Question

1131. Maximum of Absolute Value Expression

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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

Solutions

Solution 1: Mathematics + Enumeration

Let's denote $x_i = arr1[i]$, $y_i = arr2[i]$. Since the size relationship between $i$ and $j$ does not affect the value of the expression, we can assume $i \ge j$. Then the expression can be transformed into:

$$ | x_i - x_j | + | y_i - y_j | + i - j = \max \begin{cases} (x_i + y_i) - (x_j + y_j) \ (x_i - y_i) - (x_j - y_j) \ (-x_i + y_i) - (-x_j + y_j) \ (-x_i - y_i) - (-x_j - y_j) \end{cases} + i - j\ = \max \begin{cases} (x_i + y_i + i) - (x_j + y_j + j) \ (x_i - y_i + i) - (x_j - y_j + j) \ (-x_i + y_i + i) - (-x_j + y_j + j) \ (-x_i - y_i + i) - (-x_j - y_j + j) \end{cases} $$

Therefore, we just need to find the maximum value $mx$ and the minimum value $mi$ of $a \times x_i + b \times y_i + i$, where $a, b \in {-1, 1}$. The answer is the maximum value among all $mx - mi$.

The time complexity is $O(n)$, where $n$ is the length of the array. The space complexity is $O(1)$.

Similar problems:

• 1330. Reverse Subarray To Maximize Array Value

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

class Solution {
  public int maxAbsValExpr(int[] arr1, int[] arr2) {
    int[] dirs = {1, -1, -1, 1, 1};
    final int inf = 1 << 30;
    int ans = -inf;
    int n = arr1.length;
    for (int k = 0; k < 4; ++k) {
      int a = dirs[k], b = dirs[k + 1];
      int mx = -inf, mi = inf;
      for (int i = 0; i < n; ++i) {
        mx = Math.max(mx, a * arr1[i] + b * arr2[i] + i);
        mi = Math.min(mi, a * arr1[i] + b * arr2[i] + i);
        ans = Math.max(ans, mx - mi);
      }
    }
    return ans;
  }
}

class Solution {
public:
  int maxAbsValExpr(vector<int>& arr1, vector<int>& arr2) {
    int dirs[5] = {1, -1, -1, 1, 1};
    const int inf = 1 << 30;
    int ans = -inf;
    int n = arr1.size();
    for (int k = 0; k < 4; ++k) {
      int a = dirs[k], b = dirs[k + 1];
      int mx = -inf, mi = inf;
      for (int i = 0; i < n; ++i) {
        mx = max(mx, a * arr1[i] + b * arr2[i] + i);
        mi = min(mi, a * arr1[i] + b * arr2[i] + 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
}

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;
}