Maximum of Absolute Value Expression

Given two arrays of integers with equal lengths, return the maximum value of:

|arr1[i] - arr1[j]| + |arr2[i] - arr2[j]| + |i - j|

where the maximum is taken over all 0 <= i, j < arr1.length.

Example 1:

Input: arr1 = [1,2,3,4], arr2 = [-1,4,5,6]
Output: 13

Example 2:

Input: arr1 = [1,-2,-5,0,10], arr2 = [0,-2,-1,-7,-4]
Output: 20

Constraints:

• 2 <= arr1.length == arr2.length <= 40000
• -10^6 <= arr1[i], arr2[i] <= 10^6

思路：這是個數學題，如果i < j的情況下，把絕對值的公式按照4種情況展開會得到如下的情況：

(a1i + a2i - i ) - (a1j + a2j - j)

(a1i - a2i - i) - (a1j - a2j - j)

(a2i - a1i - i) -  (a2j - a1j - j)

(a1j + a2j + j) - (a1i + a2i + i)

注意前面和後面是相同的index，也就是兩者是同一個東西；如果要求最大值，那麼就是前面最大，後面最小。

那麼問題轉換爲求這四種情況下的最大值減去最小值的最大；

class Solution {
    public int maxAbsValExpr(int[] arr1, int[] arr2) {
        int min1 = Integer.MAX_VALUE;
        int min2 = Integer.MAX_VALUE;
        int min3 = Integer.MAX_VALUE;
        int min4 = Integer.MAX_VALUE;
        
        int max1 = Integer.MIN_VALUE;
        int max2 = Integer.MIN_VALUE;
        int max3 = Integer.MIN_VALUE;
        int max4 = Integer.MIN_VALUE;
        
        for(int i = 0; i < arr1.length; i++) {
            min1 = Math.min(min1, arr1[i] + arr2[i] - i);
            max1 = Math.max(max1, arr1[i] + arr2[i] - i);
            
            min2 = Math.min(min2, arr1[i] - arr2[i] - i);
            max2 = Math.max(max2, arr1[i] - arr2[i] - i);
            
            min3 = Math.min(min3, arr2[i] - arr1[i] - i);
            max3 = Math.max(max3, arr2[i] - arr1[i] - i);
            
            min4 = Math.min(min4, arr1[i] + arr2[i] + i);
            max4 = Math.max(max4, arr1[i] + arr2[i] + i);
        }
        
        int res = Integer.MIN_VALUE;
        res = Math.max( Math.max(max1 - min1, max2 - min2), Math.max(max3 - min3, max4 - min4));
        return res;
        
    }
}