Formatted question description: https://leetcode.ca/all/1131.html

1131. Maximum of Absolute Value Expression

Level

Medium

Description

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

Solution

For the expression |arr1[i] - arr1[j]| + |arr2[i] - arr2[j]| + |i - j|, there are 4 cases as follows.

  1. arr1[i] - arr1[j] >= 0 and arr2[i] - arr2[j] >= 0. The expression is equivalent to (arr1[i] + arr2[i]) - (arr1[j] + arr2[j]) + |i - j|.
  2. arr1[i] - arr1[j] >= 0 and arr2[i] - arr2[j] < 0. The expression is equivalent to (arr1[i] - arr2[i]) - (arr1[j] - arr2[j]) + |i - j|.
  3. arr1[i] - arr1[j] < 0 and arr2[i] - arr2[j] >= 0. The expression is equivalent to (-arr1[i] + arr2[i]) - (-arr1[j] + arr2[j]) + |i - j|.
  4. arr1[i] - arr1[j] < 0 and arr2[i] - arr2[j] < 0. The expression is equivalent to (-arr1[i] - arr2[i]) - (-arr1[j] - arr2[j]) + |i - j|.

Each case can be represented in the form arr[i] - arr[j] + |i - j|. In the form, either i >= j or i < j.

So loop over all possible cases and find the maximum value.

  • class Solution {
        public int maxAbsValExpr(int[] arr1, int[] arr2) {
            int length = arr1.length;
            int[][] nums = new int[4][length];
            for (int i = 0; i < length; i++) {
                nums[0][i] = arr1[i] + arr2[i];
                nums[1][i] = arr1[i] - arr2[i];
                nums[2][i] = -arr1[i] + arr2[i];
                nums[3][i] = -arr1[i] - arr2[i];
            }
            int maxValue = 0;
            for (int i = 0; i < 4; i++) {
                int curValue = maxValExpr(nums[i]);
                maxValue = Math.max(maxValue, curValue);
            }
            return maxValue;
        }
    
        public int maxValExpr(int[] arr) {
            int length = arr.length;
            int[] arr1 = new int[length];
            int[] arr2 = new int[length];
            for (int i = 0; i < length; i++) {
                arr1[i] = i + arr[i];
                arr2[i] = i - arr[i];
            }
            int maxDifference1 = maxDifference(arr1);
            int maxDifference2 = maxDifference(arr2);
            return Math.max(maxDifference1, maxDifference2);
        }
    
        public int maxDifference(int[] arr) {
            int min = Integer.MAX_VALUE;
            int length = arr.length;
            int maxDifference = Integer.MIN_VALUE;
            for (int i = 0; i < length; i++) {
                if (arr[i] < min)
                    min = arr[i];
                else if (arr[i] - min > maxDifference)
                    maxDifference = arr[i] - min;
            }
            return maxDifference;
        }
    }
    
  • // OJ: https://leetcode.com/problems/maximum-of-absolute-value-expression/
    // Time: O(N)
    // Space: O(N)
    // Ref: https://leetcode.com/problems/maximum-of-absolute-value-expression/discuss/340075/c%2B%2B-beats-100-(both-time-and-memory)-with-algorithm-and-image
    class Solution {
        int maxDiff(vector<int> &A) {
            int mx = A[0], mn = A[0];
            for (int n : A) {
                mx = max(mx, n);
                mn = min(mn, n);
            }
            return mx - mn;
        }
    public:
        int maxAbsValExpr(vector<int>& A, vector<int>& B) {
            int N = A.size();
            vector<int> a(N), b(N), c(N), d(N);
            for (int i = 0; i < A.size(); ++i) {
                a[i] = A[i] + B[i] + i;
                b[i] = A[i] + B[i] - i;
                c[i] = A[i] - B[i] + i;
                d[i] = A[i] - B[i] - i;
            }
            return max({ maxDiff(a), maxDiff(b), maxDiff(c), maxDiff(d) });
        }
    };
    
  • print("Todo!")
    

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