<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) {
        auto best = INT_MIN;
        for (auto i = -1; i <= 1; i += 2) {
            auto v = vector<int>{arr1};
            for (auto x = 0; x < arr1.size(); ++x) {
                v[x] += i * arr2[x];
            }
            for (auto j = -1; j <= 1; j += 2) {
                auto vv = vector<int>{v};
                for (auto x = 0; x < arr1.size(); ++x) {
                    vv[x] += j * x;
                }
                best =
                    std::max(best, *std::max_element(vv.begin(), vv.end()) -
                                       *std::min_element(vv.begin(), vv.end()));
            }
        }
        return best;
    }
};
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