LeetCode-in-Java
2593. Find Score of an Array After Marking All Elements
Medium
You are given an array nums consisting of positive integers.
Starting with score = 0, apply the following algorithm:
- Choose the smallest integer of the array that is not marked. If there is a tie, choose the one with the smallest index.
- Add the value of the chosen integer to
score. - Mark the chosen element and its two adjacent elements if they exist.
- Repeat until all the array elements are marked.
Return the score you get after applying the above algorithm.
Example 1:
Input: nums = [2,1,3,4,5,2]
Output: 7
Explanation: We mark the elements as follows:
- 1 is the smallest unmarked element, so we mark it and its two adjacent elements: [2,1,3,4,5,2].
- 2 is the smallest unmarked element, so we mark it and its left adjacent element: [2,1,3,4,5,2].
- 4 is the only remaining unmarked element, so we mark it: [2,1,3,4,5,2]. Our score is 1 + 2 + 4 = 7.
Example 2:
Input: nums = [2,3,5,1,3,2]
Output: 5
Explanation: We mark the elements as follows:
- 1 is the smallest unmarked element, so we mark it and its two adjacent elements: [2,3,5,1,3,2].
- 2 is the smallest unmarked element, since there are two of them, we choose the left-most one, so we mark the one at index 0 and its right adjacent element: [2,3,5,1,3,2].
- 2 is the only remaining unmarked element, so we mark it: [2,3,5,1,3,2]. Our score is 1 + 2 + 2 = 5.
Constraints:
1 <= nums.length <= 1051 <= nums[i] <= 106
Solution
import java.util.PriorityQueue;
public class Solution {
private static class Point {
int idx;
int val;
Point(int idx, int val) {
this.idx = idx;
this.val = val;
}
}
public long findScore(int[] arr) {
PriorityQueue<Point> pq =
new PriorityQueue<>((a, b) -> b.val == a.val ? a.idx - b.idx : a.val - b.val);
int n = arr.length;
boolean[] visited = new boolean[n];
for (int i = 0; i < n; i++) {
pq.add(new Point(i, arr[i]));
}
long ans = 0;
while (!pq.isEmpty()) {
Point p = pq.remove();
int idx = p.idx;
if (!visited[idx]) {
ans += p.val;
visited[idx] = true;
if (idx - 1 >= 0 && !visited[idx - 1]) {
visited[idx - 1] = true;
}
if (idx + 1 < n && !visited[idx + 1]) {
visited[idx + 1] = true;
}
}
}
return ans;
}
}