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feat: add solutions to lc problem: No.1131
No.1131. Maximum of Absolute Value Expression
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‎solution/1100-1199/1131.Maximum of Absolute Value Expression/README.md

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<!-- 这里可写通用的实现逻辑 -->
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**方法一：数学 + 枚举**
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我们不妨令 $x_i = arr1[i]$, $y_i = arr2[i]$，由于 $i$ 和 $j$ 的大小关系不影响表达式的值，我们不妨假设 $i \ge j$，那么表达式可以变为：
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$$
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| x_i - x_j | + | y_i - y_j | + i - j = \max \begin{cases} (x_i + y_i) - (x_j + y_j) \\ (x_i - y_i) - (x_j - y_j) \\ (-x_i + y_i) - (-x_j + y_j) \\ (-x_i - y_i) - (-x_j - y_j) \end{cases} + i - j\\
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= \max \begin{cases} (x_i + y_i + i) - (x_j + y_j + j) \\ (x_i - y_i + i) - (x_j - y_j + j) \\ (-x_i + y_i + i) - (-x_j + y_j + j) \\ (-x_i - y_i + i) - (-x_j - y_j + j) \end{cases}
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$$
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因此，我们只要求出 $a \times x_i + b \times y_i + i$ 的最大值 $mx$，以及最小值 $mi$，其中 $a, b \in \{-1, 1\}$。那么答案就是所有 $mx - mi$ 中的最大值。
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时间复杂度 $O(n)$，其中 $n$ 是数组长度。空间复杂度 $O(1)$。
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相似题目：
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- [1330. 翻转子数组得到最大的数组值](/solution/1300-1399/1330.Reverse%20Subarray%20To%20Maximize%20Array%20Value/README.md)
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<!-- tabs:start -->
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### **Python3**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```python
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class Solution:
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def maxAbsValExpr(self, arr1: List[int], arr2: List[int]) -> int:
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dirs = (1, -1, -1, 1, 1)
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ans = -inf
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for a, b in pairwise(dirs):
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mx, mi = -inf, inf
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for i, (x, y) in enumerate(zip(arr1, arr2)):
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mx = max(mx, a * x + b * y + i)
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mi = min(mi, a * x + b * y + i)
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ans = max(ans, mx - mi)
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return ans
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```
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### **Java**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```java
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class Solution {
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public int maxAbsValExpr(int[] arr1, int[] arr2) {
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int[] dirs = {1, -1, -1, 1, 1};
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final int inf = 1 << 30;
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int ans = -inf;
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int n = arr1.length;
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for (int k = 0; k < 4; ++k) {
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int a = dirs[k], b = dirs[k + 1];
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int mx = -inf, mi = inf;
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for (int i = 0; i < n; ++i) {
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mx = Math.max(mx, a * arr1[i] + b * arr2[i] + i);
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mi = Math.min(mi, a * arr1[i] + b * arr2[i] + i);
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ans = Math.max(ans, mx - mi);
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}
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}
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return ans;
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}
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}
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```
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### **C++**
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```cpp
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class Solution {
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public:
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int maxAbsValExpr(vector<int>& arr1, vector<int>& arr2) {
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int dirs[5] = {1, -1, -1, 1, 1};
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const int inf = 1 << 30;
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int ans = -inf;
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int n = arr1.size();
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for (int k = 0; k < 4; ++k) {
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int a = dirs[k], b = dirs[k + 1];
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int mx = -inf, mi = inf;
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for (int i = 0; i < n; ++i) {
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mx = max(mx, a * arr1[i] + b * arr2[i] + i);
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mi = min(mi, a * arr1[i] + b * arr2[i] + i);
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ans = max(ans, mx - mi);
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}
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}
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return ans;
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}
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};
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```
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### **Go**
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```go
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func maxAbsValExpr(arr1 []int, arr2 []int) int {
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dirs := [5]int{1, -1, -1, 1, 1}
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const inf = 1 << 30
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ans := -inf
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for k := 0; k < 4; k++ {
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a, b := dirs[k], dirs[k+1]
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mx, mi := -inf, inf
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for i, x := range arr1 {
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y := arr2[i]
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mx = max(mx, a*x+b*y+i)
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mi = min(mi, a*x+b*y+i)
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ans = max(ans, mx-mi)
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}
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}
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return ans
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}
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func max(a, b int) int {
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if a > b {
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return a
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}
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return b
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}
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func min(a, b int) int {
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if a < b {
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return a
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}
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return b
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}
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```
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### **TypeScript**
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```ts
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function maxAbsValExpr(arr1: number[], arr2: number[]): number {
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const dirs = [1, -1, -1, 1, 1];
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const inf = 1 << 30;
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let ans = -inf;
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for (let k = 0; k < 4; ++k) {
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const [a, b] = [dirs[k], dirs[k + 1]];
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let mx = -inf;
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let mi = inf;
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for (let i = 0; i < arr1.length; ++i) {
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const [x, y] = [arr1[i], arr2[i]];
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mx = Math.max(mx, a * x + b * y + i);
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mi = Math.min(mi, a * x + b * y + i);
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ans = Math.max(ans, mx - mi);
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}
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}
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return ans;
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}
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```
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### **...**

