README_EN.md

comments	true
difficulty	Easy
edit_url	https://github.com/doocs/leetcode/edit/main/solution/2000-2099/2078.Two%20Furthest%20Houses%20With%20Different%20Colors/README_EN.md
rating	1240
source	Weekly Contest 268 Q1
tags	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: Greedy

We can observe that if the first and last houses have different colors, the maximum distance is $n - 1$.

If the first and last houses have the same color, we can scan from the left to find the first house with a different color (let its index be $i$), and scan from the right to find the first house with a different color (let its index be $j$). The maximum distance is then $\max(n - i - 1, j)$.

The time complexity is $O(n)$, where $n$ is the number of houses. The space complexity is $O(1)$.

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 = 1, j = n - 2;
        while (colors[i] == colors[0]) {
            ++i;
        }
        while (colors[j] == colors[0]) {
            --j;
        }
        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 = 1, j = n - 2;
        while (colors[i] == colors[0]) {
            ++i;
        }
        while (colors[j] == colors[0]) {
            --j;
        }
        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)
}

TypeScript

function maxDistance(colors: number[]): number {
    const n = colors.length;
    if (colors[0] !== colors[n - 1]) {
        return n - 1;
    }
    let [i, j] = [1, n - 2];
    while (colors[i] === colors[0]) {
        i++;
    }
    while (colors[j] === colors[0]) {
        j--;
    }
    return Math.max(n - i - 1, j);
};

Rust

impl Solution {
    pub fn max_distance(colors: Vec<i32>) -> i32 {
        let n = colors.len();
        if colors[0] != colors[n - 1] {
            return (n - 1) as i32;
        }
        let mut i = 1;
        while colors[i] == colors[0] {
            i += 1;
        }
        let mut j = n - 2;
        while colors[j] == colors[0] {
            j -= 1;
        }
        std::cmp::max((n - i - 1) as i32, j as i32)
    }
}