Problem

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 themaximum 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.

Examples

Example 1

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![](https://assets.leetcode.com/uploads/2021/10/31/eg1.png)

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

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![](https://assets.leetcode.com/uploads/2021/10/31/eg2.png)

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

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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.

Solution

Method 1 – Two Pointers or Brute Force

Intuition

The answer is the maximum distance between the first house and the last house with a different color, or the last house and the first house with a different color.

Approach

  1. For i = 0, find the furthest j where colors[j] != colors[0].
  2. For i = n-1, find the furthest j where colors[j] != colors[n-1].
  3. Return the maximum of these two distances.

Code

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class Solution {
    public int maxDistance(int[] colors) {
        int n = colors.length;
        int ans = 0;
        for (int i = 0; i < n; ++i) {
            if (colors[i] != colors[0]) ans = Math.max(ans, i);
            if (colors[i] != colors[n-1]) ans = Math.max(ans, n-1-i);
        }
        return ans;
    }
}
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def maxDistance(colors):
    n = len(colors)
    ans = 0
    for i in range(n):
        if colors[i] != colors[0]:
            ans = max(ans, i)
        if colors[i] != colors[-1]:
            ans = max(ans, n-1-i)
    return ans

Complexity

  • ⏰ Time complexity: O(n) — Single pass through the array.
  • 🧺 Space complexity: O(1) — Only a few variables used.