<p>Given two arrays of integers with equal lengths, return the maximum value of:</p>

<p><code>|arr1[i] - arr1[j]| + |arr2[i] - arr2[j]| + |i - j|</code></p>

<p>where the maximum is taken over all <code>0 &lt;= i, j &lt; arr1.length</code>.</p>

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

<pre>
<strong>Input:</strong> arr1 = [1,2,3,4], arr2 = [-1,4,5,6]
<strong>Output:</strong> 13
</pre>

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

<pre>
<strong>Input:</strong> arr1 = [1,-2,-5,0,10], arr2 = [0,-2,-1,-7,-4]
<strong>Output:</strong> 20
</pre>

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

<ul>
	<li><code>2 &lt;= arr1.length == arr2.length &lt;= 40000</code></li>
	<li><code>-10^6 &lt;= arr1[i], arr2[i] &lt;= 10^6</code></li>
</ul>

class Solution {
public:
    int maxAbsValExpr(vector<int>& arr1, vector<int>& arr2) {
        int res = 0, n = arr1.size(), smallest, cur;
        for (int p : {1, -1}) {
            for (int q : {1, -1}) {
                smallest = p * arr1[0] + q * arr2[0] + 0;
                for (int i = 1; i < n; ++i) {
                    cur = p * arr1[i] + q * arr2[i] + i;
                    res = max(res, cur - smallest);
                    smallest = min(smallest, cur);
                }
            }
        }
        return res;
    }
};
#pragma GCC optimize("Ofast,unroll-loops") 
int speed_up = []{
    ios::sync_with_stdio(false);
    cin.tie(NULL);cout.tie(NULL);
    return 0;
}();
27000