Description#
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:
![](https://spcdn.pages.dev/leetcode/problems/2078.Two%20Furthest%20Houses%20With%20Different%20Colors/images/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://spcdn.pages.dev/leetcode/problems/2078.Two%20Furthest%20Houses%20With%20Different%20Colors/images/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.
Solutions#
Solution 1#
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| 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
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| 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;
}
}
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| 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;
}
};
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| 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(i-j))
}
}
}
return ans
}
func abs(x int) int {
if x >= 0 {
return x
}
return -x
}
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Solution 2#
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| class Solution:
def maxDistance(self, colors: List[int]) -> int:
n = len(colors)
if colors[0] != colors[-1]:
return n - 1
i, j = 1, n - 2
while colors[i] == colors[0]:
i += 1
while colors[j] == colors[0]:
j -= 1
return max(n - i - 1, j)
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| class Solution {
public int maxDistance(int[] colors) {
int n = colors.length;
if (colors[0] != colors[n - 1]) {
return n - 1;
}
int i = 0, j = n - 1;
while (colors[++i] == colors[0])
;
while (colors[--j] == colors[0])
;
return Math.max(n - i - 1, j);
}
}
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| class Solution {
public:
int maxDistance(vector<int>& colors) {
int n = colors.size();
if (colors[0] != colors[n - 1]) return n - 1;
int i = 0, j = n;
while (colors[++i] == colors[0])
;
while (colors[--j] == colors[0])
;
return max(n - i - 1, j);
}
};
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| func maxDistance(colors []int) int {
n := len(colors)
if colors[0] != colors[n-1] {
return n - 1
}
i, j := 1, n-2
for colors[i] == colors[0] {
i++
}
for colors[j] == colors[0] {
j--
}
return max(n-i-1, j)
}
|