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Hard
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Biweekly Contest 18 Q4
Greedy
Array
Math

1330. Reverse Subarray To Maximize Array Value

中文文档

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

Solution 1

Python3

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

Java

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;
    }
}

C++

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;
    }
};

Go

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
}

TypeScript

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;
}