<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) {
        int n=arr1.size();
        vector<int> temp(n, 0);
        vector<vector<int>> dirs = {{1, 1}, {1, -1}, {-1, 1}, {-1, -1}};
        int ans=0;
        for(auto dir: dirs){
            for(int i=0;i<n;i++){
                temp[i] = arr1[i]*dir[0]+arr2[i]*dir[1]+i;
            }
            int check = temp[0];
            for(int i=1;i<n;i++){
                ans=max(temp[i]-check, ans);
                check=min(check, temp[i]);
            }
        }
        return ans;
    }
};
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