2078\. Two Furthest Houses With Different Colors

Easy

There are `n` houses evenly lined up on the street, and each house is beautifully painted. You are given a **0-indexed** integer array `colors` of length `n`, where `colors[i]` represents the color of the <code>i<sup>th</sup></code> house.

Return _the **maximum** distance between **two** houses with **different** colors_.

The distance between the <code>i<sup>th</sup></code> and <code>j<sup>th</sup></code> houses is `abs(i - j)`, where `abs(x)` is the **absolute value** of `x`.

**Example 1:**

![](https://assets.leetcode.com/uploads/2021/10/31/eg1.png)

**Input:** colors = [**1**,1,1,**6**,1,1,1]

**Output:** 3

**Explanation:** In the above image, color 1 is blue, and color 6 is red. 

The furthest two houses with different colors are house 0 and house 3. 

House 0 has color 1, and house 3 has color 6. 

The distance between them is abs(0 - 3) = 3. 

Note that houses 3 and 6 can also produce the optimal answer. 

**Example 2:**

![](https://assets.leetcode.com/uploads/2021/10/31/eg2.png)

**Input:** colors = [**1**,8,3,8,**3**]

**Output:** 4

**Explanation:** In the above image, color 1 is blue, color 8 is yellow, and color 3 is green. 

The furthest two houses with different colors are house 0 and house 4. 

House 0 has color 1, and house 4 has color 3. 

The distance between them is abs(0 - 4) = 4. 

**Example 3:**

**Input:** colors = [**0**,**1**]

**Output:** 1

**Explanation:** The furthest two houses with different colors are house 0 and house 1. 

House 0 has color 0, and house 1 has color 1. 

The distance between them is abs(0 - 1) = 1. 

**Constraints:**

*   `n == colors.length`
*   `2 <= n <= 100`
*   `0 <= colors[i] <= 100`
*   Test data are generated such that **at least** two houses have different colors.
