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2036. Maximum Alternating Subarray Sum πŸ”’

Description

A subarray of a 0-indexed integer array is a contiguous non-empty sequence of elements within an array.

The alternating subarray sum of a subarray that ranges from index i to j (inclusive, 0 <= i <= j < nums.length) is nums[i] - nums[i+1] + nums[i+2] - ... +/- nums[j].

Given a 0-indexed integer array nums, return the maximum alternating subarray sum of any subarray of nums.

 

Example 1:

Input: nums = [3,-1,1,2]
Output: 5
Explanation:
The subarray [3,-1,1] has the largest alternating subarray sum.
The alternating subarray sum is 3 - (-1) + 1 = 5.

Example 2:

Input: nums = [2,2,2,2,2]
Output: 2
Explanation:
The subarrays [2], [2,2,2], and [2,2,2,2,2] have the largest alternating subarray sum.
The alternating subarray sum of [2] is 2.
The alternating subarray sum of [2,2,2] is 2 - 2 + 2 = 2.
The alternating subarray sum of [2,2,2,2,2] is 2 - 2 + 2 - 2 + 2 = 2.

Example 3:

Input: nums = [1]
Output: 1
Explanation:
There is only one non-empty subarray, which is [1].
The alternating subarray sum is 1.

 

Constraints:

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

Solutions

Solution 1: Dynamic Programming

We define \(f\) as the maximum sum of the alternating subarray ending with \(nums[i]\), and define \(g\) as the maximum sum of the alternating subarray ending with \(-nums[i]\). Initially, both \(f\) and \(g\) are \(-\infty\).

Next, we traverse the array \(nums\). For position \(i\), we need to maintain the values of \(f\) and \(g\), i.e., \(f = \max(g, 0) + nums[i]\), and \(g = f - nums[i]\). The answer is the maximum value among all \(f\) and \(g\).

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

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class Solution:
    def maximumAlternatingSubarraySum(self, nums: List[int]) -> int:
        ans = f = g = -inf
        for x in nums:
            f, g = max(g, 0) + x, f - x
            ans = max(ans, f, g)
        return ans

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class Solution {
    public long maximumAlternatingSubarraySum(int[] nums) {
        final long inf = 1L << 60;
        long ans = -inf, f = -inf, g = -inf;
        for (int x : nums) {
            long ff = Math.max(g, 0) + x;
            g = f - x;
            f = ff;
            ans = Math.max(ans, Math.max(f, g));
        }
        return ans;
    }
}

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class Solution {
public:
    long long maximumAlternatingSubarraySum(vector<int>& nums) {
        using ll = long long;
        const ll inf = 1LL << 60;
        ll ans = -inf, f = -inf, g = -inf;
        for (int x : nums) {
            ll ff = max(g, 0LL) + x;
            g = f - x;
            f = ff;
            ans = max({ans, f, g});
        }
        return ans;
    }
};

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func maximumAlternatingSubarraySum(nums []int) int64 {
    const inf = 1 << 60
    ans, f, g := -inf, -inf, -inf
    for _, x := range nums {
        f, g = max(g, 0)+x, f-x
        ans = max(ans, max(f, g))
    }
    return int64(ans)
}

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function maximumAlternatingSubarraySum(nums: number[]): number {
    let [ans, f, g] = [-Infinity, -Infinity, -Infinity];
    for (const x of nums) {
        [f, g] = [Math.max(g, 0) + x, f - x];
        ans = Math.max(ans, f, g);
    }
    return ans;
}

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