‎solution/1100-1199/1131.Maximum of Absolute Value Expression/README_EN.md

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### **Python3**
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```python
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class Solution:
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def maxAbsValExpr(self, arr1: List[int], arr2: List[int]) -> int:
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dirs = (1, -1, -1, 1, 1)
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ans = -inf
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for a, b in pairwise(dirs):
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mx, mi = -inf, inf
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for i, (x, y) in enumerate(zip(arr1, arr2)):
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mx = max(mx, a * x + b * y + i)
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mi = min(mi, a * x + b * y + i)
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ans = max(ans, mx - mi)
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return ans
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```
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### **Java**
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```java
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class Solution {
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public int maxAbsValExpr(int[] arr1, int[] arr2) {
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int[] dirs = {1, -1, -1, 1, 1};
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final int inf = 1 << 30;
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int ans = -inf;
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int n = arr1.length;
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for (int k = 0; k < 4; ++k) {
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int a = dirs[k], b = dirs[k + 1];
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int mx = -inf, mi = inf;
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for (int i = 0; i < n; ++i) {
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mx = Math.max(mx, a * arr1[i] + b * arr2[i] + i);
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mi = Math.min(mi, a * arr1[i] + b * arr2[i] + i);
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ans = Math.max(ans, mx - mi);
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}
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}
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return ans;
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}
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}
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```
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### **C++**
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```cpp
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class Solution {
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public:
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int maxAbsValExpr(vector<int>& arr1, vector<int>& arr2) {
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int dirs[5] = {1, -1, -1, 1, 1};
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const int inf = 1 << 30;
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int ans = -inf;
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int n = arr1.size();
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for (int k = 0; k < 4; ++k) {
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int a = dirs[k], b = dirs[k + 1];
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int mx = -inf, mi = inf;
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for (int i = 0; i < n; ++i) {
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mx = max(mx, a * arr1[i] + b * arr2[i] + i);
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mi = min(mi, a * arr1[i] + b * arr2[i] + i);
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ans = max(ans, mx - mi);
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}
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}
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return ans;
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}
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};
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```
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### **Go**
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```go
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func maxAbsValExpr(arr1 []int, arr2 []int) int {
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dirs := [5]int{1, -1, -1, 1, 1}
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const inf = 1 << 30
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ans := -inf
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for k := 0; k < 4; k++ {
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a, b := dirs[k], dirs[k+1]
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mx, mi := -inf, inf
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for i, x := range arr1 {
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y := arr2[i]
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mx = max(mx, a*x+b*y+i)
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mi = min(mi, a*x+b*y+i)
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ans = max(ans, mx-mi)
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}
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}
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return ans
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}
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func max(a, b int) int {
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if a > b {
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return a
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}
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return b
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}
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func min(a, b int) int {
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if a < b {
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return a
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}
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return b
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}
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```
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### **TypeScript**
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```ts
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function maxAbsValExpr(arr1: number[], arr2: number[]): number {
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const dirs = [1, -1, -1, 1, 1];
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const inf = 1 << 30;
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let ans = -inf;
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for (let k = 0; k < 4; ++k) {
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const [a, b] = [dirs[k], dirs[k + 1]];
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let mx = -inf;
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let mi = inf;
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for (let i = 0; i < arr1.length; ++i) {
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const [x, y] = [arr1[i], arr2[i]];
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mx = Math.max(mx, a * x + b * y + i);
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mi = Math.min(mi, a * x + b * y + i);
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ans = Math.max(ans, mx - mi);
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}
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}
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return ans;
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}
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```
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### **...**

‎solution/1100-1199/1131.Maximum of Absolute Value Expression/Solution.cpp

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

‎solution/1100-1199/1131.Maximum of Absolute Value Expression/Solution.go

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

‎solution/1100-1199/1131.Maximum of Absolute Value Expression/Solution.java

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

‎solution/1100-1199/1131.Maximum of Absolute Value Expression/Solution.py

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

‎solution/1100-1199/1131.Maximum of Absolute Value Expression/Solution.ts

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

‎solution/1300-1399/1330.Reverse Subarray To Maximize Array Value/README.md

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时间复杂度 $O(n)$，其中 $n$ 是数组 $nums$ 的长度。空间复杂度 $O(1)$。
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相似题目：
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- [1131. 绝对值表达式的最大值](/solution/1100-1199/1131.Maximum%20of%20Absolute%20Value%20Expression/README.md)
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<!-- tabs:start -->
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### **Python3**

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