[Swift]LeetCode1131. 绝对值表达式的最大值 | Maximum of Absolute Value Expression

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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

---

给你两个长度相等的整数数组，返回下面表达式的最大值：

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

其中下标 i，j 满足 0 <= i, j < arr1.length。

示例 1：

输入：arr1 = [1,2,3,4], arr2 = [-1,4,5,6]
输出：13

示例 2：

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

提示：

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

---

192ms

class Solution {
    func maxAbsValExpr(_ arr1: [Int], _ arr2: [Int]) -> Int {
        let n = arr1.count
        var closet = 0
        var cur = 0
        var res = 0
        for p in [-1,1] {
            for q in [-1, 1] {
                closet = arr1[0] * p + arr2[0] * q
                for i in 1 ..< n {
                    cur = arr1[i] * p + arr2[i] * q + i
                    res = max(res, cur - closet)
                    closet = min(closet, cur)
                }
            }
        }
        return res
    }
}

---

Runtime: 196 ms

Memory Usage: 21.3 MB

class Solution {
    func maxAbsValExpr(_ arr1: [Int], _ arr2: [Int]) -> Int {
        var n:Int = arr1.count
        var ans:Int = 0
        var maxmm:Int = Int.min / 2
        var maxmp:Int = Int.min / 2
        var maxpm:Int = Int.min / 2
        var maxpp:Int = Int.min / 2
        for i in 0..<n
        {
            maxmm = max(maxmm, -arr1[i] - arr2[i]-i)
            maxmp = max(maxmp, -arr1[i] + arr2[i]-i)
            maxpm = max(maxpm, arr1[i] - arr2[i]-i)
            maxpp = max(maxpp, arr1[i] + arr2[i]-i)
            ans = max(ans, maxmm + arr1[i] + arr2[i]+i)
            ans = max(ans, maxmp + arr1[i] - arr2[i]+i)
            ans = max(ans, maxpm - arr1[i] + arr2[i]+i)
            ans = max(ans, maxpp - arr1[i] - arr2[i]+i)
        }
        return ans        
    }
}

---

204ms

class Solution {
    func maxAbsValExpr(_ arr1: [Int], _ arr2: [Int]) -> Int {
        var ans = 0
        
        for i in [-1,1] {
            for j in [-1,1]{
                var best = arr1[0] * i + arr2[0] * j
                for k in 1..<arr1.count {
                    ans = max(ans, arr1[k]*i + arr2[k]*j + k - best)
                    best = min(best, arr1[k]*i + arr2[k]*j + k)
                }
            }
        }
        return ans
    }
}

---

224ms

class Solution {
    func maxAbsValExpr(_ arr1: [Int], _ arr2: [Int]) -> Int {
        var res = 0
        var maxmm = Int.min
        var maxmp = Int.min
        var maxpm = Int.min
        var maxpp = Int.min
        
        var i = 0
        while i < arr1.count {
            maxmm = max(maxmm, -arr1[i] - arr2[i] - i)
            maxmp = max(maxmp, -arr1[i] + arr2[i] - i)
            maxpm = max(maxpm, arr1[i] - arr2[i] - i)
            maxpp = max(maxpp, arr1[i] + arr2[i] - i)
            res = max(res, maxmm + arr1[i] + arr2[i] + i)
            res = max(res, maxmp + arr1[i] - arr2[i] + i)
            res = max(res, maxpm - arr1[i] + arr2[i] + i)
            res = max(res, maxpp - arr1[i] - arr2[i] + i)
            
            i += 1
        }
        
        return res;
    }
}

---

252ms

class Solution {
    func maxAbsValExpr(_ arr1: [Int], _ arr2: [Int]) -> Int {
        var aux = [Int](repeating: 0, count: arr1.count)
        var result = Int.min
        for j in [-1, 1] {
            for k in [-1, 1] {
                
                for i in stride(from: 0, to: arr1.count, by: 1) {
                    aux[i] = arr1[i] * j + arr2[i] * k + i
                }
                let iPart = aux.max()!
                let jPart = aux.min()!
                result = max(result, iPart - jPart)
            }
        }
        return result
    }
}