<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) {
        vector<int> a, b, c, d;
        for (int i = 0; i < arr1.size(); ++i) {
            a.push_back(arr1[i] + arr2[i] + i);
            b.push_back(arr1[i] - arr2[i] + i);
            c.push_back(-arr1[i] + arr2[i] + i);
            d.push_back(-arr1[i] - arr2[i] + i);
            //e.push_back(-arr1[i] - arr2[i] + i);
        }
        sort(begin(a), end(a));
        sort(begin(b), end(b));
        sort(begin(c), end(c));
        sort(begin(d), end(d));
        return max({a.back() - a[0], b.back() - b[0], c.back() - c[0], d.back() - d[0]});    
    }
};
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