记元素的和为 4 + 3 = 7 。
\n\n
提示：
\n\nn == nums.lengthm == queries.length1 <= m <= n <= 1051 <= nums[i] <= 105queries[i].length == 20 <= indexi, ki <= n - 1nums to be able to find the smallest unmarked elements quickly in each query."
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