LeetCode: Maximum of Absolute Value Expression

// https://code.dennyzhang.com/maximum-of-absolute-value-expression
// Basic Ideas:
//
//  Notice
//  1. To get max(|f(i)-f(j)|), find the smallest and biggest of f(i)
//  2. abs(A) + abs(B) = max(A+B, A-B, -A+B, -A-B)
//
//  max(|arr1[i] - arr1[j]| + |arr2[i] - arr2[j]| + |i - j|)
//      max(|(arr1[i]+arr2[i]+i)-(arr1[j]+arr2[j]+j|),
//          |(arr1[i]+arr2[i]-i)-(arr1[j]+arr2[j]-j)|,
//          |(arr1[i]-arr2[i]+i)-(arr1[j]-arr2[j]+j)|,
//          |(arr1[i]-arr2[i]-i)-(arr1[j]-arr2[j]-j)|)
//
//  Since we only need to choose two elements, we can sort the arrays.
//  But this turns out will not help to speed up the search
//
// Complexity: Time O(n), Space O(1)
// Basic Ideas: math
//
//  max(|arr1[i] - arr1[j]| + |arr2[i] - arr2[j]| + |i - j|)
//      max(|(arr1[i]+arr2[i]+i)-(arr1[j]+arr2[j]+j|),
//          |(arr1[i]+arr2[i]-i)-(arr1[j]+arr2[j]-j)|,
//          |(arr1[i]-arr2[i]+i)-(arr1[j]-arr2[j]+j)|,
//          |(arr1[i]-arr2[i]-i)-(arr1[j]-arr2[j]-j)|)
//
// Complexity: Time O(n), Space O(1)
func maxAbsValExpr(arr1 []int, arr2 []int) int {
    smalls := []int{1<<32-1, 1<<32-1, 1<<32-1, 1<<32-1}
    bigs := []int{-1<<32, -1<<32, -1<<32, -1<<32}
    for i, _ := range arr1 {
        for j, v := range []int{arr1[i]+arr2[i]+i, 
                               arr1[i]+arr2[i]-i,
                               arr1[i]-arr2[i]+i,
                                arr1[i]-arr2[i]-i} {
            if v < smalls[j] {
                smalls[j] = v
            }
            if v > bigs[j] {
                bigs[j] = v
            }
        }
    }
    res := 0
    for j:=0; j<=3; j++ {
        if bigs[j]-smalls[j] > res {
            res = bigs[j]-smalls[j]
        }
    }
    return res
}