# 3080. Mark Elements on Array by Performing Queries

## Description

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.

You need to apply m queries on the array in order, where on the ith query you do the following:

• Mark the element at index indexi if it is not already marked.
• Then mark 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.

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 nums = [1,2,2,1,2,3,1]. 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 nums = [1,2,2,1,2,3,1]. 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 nums = [1,2,2,1,2,3,1]. 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 nums = [1,4,2,3], and the sum of unmarked elements is 4 + 3 = 7.

Constraints:

• n == nums.length
• m == queries.length
• 1 <= m <= n <= 105
• 1 <= nums[i] <= 105
• queries[i].length == 2
• 0 <= indexi, ki <= n - 1

## Solutions

### Solution 1: Sorting + Simulation

First, we calculate the sum $s$ of the array $nums$. We define an array $mark$ to indicate whether the elements in the array have been marked, initializing all elements as unmarked.

Then, we create an array $arr$, where each element is a tuple $(x, i)$, indicating that the $i$-th element in the array has a value of $x$. We sort the array $arr$ by the value of the elements. If the values are equal, we sort them in ascending order of the index.

Next, we traverse the array $queries$. For each query $[index, k]$, we first check whether the element at index $index$ has been marked. If it has not been marked, we mark it and subtract the value of the element at index $index$ from $s$. Then, we traverse the array $arr$. For each element $(x, i)$, if element $i$ has not been marked, we mark it and subtract the value $x$ corresponding to element $i$ from $s$, until $k$ is $0$ or the array $arr$ is fully traversed. Then, we add $s$ to the answer array.

After traversing all the queries, we get the answer array.

The time complexity is $O(n \times \log n)$, and the space complexity is $O(n)$. Here, $n$ is the length of the array $nums$.

• class Solution {
public long[] unmarkedSumArray(int[] nums, int[][] queries) {
int n = nums.length;
long s = Arrays.stream(nums).asLongStream().sum();
boolean[] mark = new boolean[n];
int[][] arr = new int[n][0];
for (int i = 0; i < n; ++i) {
arr[i] = new int[] {nums[i], i};
}
Arrays.sort(arr, (a, b) -> a[0] == b[0] ? a[1] - b[1] : a[0] - b[0]);
int m = queries.length;
long[] ans = new long[m];
for (int i = 0, j = 0; i < m; ++i) {
int index = queries[i][0], k = queries[i][1];
if (!mark[index]) {
mark[index] = true;
s -= nums[index];
}
for (; k > 0 && j < n; ++j) {
if (!mark[arr[j][1]]) {
mark[arr[j][1]] = true;
s -= arr[j][0];
--k;
}
}
ans[i] = s;
}
return ans;
}
}

• class Solution {
public:
vector<long long> unmarkedSumArray(vector<int>& nums, vector<vector<int>>& queries) {
int n = nums.size();
long long s = accumulate(nums.begin(), nums.end(), 0LL);
vector<bool> mark(n);
vector<pair<int, int>> arr;
for (int i = 0; i < n; ++i) {
arr.emplace_back(nums[i], i);
}
sort(arr.begin(), arr.end());
vector<long long> ans;
int m = queries.size();
for (int i = 0, j = 0; i < m; ++i) {
int index = queries[i][0], k = queries[i][1];
if (!mark[index]) {
mark[index] = true;
s -= nums[index];
}
for (; k && j < n; ++j) {
if (!mark[arr[j].second]) {
mark[arr[j].second] = true;
s -= arr[j].first;
--k;
}
}
ans.push_back(s);
}
return ans;
}
};

• class Solution:
def unmarkedSumArray(self, nums: List[int], queries: List[List[int]]) -> List[int]:
n = len(nums)
s = sum(nums)
mark = [False] * n
arr = sorted((x, i) for i, x in enumerate(nums))
j = 0
ans = []
for index, k in queries:
if not mark[index]:
mark[index] = True
s -= nums[index]
while k and j < n:
if not mark[arr[j][1]]:
mark[arr[j][1]] = True
s -= arr[j][0]
k -= 1
j += 1
ans.append(s)
return ans


• func unmarkedSumArray(nums []int, queries [][]int) []int64 {
n := len(nums)
var s int64
for _, x := range nums {
s += int64(x)
}
mark := make([]bool, n)
arr := make([][2]int, 0, n)
for i, x := range nums {
arr = append(arr, [2]int{x, i})
}
sort.Slice(arr, func(i, j int) bool {
if arr[i][0] == arr[j][0] {
return arr[i][1] < arr[j][1]
}
return arr[i][0] < arr[j][0]
})
ans := make([]int64, len(queries))
j := 0
for i, q := range queries {
index, k := q[0], q[1]
if !mark[index] {
mark[index] = true
s -= int64(nums[index])
}
for ; k > 0 && j < n; j++ {
if !mark[arr[j][1]] {
mark[arr[j][1]] = true
s -= int64(arr[j][0])
k--
}
}
ans[i] = s
}
return ans
}

• function unmarkedSumArray(nums: number[], queries: number[][]): number[] {
const n = nums.length;
let s = nums.reduce((acc, x) => acc + x, 0);
const mark: boolean[] = Array(n).fill(false);
const arr = nums.map((x, i) => [x, i]);
arr.sort((a, b) => (a[0] === b[0] ? a[1] - b[1] : a[0] - b[0]));
let j = 0;
const ans: number[] = [];
for (let [index, k] of queries) {
if (!mark[index]) {
mark[index] = true;
s -= nums[index];
}
for (; k && j < n; ++j) {
if (!mark[arr[j][1]]) {
mark[arr[j][1]] = true;
s -= arr[j][0];
--k;
}
}
ans.push(s);
}
return ans;
}