# [LeetCode]#2078. Two Furthest Houses With Different Colors

--

**Environment: Python 3.8**

**Key technique: **enumerate, if

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 `ith`

house.

Return *the **maximum** distance between **two** houses with **different** colors*.

The distance between the `ith`

and `jth`

houses is `abs(i - j)`

, where `abs(x)`

is the **absolute value** of `x`

.

**Example 1:**

**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.

**Analysis:**

- Use enumerate find location and number in list.
- Check location 0 number is equal or not.
- If yes, distance is 0.
- If yes location index -0.
- output max one.

**Solution:**

`class Solution:`

def maxDistance(self, colors):

d=0

for i,j in enumerate(colors):

for a,b in enumerate(colors):

if j!=b:

if abs(i-a)>d:

d=abs(i-a)

return d

**Submissions:**

**Reference:**