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Easy
1240
Weekly Contest 268 Q1
Greedy
Array

2078. Two Furthest Houses With Different Colors

中文文档

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:

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

Solution 1

Python3

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

Java

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;
    }
}

C++

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;
    }
};

Go

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
}

Solution 2

Python3

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)

Java

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);
    }
}

C++

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);
    }
};

Go

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)
}