K-Concatenation Maximum Sum
<p>Given an integer array <code>arr</code> and an integer <code>k</code>, modify the array by repeating it <code>k</code> times.</p>

<p>For example, if <code>arr = [1, 2]</code> and <code>k = 3 </code>then the modified array will be <code>[1, 2, 1, 2, 1, 2]</code>.</p>

<p>Return the maximum sub-array sum in the modified array. Note that the length of the sub-array can be <code>0</code> and its sum in that case is <code>0</code>.</p>

<p>As the answer can be very large, return the answer <strong>modulo</strong> <code>10<sup>9</sup> + 7</code>.</p>

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

<pre>
<strong>Input:</strong> arr = [1,2], k = 3
<strong>Output:</strong> 9
</pre>

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

<pre>
<strong>Input:</strong> arr = [1,-2,1], k = 5
<strong>Output:</strong> 2
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> arr = [-1,-2], k = 7
<strong>Output:</strong> 0
</pre>

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

<ul>
	<li><code>1 &lt;= arr.length &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= k &lt;= 10<sup>5</sup></code></li>
	<li><code>-10<sup>4</sup> &lt;= arr[i] &lt;= 10<sup>4</sup></code></li>
</ul>
Array; Dynamic Programming
class Solution:
    def kConcatenationMaxSum(self, arr: List[int], k: int) -> int:
        s = mx_pre = mi_pre = mx_sub = 0
        for x in arr:
            s += x
            mx_pre = max(mx_pre, s)
            mi_pre = min(mi_pre, s)
            mx_sub = max(mx_sub, s - mi_pre)
        ans = mx_sub
        mod = 10**9 + 7
        if k == 1:
            return ans % mod
        mx_suf = s - mi_pre
        ans = max(ans, mx_pre + mx_suf)
        if s > 0:
            ans = max(ans, (k - 2) * s + mx_pre + mx_suf)
        return ans % mod
