3080 - Mark Elements on Array by Performing Queries\.

Medium

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 <code>queriesi = index_i, k_i</code>.

Initially all elements of the array are unmarked.

You need to apply m queries on the array in order, where on the <code>i^th</code> query you do the following:

• Mark the element at index <code>index_i</code> if it is not already marked.

• Then mark <code>k_i</code> 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 <code>k_i</code> 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 <code>i^th</code> query.

Example 1:

Input: nums = 1,2,2,1,2,3,1, queries = \[\[1,2],3,3,4,2]

Output: 8,3,0

Explanation:

We do the following queries on the array:

• Mark the element at index 1, and 2 of the smallest unmarked elements with the smallest indices if they exist, the marked elements now are <code>nums = **<ins>1</ins>** ,<ins> **2** </ins>,2,<ins> **1** </ins>,2,3,1</code>. The sum of unmarked elements is 2 + 2 + 3 + 1 = 8.

• Mark the element at index 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 <code>nums = **<ins>1</ins>** ,<ins> **2** </ins>,<ins> **2** </ins>,<ins> **1** </ins>,<ins> **2** </ins>,3, **<ins>1</ins>** </code>. The sum of unmarked elements is 3.

• Mark the element at index 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 <code>nums = **<ins>1</ins>** ,<ins> **2** </ins>,<ins> **2** </ins>,<ins> **1** </ins>,<ins> **2** </ins>, **<ins>3</ins>** ,<ins> **1** </ins></code>. The sum of unmarked elements is 0.

Example 2:

Input: nums = 1,4,2,3, queries = \[\[0,1]]

Output: 7

Explanation: 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 <code>nums = **<ins>1</ins>** ,4,<ins> **2** </ins>,3</code>, and the sum of unmarked elements is 4 + 3 = 7.

Constraints:

• n == nums.length

• m == queries.length

• <code>1 <= m <= n <= 10⁵</code>

• <code>1 <= numsi<= 10⁵</code>

• queries[i].length == 2

• <code>0 <= index_i, k_i<= n - 1</code>