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Leetcode 2765. Longest Alternating Subarray

Problem Statement

You need to determine the longest alternating subarray in a given integer array. An alternating subarray is defined as a subarray where the elements alternately increase and decrease.

For example, in the array [1, 3, 2, 4], the longest alternating subarray is [1, 3, 2, 4].

Clarifying Questions

To fully understand the problem, let’s consider some clarifying questions:

  1. What is the definition of an “alternating subarray”?
    • It’s a subarray where elements alternately increase and decrease in value.
  2. Can the elements in the array be negative or zero?
    • Yes, the array can contain negative numbers, positive numbers, and zeros.
  3. Is a single element considered an alternating subarray?
    • No, we need at least two elements to alternate.
  4. Are there any constraints on the size of the array?
    • The typical constraints of a LeetCode problem would apply unless specified otherwise.

Strategy

To find the longest alternating subarray, we can use the following strategy:

  1. Iterate through the array while maintaining a current alternating subarray.
  2. Keep a counter to track the length of the current alternating subarray.
  3. Use another variable to keep track of the maximum length found.
  4. Compare current and previous elements to check if they alternate.
  5. If they do alternate, increase the counter.
  6. If they do not alternate, reset the counter.
  7. Throughout the iteration, update the maximum length whenever the current length surpasses it.

Code

Here’s the Java implementation based on the above strategy:

public class Solution {
    public int longestAlternatingSubarray(int[] nums) {
        if (nums == null || nums.length == 0) {
            return 0;
        }
        
        int maxLength = 1; // at least one element
        int currentLength = 1;

        for (int i = 1; i < nums.length; i++) {
            // Check if the current element alternates with the previous one
            if ((nums[i] > nums[i - 1] && (i <= 1 || nums[i - 1] < nums[i - 2])) || 
                (nums[i] < nums[i - 1] && (i <= 1 || nums[i - 1] > nums[i - 2]))) {
                currentLength++;
            } else {
                currentLength = 2; // reset to at least the last two elements
            }
            
            maxLength = Math.max(maxLength, currentLength);
        }

        return maxLength;
    }

    public static void main(String[] args) {
        Solution solution = new Solution();
        int[] nums = {1, 3, 2, 4, 3, 5};
        System.out.println(solution.longestAlternatingSubarray(nums)); // Output: 6
    }
}

Time Complexity

The time complexity of this solution is O(n), where n is the length of the input array. This is because we only make a single pass through the array, checking elements and updating the current and maximum lengths as we go.

The space complexity is O(1), as we are using a constant amount of extra space regardless of the input size.

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