Maximum Absolute Difference

/*You are given an array of N integers, A1, A2 ,…, AN. Return maximum value of f(i, j) for all 1 ≤ i, j ≤ N.
f(i, j) is defined as |A[i] - A[j]| + |i - j|, where |x| denotes absolute value of x.

For example,

A=[1, 3, -1]

f(1, 1) = f(2, 2) = f(3, 3) = 0
f(1, 2) = f(2, 1) = |1 - 3| + |1 - 2| = 3
f(1, 3) = f(3, 1) = |1 - (-1)| + |1 - 3| = 4
f(2, 3) = f(3, 2) = |3 - (-1)| + |2 - 3| = 5

So, we return 5.*/
    
    //this one is 0(n^2) solution
    int f_max = 0;
    for(int i=0 ;i< A.size() ; i++){
        for(int j=i+1 ; j<A.size() ; j++){
           int f = abs(A[i]-A[j])+(j-i);
           if(f > f_max) f_max = f;
        }
    }
    return f_max;


    //this one is 0(n) solution
    
int Solution::maxArr(vector<int> &A) {
  int max1 = INT_MIN, max2 = INT_MIN, max3 = INT_MIN, max4 = INT_MIN;
    int ans = INT_MIN;
    int size = A.size();
    for (auto i = 0; i<size; ++i)
    {
        max1 = max(max1, A[i] + i);
        max2 = max(max2, -A[i] + i);
        max3 = max(max3, A[i] - i);
        max4 = max(max4, -A[i] - i);
    }
    for (auto i = 0; i<size; ++i)
    {
        ans = max(ans, max1 - A[i] - i);
        ans = max(ans, max2 + A[i] - i);
        ans = max(ans, max3 - A[i] + i);
        ans = max(ans, max4 + A[i] + i);
    }
    return ans;
}