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2765. Longest Alternating Subarray

Description

You are given a 0-indexed integer array nums. A subarray s of length m is called alternating if:

  • m is greater than 1.
  • s1 = s0 + 1.
  • The 0-indexed subarray s looks like [s0, s1, s0, s1,...,s(m-1) % 2]. In other words, s1 - s0 = 1, s2 - s1 = -1, s3 - s2 = 1, s4 - s3 = -1, and so on up to s[m - 1] - s[m - 2] = (-1)m.

Return the maximum length of all alternating subarrays present in nums or -1 if no such subarray exists.

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

 

Example 1:

Input: nums = [2,3,4,3,4]
Output: 4
Explanation: The alternating subarrays are [3,4], [3,4,3], and [3,4,3,4]. The longest of these is [3,4,3,4], which is of length 4.

Example 2:

Input: nums = [4,5,6]
Output: 2
Explanation: [4,5] and [5,6] are the only two alternating subarrays. They are both of length 2.

 

Constraints:

  • 2 <= nums.length <= 100
  • 1 <= nums[i] <= 104

Solutions

Solution 1: Enumeration

We can enumerate the left endpoint $i$ of the subarray, and for each $i$, we need to find the longest subarray that satisfies the condition. We can start traversing to the right from $i$, and each time we encounter adjacent elements whose difference does not satisfy the alternating condition, we find a subarray that satisfies the condition. We can use a variable $k$ to record whether the difference of the current element should be $1$ or $-1$. If the difference of the current element should be $-k$, then we take the opposite of $k$. When we find a subarray $nums[i..j]$ that satisfies the condition, we update the answer to $\max(ans, j - i + 1)$.

The time complexity is $O(n^2)$, where $n$ is the length of the array. We need to enumerate the left endpoint $i$ of the subarray, and for each $i$, we need $O(n)$ time to find the longest subarray that satisfies the condition. The space complexity is $O(1)$.

  • class Solution {
    public int alternatingSubarray(int[] nums) {
        int ans = -1, n = nums.length;
        for (int i = 0; i < n; ++i) {
            int k = 1;
            int j = i;
            for (; j + 1 < n && nums[j + 1] - nums[j] == k; ++j) {
                k *= -1;
            }
            if (j - i + 1 > 1) {
                ans = Math.max(ans, j - i + 1);
            }
        }
        return ans;
    }
}

  • class Solution {
public:
    int alternatingSubarray(vector<int>& nums) {
        int ans = -1, n = nums.size();
        for (int i = 0; i < n; ++i) {
            int k = 1;
            int j = i;
            for (; j + 1 < n && nums[j + 1] - nums[j] == k; ++j) {
                k *= -1;
            }
            if (j - i + 1 > 1) {
                ans = max(ans, j - i + 1);
            }
        }
        return ans;
    }
};

  • class Solution:
    def alternatingSubarray(self, nums: List[int]) -> int:
        ans, n = -1, len(nums)
        for i in range(n):
            k = 1
            j = i
            while j + 1 < n and nums[j + 1] - nums[j] == k:
                j += 1
                k *= -1
            if j - i + 1 > 1:
                ans = max(ans, j - i + 1)
        return ans


  • func alternatingSubarray(nums []int) int {
	ans, n := -1, len(nums)
	for i := range nums {
		k := 1
		j := i
		for ; j+1 < n && nums[j+1]-nums[j] == k; j++ {
			k *= -1
		}
		if t := j - i + 1; t > 1 && ans < t {
			ans = t
		}
	}
	return ans
}

  • function alternatingSubarray(nums: number[]): number {
    let ans = -1;
    const n = nums.length;
    for (let i = 0; i < n; ++i) {
        let k = 1;
        let j = i;
        for (; j + 1 < n && nums[j + 1] - nums[j] === k; ++j) {
            k *= -1;
        }
        if (j - i + 1 > 1) {
            ans = Math.max(ans, j - i + 1);
        }
    }
    return ans;
}

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