<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 {
    int dif(vector<int> s){
        int mn= INT_MAX, mx=INT_MIN;
        for(auto a :s){
            mn=min(mn, a);
            mx=max(mx, a);
        }
        return mx-mn;
    }
public:
    int maxAbsValExpr(vector<int>& a, vector<int>& b) {
        vector<int> s1,s2,d1,d2;
        for(int i=0; i < a.size();i++){
            s1.push_back(a[i]+b[i]+i);
            s2.push_back(a[i]+b[i]-i);
            d1.push_back(a[i]-b[i]+i);
            d2.push_back(a[i]-b[i]-i);
        }
        int dif1=dif(s1);
        int dif2=dif(s2);
        int dif3=dif(d1);
        int dif4=dif(d2);
        return max(max(dif1,dif2),max(dif3,dif4));
    }
};
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