<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 f(vector<int>& a, int n) { 
        int max_sum = a[0], min_sum = a[0];  
        for (int i = 0; i < n; i++){ 
            max_sum = max(max_sum, a[i]);
            min_sum = min(min_sum, a[i]);
        } 
        return (max_sum - min_sum); 
    } 
    
    int maxAbsValExpr(vector<int>& a, vector<int>& b) {
        int n = (int)a.size();
        vector<int> sum1(n,0), diff1(n,0), sum2(n,0), diff2(n,0);
        for(int i=0;i<n;i++) {
            sum1[i] = a[i] + b[i] + i;
            diff1[i] = a[i] - b[i] + i;
            sum2[i] = a[i] + b[i] - i;
            diff2[i] = a[i] - b[i] - i;
        }
        return max({f(sum1, n), f(sum2, n), f(diff1, n), f(diff2, n)});
    }
};
29000