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2078. Two Furthest Houses With Different Colors

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.

Β 

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.

Β 

Constraints:

  • n ==Β 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 \(O(1)\).

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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)

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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);
    }
}

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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);
    }
};

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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)
}

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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);
}

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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)
    }
}

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