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
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