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Abhishek Chaudhary

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# Two Furthest Houses With Different Colors

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.

Example 2:

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.

SOLUTION:

class Solution:
def maxDistance(self, colors: List[int]) -> int:
indexes = {}
for i, c in enumerate(colors):
if c in indexes:
indexes[c] = (min(i, indexes[c][0]), max(i, indexes[c][1]))
else:
indexes[c] = (i, i)
mdiff = 1
n = len(indexes)
colors = list(indexes.keys())
for i in range(n):
for j in range(i + 1, n):
mdiff = max(mdiff, abs(indexes[colors[i]][1] - indexes[colors[j]][0]), abs(indexes[colors[j]][1] - indexes[colors[i]][0]))
return mdiff