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
func kConcatenationMaxSum(arr []int, k int) int {
	var s, mxPre, miPre, mxSub int
	for _, x := range arr {
		s += x
		mxPre = max(mxPre, s)
		miPre = min(miPre, s)
		mxSub = max(mxSub, s-miPre)
	}
	const mod = 1e9 + 7
	ans := mxSub
	if k == 1 {
		return ans % mod
	}
	mxSuf := s - miPre
	ans = max(ans, mxSuf+mxPre)
	if s > 0 {
		ans = max(ans, mxSuf+(k-2)*s+mxPre)
	}
	return ans % mod
}