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Java-based LeetCode algorithm problem solutions, regularly updated
package g2501_2600.s2551_put_marbles_in_bags;
// #Hard #Array #Sorting #Greedy #Heap_Priority_Queue
// #2023_08_15_Time_33_ms_(99.82%)_Space_55.1_MB_(97.58%)
import java.util.Arrays;
/**
* 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 ith
marble and jth
marble are in a bag, then all marbles with an index between the ith
and jth
indices should also be in that same bag.
* * If a bag consists of all the marbles with an index from `i` to `j` inclusively, then the cost of the bag is `weights[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 <= 105
* * 1 <= weights[i] <= 109
**/
public class Solution {
public long putMarbles(int[] weights, int k) {
long minAns = weights[0] + (long) weights[weights.length - 1];
long maxAns = weights[0] + (long) weights[weights.length - 1];
long[] arr = new long[weights.length - 1];
for (int i = 1; i < weights.length; i++) {
arr[i - 1] = weights[i] + (long) weights[i - 1];
}
Arrays.sort(arr);
for (int i = 0; i < k - 1; i++) {
minAns = minAns + arr[i];
}
for (int i = arr.length - 1; i > arr.length - k; i--) {
maxAns = maxAns + arr[i];
}
return maxAns - minAns;
}
}