LeetCode in Kotlin
2551. Put Marbles in Bags
Hard
You have k bags. You are given a 0-indexed integer array weights where weights[i] is the weight of the ith marble. You are also given the integer k.
Divide the marbles into the k bags according to the following rules:
- No bag is empty.
- If the
ithmarble andjthmarble are in a bag, then all marbles with an index between theithandjthindices should also be in that same bag. - If a bag consists of all the marbles with an index from
itojinclusively, then the cost of the bag isweights[i] + weights[j].
The score after distributing the marbles is the sum of the costs of all the k bags.
Return the difference between the maximum and minimum scores among marble distributions.
Example 1:
Input: weights = [1,3,5,1], k = 2
Output: 4
Explanation:
The distribution [1],[3,5,1] results in the minimal score of (1+1) + (3+1) = 6.
The distribution [1,3],[5,1], results in the maximal score of (1+3) + (5+1) = 10.
Thus, we return their difference 10 - 6 = 4.
Example 2:
Input: weights = [1, 3], k = 2
Output: 0
Explanation:
The only distribution possible is [1],[3].
Since both the maximal and minimal score are the same, we return 0.
Constraints:
1 <= k <= weights.length <= 1051 <= weights[i] <= 109
Solution
import java.util.PriorityQueue
class Solution {
fun putMarbles(weights: IntArray, k: Int): Long {
// Map<Pair<Integer, Integer>, long[]> memo = new HashMap<>();
// long[] res = dfs(weights, 0, k, memo);
// return res[1] - res[0];
if (k == 1 || k == weights.size) return 0
val min = PriorityQueue<Long>()
val max = PriorityQueue { a: Long?, b: Long? ->
java.lang.Long.compare(
b!!, a!!
)
}
for (i in 0 until weights.size - 1) {
val sum = weights[i].toLong() + weights[i + 1]
min.offer(sum)
max.offer(sum)
if (min.size == k) min.poll()
if (max.size == k) max.poll()
}
var res: Long = 0
while (max.isNotEmpty()) {
res += min.poll() - max.poll()!!
}
return res
}
}