LeetCode #3830 — HARD

Longest Alternating Subarray After Removing At Most One Element

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

The Problem

Problem Statement

You are given an integer array nums.

A subarray nums[l..r] is alternating if one of the following holds:

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

In other words, if we compare adjacent elements in the subarray, then the comparisons alternate between strictly greater and strictly smaller.

You can remove at most one element from nums. Then, you select an alternating subarray from nums.

Return an integer denoting the maximum length of the alternating subarray you can select.

A subarray of length 1 is considered alternating.

Example 1:

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

Output: 4

Explanation:

Choose not to remove elements.
Select the entire array [2, 1, 3, 2], which is alternating because 2 > 1 < 3 > 2.

Example 2:

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

Output: 4

Explanation:

Choose to remove nums[3] i.e., [3, 2, 1, 2, 3, 2, 1]. The array becomes [3, 2, 1, 3, 2, 1].
Select the subarray [3, 2, 1, 3, 2, 1].

Example 3:

Input: nums = [100000,100000]

Output: 1

Explanation:

Choose not to remove elements.
Select the subarray [100000, 100000].

Constraints:

2 <= nums.length <= 10⁵
1 <= nums[i] <= 10⁵

Step 01

Brute Force Baseline

Problem summary: You are given an integer array nums. A subarray nums[l..r] is alternating if one of the following holds: nums[l] < nums[l + 1] > nums[l + 2] < nums[l + 3] > ... nums[l] > nums[l + 1] < nums[l + 2] > nums[l + 3] < ... In other words, if we compare adjacent elements in the subarray, then the comparisons alternate between strictly greater and strictly smaller. You can remove at most one element from nums. Then, you select an alternating subarray from nums. Return an integer denoting the maximum length of the alternating subarray you can select. A subarray of length 1 is considered alternating.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Dynamic Programming

Example 1

[2,1,3,2]

Example 2

[3,2,1,2,3,2,1]

Example 3

[100000,100000]

Step 02

Core Insight

What unlocks the optimal approach

Define <code>left[i][d]</code> as the maximum length of an alternating subarray ending at index <code>i</code>, where <code>d = 0</code> means the last comparison is <code><</code> and <code>d = 1</code> means the last comparison is <code>></code>. Define <code>right[i][d]</code> similarly for subarrays starting at <code>i</code>.
Fill <code>left</code> from left to right and <code>right</code> from right to left; if adjacent values are equal, the alternating chain must restart since <code>==</code> is invalid.
Try removing each index <code>r</code>: if <code>nums[r - 1] < nums[r + 1]</code>, the two sides can connect with pattern <code>< ></code>, giving length <code>left[r - 1][0] + right[r + 1][1]</code>; if <code>nums[r - 1] > nums[r + 1]</code>, connect with pattern <code>> <</code>, giving <code>left[r - 1][1] + right[r + 1][0]</code>.
Also consider not removing any element by taking the maximum value over all <code>left[i][d]</code>.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Largest constraint values

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #3830: Longest Alternating Subarray After Removing At Most One Element
class Solution {
    public int longestAlternating(int[] nums) {
        int n = nums.length;
        int[] l1 = new int[n];
        int[] l2 = new int[n];
        int[] r1 = new int[n];
        int[] r2 = new int[n];

        for (int i = 0; i < n; i++) {
            l1[i] = 1;
            l2[i] = 1;
            r1[i] = 1;
            r2[i] = 1;
        }

        int ans = 0;

        for (int i = 1; i < n; i++) {
            if (nums[i - 1] < nums[i]) {
                l1[i] = l2[i - 1] + 1;
            } else if (nums[i - 1] > nums[i]) {
                l2[i] = l1[i - 1] + 1;
            }
            ans = Math.max(ans, l1[i]);
            ans = Math.max(ans, l2[i]);
        }

        for (int i = n - 2; i >= 0; i--) {
            if (nums[i + 1] > nums[i]) {
                r1[i] = r2[i + 1] + 1;
            } else if (nums[i + 1] < nums[i]) {
                r2[i] = r1[i + 1] + 1;
            }
        }

        for (int i = 1; i < n - 1; i++) {
            if (nums[i - 1] < nums[i + 1]) {
                ans = Math.max(ans, l2[i - 1] + r2[i + 1]);
            } else if (nums[i - 1] > nums[i + 1]) {
                ans = Math.max(ans, l1[i - 1] + r1[i + 1]);
            }
        }

        return ans;
    }
}

// Accepted solution for LeetCode #3830: Longest Alternating Subarray After Removing At Most One Element
func longestAlternating(nums []int) int {
	n := len(nums)
	l1 := make([]int, n)
	l2 := make([]int, n)
	r1 := make([]int, n)
	r2 := make([]int, n)

	for i := 0; i < n; i++ {
		l1[i] = 1
		l2[i] = 1
		r1[i] = 1
		r2[i] = 1
	}

	ans := 0

	for i := 1; i < n; i++ {
		if nums[i-1] < nums[i] {
			l1[i] = l2[i-1] + 1
		} else if nums[i-1] > nums[i] {
			l2[i] = l1[i-1] + 1
		}

