<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 maxarray(int a[],int n) {
        
        int max = a[0];
        int min = a[0]; 
        for(int i=1;i< n ;i++) { 
            if(a[i]>max) max = a[i] ;

            if(a[i]<min) min = a[i] ;
        }
        return max - min ; 
    }
    int maxAbsValExpr(vector<int>& arr1, vector<int>& arr2) {
        int sumarr1[arr1.size()];
        int sumarr2[arr1.size()];
        int sumarr3[arr1.size()];
        int sumarr4[arr1.size()];
        int len = arr1.size();
        for(int i=0;i<arr1.size() ; i++ ) {
            sumarr1[i] = arr1[i] + arr2[i] + i ;
            sumarr2[i] = arr1[i] + arr2[i] - i ;
            sumarr3[i] = arr1[i] - arr2[i] + i ;
            sumarr4[i] = arr1[i] - arr2[i] - i ;
        }
         
        
        return max(max(maxarray(sumarr1,len),maxarray(sumarr2,len)),max(maxarray(sumarr3,len),maxarray(sumarr4,len))); 
    } 
};
27800