Formatted question description: https://leetcode.ca/all/2078.html

# 2078. Two Furthest Houses With Different Colors (Easy)

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.

Similar Questions:

## Solution 1. Map

For each A[i]:

• If this is the first occurrence of this value, store A[i] -> i in a map m.
• Loop through each num, index pair the map m and calculate the maximum i - index value if num != A[i].
// OJ: https://leetcode.com/problems/two-furthest-houses-with-different-colors/
// Time: O(NM) where N is the length of A and M is the range of numbers in A.
// Space: O(M)
class Solution {
public:
int maxDistance(vector<int>& A) {
unordered_map<int, int> m; // first occurrence index
int ans = 0;
for (int i = 0; i < A.size(); ++i) {
if (m.count(A[i]) == 0) m[A[i]] = i;
for (auto &[c, j] : m) {
if (c != A[i]) {
ans = max(ans, i - j);
}
}
}
return ans;
}
};


Or use array.

// OJ: https://leetcode.com/problems/two-furthest-houses-with-different-colors/
// Time: O(NM) where N is the length of A and M is the range of numbers in A.
// Space: O(M)
class Solution {
public:
int maxDistance(vector<int>& A) {
int ans = 0, index[101] = {[0 ... 100] = -1};
for (int i = 0; i < A.size(); ++i) {
if (index[A[i]] == -1) index[A[i]] = i;
for (int j = 0; j <= 100; ++j) {
if (index[j] != -1 && j != A[i]) {
ans = max(ans, i - index[j]);
}
}
}
return ans;
}
};


## Solution 2. Mono-increasing Index Array

For each A[i]:

• If this is the first occurrence of this value, push i into a vector<int> index.
• Loop through each index value j in index array, and update answer with i - j for the first A[j] != A[i]. This step at most looks at two indices, so it’s O(1) time.
// OJ: https://leetcode.com/problems/two-furthest-houses-with-different-colors/
// Time: O(N) where N is the length of A and M is the range of numbers in A.
// Space: O(M)
class Solution {
public:
int maxDistance(vector<int>& A) {
vector<int> index; // first occurrence index
int ans = 0, seen[101] = {};
for (int i = 0; i < A.size(); ++i) {
if (!seen[i]) {
seen[i] = 1;
index.push_back(i);
}
for (int j : index) {
if (A[j] != A[i]) {
ans = max(ans, i - j);
break;
}
}
}
return ans;
}
};


## Discuss

https://leetcode.com/problems/two-furthest-houses-with-different-colors/discuss/1589010/C%2B%2B-O(N)-Time-One-Pass