LeetCode in Kotlin
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
class Solution {
fun findScore(nums: IntArray): Long {
var ans: Long = 0
val pq = PriorityQueue { a: IntArray, b: IntArray ->
if (a[0] == b[0]
) a[1] - b[1] else a[0] - b[0]
}
val vis = BooleanArray(nums.size)
for (i in nums.indices) {
pq.offer(intArrayOf(nums[i], i))
}
while (pq.isNotEmpty()) {
val it = pq.poll()
if (!vis[it[1]]) {
vis[it[1]] = true
ans += it[0].toLong()
if (it[1] > 0) {
vis[it[1] - 1] = true
}
if (it[1] < nums.size - 1) {
vis[it[1] + 1] = true
}
}
}
return ans
}
}