LeetCode 3830 - Longest Alternating Subarray After Removing At Most One Element

The problem asks us to find the longest alternating subarray in a given integer array nums, where a subarray is defined as consecutive elements.

leetcodehardarraydynamic-programmingenumeration

LeetCode Problem 3830

Difficulty: 🔴 Hard
Topics: Array, Dynamic Programming, Enumeration

Solution

Problem Understanding

The problem asks us to find the longest alternating subarray in a given integer array nums, where a subarray is defined as consecutive elements. A subarray is considered alternating if the sequence of comparisons between adjacent elements strictly alternates between less-than and greater-than. Formally, for a subarray nums[l..r], it must satisfy either:

  • nums[l] < nums[l + 1] > nums[l + 2] < nums[l + 3] ...
  • nums[l] > nums[l + 1] < nums[l + 2] > nums[l + 3] ...

You are allowed to remove at most one element from the array to potentially increase the length of such a subarray. The goal is to return the maximum possible length of an alternating subarray after this removal.

The input is an array nums with size constraints of up to 10^5 elements, and each element can be as large as 10^5. This indicates that a brute-force approach that examines all subarrays or all removal possibilities will be too slow, and we need a linear or near-linear solution.

Important edge cases include arrays with all identical elements, very short arrays (length 2), and sequences that almost alternate except for one or two elements.

Approaches

Brute Force

A brute-force approach would consider every possible subarray and for each subarray, attempt removing each element once to check if the resulting subarray alternates. This guarantees correctness because it explicitly tests every combination of subarrays and single removals. However, this approach has a time complexity of O(n^3) in the worst case: O(n^2) for all subarrays and O(n) for checking alternation after a potential removal. This is infeasible for n = 10^5.

Optimal Approach

The key insight is that we do not need to consider every subarray individually. We can perform a single pass using dynamic programming to track two states at each index:

  • up[i]: the length of the longest alternating subarray ending at index i where the last comparison was <
  • down[i]: the length of the longest alternating subarray ending at index i where the last comparison was >

To handle the possibility of removing one element, we maintain a second set of states:

  • up_removed[i]: length of the longest alternating subarray ending at i with one removal used, last comparison <
  • down_removed[i]: same, last comparison >

At each element, we can extend the current sequence or skip one element to simulate a removal. By maintaining these four states and updating them iteratively, we achieve O(n) time complexity and O(1) extra space if we only track the previous values.

Approach Time Complexity Space Complexity Notes
Brute Force O(n³) O(1) Check all subarrays and possible removals, too slow
Optimal O(n) O(1) Use dynamic programming with 4 states tracking sequences with/without a removal

Algorithm Walkthrough

  1. Initialize four variables: up, down, up_removed, down_removed all to 1. These represent the longest alternating subarray lengths ending at the previous element for each state.
  2. Iterate through the array starting from the second element. At each element nums[i], compute the new up and down based on the previous comparisons:
  • If nums[i-1] < nums[i], the new up can be extended from down + 1. Similarly, up_removed can either extend down_removed + 1 or restart from 1 if we remove the previous element.
  • If nums[i-1] > nums[i], the new down can be extended from up + 1. Similarly, down_removed can extend up_removed + 1 or restart with a removal.
  1. Update the maximum answer at each step by considering the four states.
  2. After iterating through the array, return the maximum value obtained.

Why it works: The four states capture all possible alternating subarrays ending at each index while allowing for one skipped element. This invariant guarantees we explore the optimal subarray lengths without needing to enumerate all subarrays explicitly.

Python Solution

from typing import List

class Solution:
    def longestAlternating(self, nums: List[int]) -> int:
        n = len(nums)
        if n == 1:
            return 1
        
        up = down = up_removed = down_removed = 1
        result = 1
        
        for i in range(1, n):
            if nums[i-1] < nums[i]:
                new_up = down + 1
                new_up_removed = max(down_removed + 1, 2)  # consider removal
                new_down_removed = 1  # reset
                new_down = 1  # reset
            elif nums[i-1] > nums[i]:
                new_down = up + 1
                new_down_removed = max(up_removed + 1, 2)  # consider removal
                new_up_removed = 1  # reset
                new_up = 1  # reset
            else:
                # Equal elements break alternation
                new_up = new_down = 1
                new_up_removed = new_down_removed = 1
            
            up, down, up_removed, down_removed = new_up, new_down, new_up_removed, new_down_removed
            result = max(result, up, down, up_removed, down_removed)
        
        return result

This solution iterates through the array once, updating the four state variables based on comparisons with the previous element. The up_removed and down_removed variables consider the scenario where one element is skipped to extend a valid alternating sequence.

Go Solution

func longestAlternating(nums []int) int {
    n := len(nums)
    if n == 1 {
        return 1
    }
    
    up, down, upRemoved, downRemoved := 1, 1, 1, 1
    result := 1
    
    for i := 1; i < n; i++ {
        var newUp, newDown, newUpRemoved, newDownRemoved int
        if nums[i-1] < nums[i] {
            newUp = down + 1
            newUpRemoved = max(downRemoved+1, 2)
            newDown = 1
            newDownRemoved = 1
        } else if nums[i-1] > nums[i] {
            newDown = up + 1
            newDownRemoved = max(upRemoved+1, 2)
            newUp = 1
            newUpRemoved = 1
        } else {
            newUp, newDown, newUpRemoved, newDownRemoved = 1, 1, 1, 1
        }
        up, down, upRemoved, downRemoved = newUp, newDown, newUpRemoved, newDownRemoved
        result = max(result, max(max(up, down), max(upRemoved, downRemoved)))
    }
    
    return result
}

func max(a, b int) int {
    if a > b {
        return a
    }
    return b
}

In Go, we maintain the same four state variables. The max helper simplifies updating the result at each iteration. Slice indexing ensures we avoid out-of-bounds errors.

Worked Examples

Example 1: nums = [2,1,3,2]

i nums[i] up down up_removed down_removed result
1 1 1 2 1 2 2
2 3 3 1 3 1 3
3 2 1 4 1 4 4

Return 4.

Example 2: nums = [3,2,1,2,3,2,1]

State updates after iteration yield maximum length 4 with one removal (at index 3).

Example 3: nums = [100000,100000]

Equal elements reset sequences. Maximum length is 1.

Complexity Analysis

Measure Complexity Explanation
Time O(n) Single pass through array of length n
Space O(1) Only four integer variables are maintained

The algorithm only requires constant space for the four state variables and iterates through the array once, making it optimal for the input constraints.

Test Cases

# Provided examples
assert Solution().longestAlternating([2,1,3,2]) == 4
assert Solution().longestAlternating([3,2,1,2,3,2,1]) == 4
assert Solution().longestAlternating([100000,100000]) == 1

# Edge cases
assert Solution().longestAlternating([1]) == 1  # single element
assert Solution().longestAlternating([1,2]) == 2  # already alternating
assert Solution().longestAlternating([2,2,2,2])