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;
        }
    };
    
  • // Todo
    
  • # 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
    
    

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;
        }
    };
    
  • // Todo
    
  • # 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
    
    

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;
        }
    };
    
  • // Todo
    
  • # 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
    
    

Discuss

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

All Problems

All Solutions