		ans = max(ans, l1[i], l2[i])
	}

	for i := n - 2; i >= 0; i-- {
		if nums[i+1] > nums[i] {
			r1[i] = r2[i+1] + 1
		} else if nums[i+1] < nums[i] {
			r2[i] = r1[i+1] + 1
		}
	}

	for i := 1; i < n-1; i++ {
		if nums[i-1] < nums[i+1] {
			if l2[i-1]+r2[i+1] > ans {
				ans = l2[i-1] + r2[i+1]
			}
		} else if nums[i-1] > nums[i+1] {
			if l1[i-1]+r1[i+1] > ans {
				ans = l1[i-1] + r1[i+1]
			}
		}
	}

	return ans
}

# Accepted solution for LeetCode #3830: Longest Alternating Subarray After Removing At Most One Element
class Solution:
    def longestAlternating(self, nums: List[int]) -> int:
        n = len(nums)
        l1 = [1] * n
        l2 = [1] * n
        r1 = [1] * n
        r2 = [1] * n
        ans = 0
        for i in range(1, n):
            if nums[i - 1] < nums[i]:
                l1[i] = l2[i - 1] + 1
            elif nums[i - 1] > nums[i]:
                l2[i] = l1[i - 1] + 1
            ans = max(ans, l1[i], l2[i])
        for i in range(n - 2, -1, -1):
            if nums[i + 1] > nums[i]:
                r1[i] = r2[i + 1] + 1
            elif nums[i + 1] < nums[i]:
                r2[i] = r1[i + 1] + 1
        for i in range(1, n - 1):
            if nums[i - 1] < nums[i + 1]:
                ans = max(ans, l2[i - 1] + r2[i + 1])
            elif nums[i - 1] > nums[i + 1]:
                ans = max(ans, l1[i - 1] + r1[i + 1])
        return ans

// Accepted solution for LeetCode #3830: Longest Alternating Subarray After Removing At Most One Element
impl Solution {
    pub fn longest_alternating(nums: Vec<i32>) -> i32 {
        let n = nums.len();
        let mut l1 = vec![1; n];
        let mut l2 = vec![1; n];
        let mut r1 = vec![1; n];
        let mut r2 = vec![1; n];

        let mut ans = 0;

        for i in 1..n {
            if nums[i - 1] < nums[i] {
                l1[i] = l2[i - 1] + 1;
            } else if nums[i - 1] > nums[i] {
                l2[i] = l1[i - 1] + 1;
            }
            ans = ans.max(l1[i]);
            ans = ans.max(l2[i]);
        }

        for i in (0..n - 1).rev() {
            if nums[i + 1] > nums[i] {
                r1[i] = r2[i + 1] + 1;
            } else if nums[i + 1] < nums[i] {
                r2[i] = r1[i + 1] + 1;
            }
        }

        for i in 1..n - 1 {
            if nums[i - 1] < nums[i + 1] {
                ans = ans.max(l2[i - 1] + r2[i + 1]);
            } else if nums[i - 1] > nums[i + 1] {
                ans = ans.max(l1[i - 1] + r1[i + 1]);
            }
        }

        ans
    }
}

// Accepted solution for LeetCode #3830: Longest Alternating Subarray After Removing At Most One Element
function longestAlternating(nums: number[]): number {
    const n = nums.length;
    const l1 = new Array<number>(n).fill(1);
    const l2 = new Array<number>(n).fill(1);
    const r1 = new Array<number>(n).fill(1);
    const r2 = new Array<number>(n).fill(1);

    let ans = 0;

    for (let i = 1; i < n; i++) {
        if (nums[i - 1] < nums[i]) {
            l1[i] = l2[i - 1] + 1;
        } else if (nums[i - 1] > nums[i]) {
            l2[i] = l1[i - 1] + 1;
        }
        ans = Math.max(ans, l1[i]);
        ans = Math.max(ans, l2[i]);
    }

    for (let i = n - 2; i >= 0; i--) {
        if (nums[i + 1] > nums[i]) {
            r1[i] = r2[i + 1] + 1;
        } else if (nums[i + 1] < nums[i]) {
            r2[i] = r1[i + 1] + 1;
        }
    }

    for (let i = 1; i < n - 1; i++) {
        if (nums[i - 1] < nums[i + 1]) {
            ans = Math.max(ans, l2[i - 1] + r2[i + 1]);
        } else if (nums[i - 1] > nums[i + 1]) {
            ans = Math.max(ans, l1[i - 1] + r1[i + 1]);
        }
    }

    return ans;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(n)

Approach Breakdown

RECURSIVE

O(2ⁿ) time

O(n) space

Pure recursion explores every possible choice at each step. With two choices per state (take or skip), the decision tree has 2ⁿ leaves. The recursion stack uses O(n) space. Many subproblems are recomputed exponentially many times.

DYNAMIC PROGRAMMING

O(n × m) time

O(n × m) space

Each cell in the DP table is computed exactly once from previously solved subproblems. The table dimensions determine both time and space. Look for the state variables — each unique combination of state values is one cell. Often a rolling array can reduce space by one dimension.

Shortcut: Count your DP state dimensions → that’s your time. Can you drop one? That’s your space optimization.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

State misses one required dimension

Wrong move: An incomplete state merges distinct subproblems and caches incorrect answers.

Usually fails on: Correctness breaks on cases that differ only in hidden state.

Fix: Define state so each unique subproblem maps to one DP cell.