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2078. Two Furthest Houses With Different Colors
Description
There are n
houses evenly lined up on the street, and each house is beautifully painted. You are given a 0indexed integer array colors
of length n
, where colors[i]
represents the color of the i^{th}
house.
Return the maximum distance between two houses with different colors.
The distance between the i^{th}
and j^{th}
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.
Solutions

class Solution { public int maxDistance(int[] colors) { int ans = 0, n = colors.length; for (int i = 0; i < n; ++i) { for (int j = i + 1; j < n; ++j) { if (colors[i] != colors[j]) { ans = Math.max(ans, Math.abs(i  j)); } } } return ans; } }

class Solution { public: int maxDistance(vector<int>& colors) { int ans = 0, n = colors.size(); for (int i = 0; i < n; ++i) for (int j = i + 1; j < n; ++j) if (colors[i] != colors[j]) ans = max(ans, abs(i  j)); return ans; } };

class Solution: def maxDistance(self, colors: List[int]) > int: ans, n = 0, len(colors) for i in range(n): for j in range(i + 1, n): if colors[i] != colors[j]: ans = max(ans, abs(i  j)) return ans

func maxDistance(colors []int) int { ans, n := 0, len(colors) for i := 0; i < n; i++ { for j := i + 1; j < n; j++ { if colors[i] != colors[j] { ans = max(ans, abs(ij)) } } } return ans } func abs(x int) int { if x >= 0 { return x } return x }