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 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:
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.
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:
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
.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:
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
@SuppressWarnings({"java:S1871", "java:S6541"})
public class Solution {
public long[] unmarkedSumArray(int[] nums, int[][] queries) {
int l = nums.length;
int[] orig = new int[l];
for (int i = 0; i < l; i++) {
orig[i] = i;
}
int x = 1;
while (x < l) {
int[] temp = new int[l];
int[] teor = new int[l];
int y = 0;
while (y < l) {
int s1 = 0;
int s2 = 0;
while (s1 + s2 < 2 * x && y + s1 + s2 < l) {
if (s2 >= x || y + x + s2 >= l) {
temp[y + s1 + s2] = nums[y + s1];
teor[y + s1 + s2] = orig[y + s1];
s1++;
} else if (s1 >= x) {
temp[y + s1 + s2] = nums[y + x + s2];
teor[y + s1 + s2] = orig[y + x + s2];
s2++;
} else if (nums[y + s1] <= nums[y + x + s2]) {
temp[y + s1 + s2] = nums[y + s1];
teor[y + s1 + s2] = orig[y + s1];
s1++;
} else {
temp[y + s1 + s2] = nums[y + x + s2];
teor[y + s1 + s2] = orig[y + x + s2];
s2++;
}
}
y += 2 * x;
}
for (int i = 0; i < l; i++) {
nums[i] = temp[i];
orig[i] = teor[i];
}
x *= 2;
}
int[] change = new int[l];
for (int i = 0; i < l; i++) {
change[orig[i]] = i;
}
boolean[] mark = new boolean[l];
int m = queries.length;
int st = 0;
long sum = 0;
for (int num : nums) {
sum += num;
}
long[] out = new long[m];
for (int i = 0; i < m; i++) {
int a = queries[i][0];
if (!mark[change[a]]) {
mark[change[a]] = true;
sum -= nums[change[a]];
}
int b = queries[i][1];
int many = 0;
while (many < b) {
if (st == l) {
out[i] = sum;
break;
}
if (!mark[st]) {
mark[st] = true;
sum -= nums[st];
many++;
}
st++;
}
out[i] = sum;
}
return out;
}
}