Welcome to Subscribe On Youtube

2036. Maximum Alternating Subarray Sum

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 $O(1)$.

  • 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;
    }
}

  • 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;
    }
};

  • 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


  • 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)
}

  • 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;
}

All Problems

All Solutions