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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;
}
};

• 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)

############

# 2078. Two Furthest Houses With Different Colors
# https://leetcode.com/problems/two-furthest-houses-with-different-colors/

class Solution:
def maxDistance(self, colors: List[int]) -> int:
n = len(colors)
res = 0

for i in range(n):
for j in range(i + 1, n):
if colors[i] != colors[j]:
res = max(res, abs(j - i))

return res


• 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);
}
}

• 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)
}

func max(a, b int) int {
if a > b {
return a
}
return b
}


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;
}
};

• 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)

############

# 2078. Two Furthest Houses With Different Colors
# https://leetcode.com/problems/two-furthest-houses-with-different-colors/

class Solution:
def maxDistance(self, colors: List[int]) -> int:
n = len(colors)
res = 0

for i in range(n):
for j in range(i + 1, n):
if colors[i] != colors[j]:
res = max(res, abs(j - i))

return res


• 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);
}
}

• 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)
}

func max(a, b int) int {
if a > b {
return a
}
return b
}


## 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;
}
};

• 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)

############

# 2078. Two Furthest Houses With Different Colors
# https://leetcode.com/problems/two-furthest-houses-with-different-colors/

class Solution:
def maxDistance(self, colors: List[int]) -> int:
n = len(colors)
res = 0

for i in range(n):
for j in range(i + 1, n):
if colors[i] != colors[j]:
res = max(res, abs(j - i))

return res


• 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);
}
}

• 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)
}

func max(a, b int) int {
if a > b {
return a
}
return b
}


## Discuss

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