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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) });
    }
};

  • class Solution:
    def maxAbsValExpr(self, arr1: List[int], arr2: List[int]) -> int:
        dirs = (1, -1, -1, 1, 1)
        ans = -inf
        for a, b in pairwise(dirs):
            mx, mi = -inf, inf
            for i, (x, y) in enumerate(zip(arr1, arr2)):
                mx = max(mx, a * x + b * y + i)
                mi = min(mi, a * x + b * y + i)
                ans = max(ans, mx - mi)
        return ans


  • func maxAbsValExpr(arr1 []int, arr2 []int) int {
	dirs := [5]int{1, -1, -1, 1, 1}
	const inf = 1 << 30
	ans := -inf
	for k := 0; k < 4; k++ {
		a, b := dirs[k], dirs[k+1]
		mx, mi := -inf, inf
		for i, x := range arr1 {
			y := arr2[i]
			mx = max(mx, a*x+b*y+i)
			mi = min(mi, a*x+b*y+i)
			ans = max(ans, mx-mi)
		}
	}
	return ans
}

func max(a, b int) int {
	if a > b {
		return a
	}
	return b
}

func min(a, b int) int {
	if a < b {
		return a
	}
	return b
}

  • function maxAbsValExpr(arr1: number[], arr2: number[]): number {
    const dirs = [1, -1, -1, 1, 1];
    const inf = 1 << 30;
    let ans = -inf;
    for (let k = 0; k < 4; ++k) {
        const [a, b] = [dirs[k], dirs[k + 1]];
        let mx = -inf;
        let mi = inf;
        for (let i = 0; i < arr1.length; ++i) {
            const [x, y] = [arr1[i], arr2[i]];
            mx = Math.max(mx, a * x + b * y + i);
            mi = Math.min(mi, a * x + b * y + i);
            ans = Math.max(ans, mx - mi);
        }
    }
    return ans;
}

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