## Solution

#### Approach 1: Scan Through Blocks

Intuition

Equivalently, we want the longest subarray with at most two "types" (values of tree[i]).

Instead of considering each element individually, we can consider blocks of adjacent elements of the same type.

For example, instead of tree = [1, 1, 1, 1, 2, 2, 3, 3, 3], we can say this is blocks = [(1, weight = 4), (2, weight = 2), (3, weight = 3)].

Now say we brute forced, scanning from left to right. We'll have something like blocks = [1, _2_, 1, 2, 1, 2, _1_, 3, ...] (with various weights).

The key insight is that when we encounter a 3, we do not need to start from the second element 2 (marked _2_ for convenience); we can start from the first element (_1_) before the 3. This is because if we started two or more elements before, the sequence must have types 1 and 2, and that sequence is going to end at the 3, and thus be shorter than anything we've already considered.

Since every starting point (that is the left-most index of a block) was considered, this solution is correct.

Algorithm

As the notation and strategy around implementing this differs between Python and Java, please see the inline comments for more details.

Complexity Analysis

• Time Complexity: , where is the length of tree.

• Space Complexity: .

#### Approach 2: Sliding Window

Intuition

As in Approach 1, we want the longest subarray with at most two different "types" (values of tree[i]). Call these subarrays valid.

Say we consider all valid subarrays that end at index j. There must be one with the smallest possible starting index i: lets say opt(j) = i.

Now the key idea is that opt(j) is a monotone increasing function. This is because any subarray of a valid subarray is valid.

Algorithm

Let's perform a sliding window, keeping the loop invariant that i will be the smallest index for which [i, j] is a valid subarray.

We'll maintain count, the count of all the elements in the subarray. This allows us to quickly query whether there are 3 types in the subarray or not.

Complexity Analysis

• Time Complexity: , where is the length of tree.

• Space Complexity: .

Analysis written by: @awice.