Best Sightseeing Pair
<p>You are given an integer array <code>values</code> where values[i] represents the value of the <code>i<sup>th</sup></code> sightseeing spot. Two sightseeing spots <code>i</code> and <code>j</code> have a <strong>distance</strong> <code>j - i</code> between them.</p>

<p>The score of a pair (<code>i &lt; j</code>) of sightseeing spots is <code>values[i] + values[j] + i - j</code>: the sum of the values of the sightseeing spots, minus the distance between them.</p>

<p>Return <em>the maximum score of a pair of sightseeing spots</em>.</p>

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

<pre>
<strong>Input:</strong> values = [8,1,5,2,6]
<strong>Output:</strong> 11
<strong>Explanation:</strong> i = 0, j = 2, values[i] + values[j] + i - j = 8 + 5 + 0 - 2 = 11
</pre>

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

<pre>
<strong>Input:</strong> values = [1,2]
<strong>Output:</strong> 2
</pre>

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

<ul>
	<li><code>2 &lt;= values.length &lt;= 5 * 10<sup>4</sup></code></li>
	<li><code>1 &lt;= values[i] &lt;= 1000</code></li>
</ul>
Array; Dynamic Programming
func maxScoreSightseeingPair(values []int) (ans int) {
	mx := 0
	for j, x := range values {
		ans = max(ans, mx+x-j)
		mx = max(mx, x+j)
	}
	return
}
