<p>You have <code>k</code> bags. You are given a <strong>0-indexed</strong> integer array <code>weights</code> where <code>weights[i]</code> is the weight of the <code>i<sup>th</sup></code> marble. You are also given the integer <code>k.</code></p>

<p>Divide the marbles into the <code>k</code> bags according to the following rules:</p>

<ul>
	<li>No bag is empty.</li>
	<li>If the <code>i<sup>th</sup></code> marble and <code>j<sup>th</sup></code> marble are in a bag, then all marbles with an index between the <code>i<sup>th</sup></code> and <code>j<sup>th</sup></code> indices should also be in that same bag.</li>
	<li>If a bag consists of all the marbles with an index from <code>i</code> to <code>j</code> inclusively, then the cost of the bag is <code>weights[i] + weights[j]</code>.</li>
</ul>

<p>The <strong>score</strong> after distributing the marbles is the sum of the costs of all the <code>k</code> bags.</p>

<p>Return <em>the <strong>difference</strong> between the <strong>maximum</strong> and <strong>minimum</strong> scores among marble distributions</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> weights = [1,3,5,1], k = 2
<strong>Output:</strong> 4
<strong>Explanation:</strong> 
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.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> weights = [1, 3], k = 2
<strong>Output:</strong> 0
<strong>Explanation:</strong> The only distribution possible is [1],[3]. 
Since both the maximal and minimal score are the same, we return 0.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= k &lt;= weights.length &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= weights[i] &lt;= 10<sup>9</sup></code></li>
</ul>

class Solution {
 public:
  long long putMarbles(const vector<int> &weights, const int k) {
    const int n = static_cast<int>(weights.size());
    if (k == 1 || k == n) {
      return 0;
    }

    int pair_sums[n - 1];
    memset(pair_sums, 0, sizeof(pair_sums));
    for (int i = 0; i < n - 1; ++i) {
      pair_sums[i] = weights[i] + weights[i + 1];
    }
    
    const int pairs = (k - 1) - max(0, ((k - 1) << 1) - (n - 1));
    nth_element(pair_sums, pair_sums + pairs, pair_sums + (n - 1));
    const long long small = accumulate(pair_sums, pair_sums + pairs, 0LL);
    nth_element(pair_sums, pair_sums + pairs, pair_sums + (n - 1), greater<>{});
    const long long great = accumulate(pair_sums, pair_sums + pairs, 0LL);
    return great - small;
  }
};
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