<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:
    def putMarbles(self, weights: List[int], k: int) -> int:
        #first and last element always in the sum
        #then pick k-1 pairs
        pairs = []
        for i in range(len(weights)-1):
            pairs.append((weights[i], weights[i+1]))
        pairs.sort(key=lambda x: x[0] + x[1])
        min_ans = weights[0] + weights[-1] + sum([a[0]+a[1] for a in pairs[:k-1]])
        max_ans = weights[0] + weights[-1] + sum([a[0] + a[1] for a in pairs[len(pairs)-k+1:]])
        return max_ans - min_ans
        """
        #dp twice for min and max
        #dp[a][b] = best score for weights[:a+1] with b bags
        min_dp = [[float("inf")] * (k+1) for _ in range(len(weights))]
        max_dp = [[0] * (k+1) for _ in range(len(weights))]
        for i in range(len(weights)):
            max_dp[i][0] = 0
            max_dp[i][1] = weights[0] + weights[i]
            min_dp[i][0] = 0
            min_dp[i][1] = weights[0] + weights[i]
        min_dp[0][1] = 2 * weights[0]
        max_dp[0][1] = 2 * weights[0]
        for b in range(2, k+1):
            for a in range(1, len(weights)):
                if a+1 >= b:
                    for i in range(a):
                        min_dp[a][b] = min(min_dp[a][b], weights[a] + weights[i+1] + min_dp[i][b-1])
                        max_dp[a][b] = max(max_dp[a][b], weights[a] + weights[i+1] + max_dp[i][b-1])
        
        return max_dp[-1][-1] - min_dp[-1][-1]  
        """
                

        
627