# 1330. Reverse Subarray To Maximize Array Value

## Description

You are given an integer array nums. The value of this array is defined as the sum of |nums[i] - nums[i + 1]| for all 0 <= i < nums.length - 1.

You are allowed to select any subarray of the given array and reverse it. You can perform this operation only once.

Find maximum possible value of the final array.

Example 1:

Input: nums = [2,3,1,5,4]
Output: 10
Explanation: By reversing the subarray [3,1,5] the array becomes [2,5,1,3,4] whose value is 10.


Example 2:

Input: nums = [2,4,9,24,2,1,10]
Output: 68


Constraints:

• 1 <= nums.length <= 3 * 104
• -105 <= nums[i] <= 105

## Solutions

• class Solution {
public int maxValueAfterReverse(int[] nums) {
int n = nums.length;
int s = 0;
for (int i = 0; i < n - 1; ++i) {
s += Math.abs(nums[i] - nums[i + 1]);
}
int ans = s;
for (int i = 0; i < n - 1; ++i) {
ans = Math.max(
ans, s + Math.abs(nums[0] - nums[i + 1]) - Math.abs(nums[i] - nums[i + 1]));
ans = Math.max(
ans, s + Math.abs(nums[n - 1] - nums[i]) - Math.abs(nums[i] - nums[i + 1]));
}
int[] dirs = {1, -1, -1, 1, 1};
final int inf = 1 << 30;
for (int k = 0; k < 4; ++k) {
int k1 = dirs[k], k2 = dirs[k + 1];
int mx = -inf, mi = inf;
for (int i = 0; i < n - 1; ++i) {
int a = k1 * nums[i] + k2 * nums[i + 1];
int b = Math.abs(nums[i] - nums[i + 1]);
mx = Math.max(mx, a - b);
mi = Math.min(mi, a + b);
}
ans = Math.max(ans, s + Math.max(0, mx - mi));
}
return ans;
}
}

• class Solution {
public:
int maxValueAfterReverse(vector<int>& nums) {
int n = nums.size();
int s = 0;
for (int i = 0; i < n - 1; ++i) {
s += abs(nums[i] - nums[i + 1]);
}
int ans = s;
for (int i = 0; i < n - 1; ++i) {
ans = max(ans, s + abs(nums[0] - nums[i + 1]) - abs(nums[i] - nums[i + 1]));
ans = max(ans, s + abs(nums[n - 1] - nums[i]) - abs(nums[i] - nums[i + 1]));
}
int dirs[5] = {1, -1, -1, 1, 1};
const int inf = 1 << 30;
for (int k = 0; k < 4; ++k) {
int k1 = dirs[k], k2 = dirs[k + 1];
int mx = -inf, mi = inf;
for (int i = 0; i < n - 1; ++i) {
int a = k1 * nums[i] + k2 * nums[i + 1];
int b = abs(nums[i] - nums[i + 1]);
mx = max(mx, a - b);
mi = min(mi, a + b);
}
ans = max(ans, s + max(0, mx - mi));
}
return ans;
}
};

• class Solution:
def maxValueAfterReverse(self, nums: List[int]) -> int:
ans = s = sum(abs(x - y) for x, y in pairwise(nums))
for x, y in pairwise(nums):
ans = max(ans, s + abs(nums[0] - y) - abs(x - y))
ans = max(ans, s + abs(nums[-1] - x) - abs(x - y))
for k1, k2 in pairwise((1, -1, -1, 1, 1)):
mx, mi = -inf, inf
for x, y in pairwise(nums):
a = k1 * x + k2 * y
b = abs(x - y)
mx = max(mx, a - b)
mi = min(mi, a + b)
ans = max(ans, s + max(mx - mi, 0))
return ans


• func maxValueAfterReverse(nums []int) int {
s, n := 0, len(nums)
for i, x := range nums[:n-1] {
y := nums[i+1]
s += abs(x - y)
}
ans := s
for i, x := range nums[:n-1] {
y := nums[i+1]
ans = max(ans, s+abs(nums[0]-y)-abs(x-y))
ans = max(ans, s+abs(nums[n-1]-x)-abs(x-y))
}
dirs := [5]int{1, -1, -1, 1, 1}
const inf = 1 << 30
for k := 0; k < 4; k++ {
k1, k2 := dirs[k], dirs[k+1]
mx, mi := -inf, inf
for i, x := range nums[:n-1] {
y := nums[i+1]
a := k1*x + k2*y
b := abs(x - y)
mx = max(mx, a-b)
mi = min(mi, a+b)
}
ans = max(ans, s+max(mx-mi, 0))
}
return ans
}

func abs(x int) int {
if x < 0 {
return -x
}
return x
}

• function maxValueAfterReverse(nums: number[]): number {
const n = nums.length;
let s = 0;
for (let i = 0; i < n - 1; ++i) {
s += Math.abs(nums[i] - nums[i + 1]);
}
let ans = s;
for (let i = 0; i < n - 1; ++i) {
const d = Math.abs(nums[i] - nums[i + 1]);
ans = Math.max(ans, s + Math.abs(nums[0] - nums[i + 1]) - d);
ans = Math.max(ans, s + Math.abs(nums[n - 1] - nums[i]) - d);
}
const dirs = [1, -1, -1, 1, 1];
const inf = 1 << 30;
for (let k = 0; k < 4; ++k) {
let mx = -inf;
let mi = inf;
for (let i = 0; i < n - 1; ++i) {
const a = dirs[k] * nums[i] + dirs[k + 1] * nums[i + 1];
const b = Math.abs(nums[i] - nums[i + 1]);
mx = Math.max(mx, a - b);
mi = Math.min(mi, a + b);
}
ans = Math.max(ans, s + Math.max(0, mx - mi));
}
return ans;
}