#2078

Easy

auto_awesomeGreedy choice plus invariant validation

LeetCode Problem Workspace

Two Furthest Houses With Different Colors

Maximize the distance between two houses with different colors by using a greedy approach and validating the choice.

Problem Statement

You are given an integer array 'colors' representing the colors of n houses lined up on a street. Each house has a distinct color, and you need to find the maximum distance between two houses with different colors.

The distance between two houses is measured as the absolute difference between their indices in the array. You need to return the largest such distance between two houses having different colors.

Examples

Example 1

Input: colors = [1,1,1,6,1,1,1]

Output: 3

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

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

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 Approach

Greedy Approach

Use a greedy approach to select the first and last house with different colors. The maximum distance between them will give the solution. This approach works because we only need to check the first and last occurrences of distinct colors to achieve the largest gap.

Invariant Validation

As you iterate through the array, ensure that you are maintaining the correct invariant: two houses with different colors. Check for the largest possible gap by keeping track of the first and last houses with distinct colors.

Combination Checking

Given the problem's constraints, we can efficiently check combinations of house pairs. Try checking the distance between the first house and any house of a different color, and similarly, check from the last house back to any house of a different color.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time complexity is O(n), where n is the number of houses. Since we are iterating through the array only once and using constant extra space, the space complexity is O(1).

What Interviewers Usually Probe

Evaluate the candidate's understanding of greedy algorithms and invariant validation.
Check if the candidate recognizes the simplicity of the problem due to the small input size.
Test the candidate's ability to optimize the approach without unnecessary calculations.

Common Pitfalls or Variants

Common pitfalls

Overcomplicating the solution by checking all possible house pairs rather than focusing on the first and last occurrences of different colors.
Failing to correctly handle cases where the first and last houses with distinct colors are not at the extremes.
Misunderstanding the constraints and trying inefficient solutions despite small input sizes.

Follow-up variants

What if the input size increased to 1000 or more? Consider adjusting the approach to handle larger inputs efficiently.
What if the problem asks for the smallest distance instead of the largest? Modify the greedy approach accordingly.
What if the colors array contains many duplicates? Ensure your solution still adheres to the pattern of maximizing the distance between different colors.

FAQ

What is the time complexity of the Two Furthest Houses With Different Colors problem?

The time complexity is O(n), where n is the number of houses, since we only need to iterate through the array once.

How do I solve the Two Furthest Houses With Different Colors problem?

Use a greedy approach by selecting the first and last house with different colors, and calculate the maximum distance between them.

Can I check all combinations of house pairs to solve the problem?

While you could check all combinations, a greedy approach focusing on the first and last houses with distinct colors is more efficient.

What are common mistakes when solving the Two Furthest Houses With Different Colors?

Common mistakes include checking all house pairs unnecessarily, and failing to correctly handle the extreme cases of the array.

How does the greedy approach work in the Two Furthest Houses With Different Colors problem?

The greedy approach works by selecting the first and last house with different colors, ensuring the maximum distance between them.

terminal

Solution

Solution 1: Greedy

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

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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)

Continue Practicing

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