How do you solve Two Furthest Houses With Different Colors on LeetCode?

Two Furthest Houses With Different Colors (LeetCode #2078) can be solved using Brute Force. The optimal approach is Brute Force with O(n²) time complexity and O(1) space complexity. Key insight: The brute-force approach checks every possible pair of houses to find the maximum distance between two houses of different colors. This is straightforward but inefficient for larger inputs.

What algorithm or data structure is used to solve Two Furthest Houses With Different Colors?

Two Furthest Houses With Different Colors uses the following concepts: Array, Greedy. The recommended patterns to study are: .

Two Furthest Houses With Different Colors — LeetCode #2078 (Easy)

Tags: Array, Greedy

Related patterns:

Brute Force approach

Time complexity: O(n²). Space complexity: O(1).

The brute-force approach checks every possible pair of houses to find the maximum distance between two houses of different colors. This is straightforward but inefficient for larger inputs.

The time complexity is O(n²) because we are using two nested loops to check every pair of houses. The space complexity is O(1) as we are using a constant amount of space.

Step 1: Initialize a variable to keep track of the maximum distance found.
Step 2: Use two nested loops to iterate through all pairs of houses.
Step 3: For each pair, check if the colors are different. If they are, calculate the distance and update the maximum distance if this distance is greater.

For colors = [1, 1, 1, 6, 1, 1, 1]: 1. i = 0, j = 1 -> colors[0] == colors[1] (skip) 2. i = 0, j = 2 -> colors[0] == colors[2] (skip) 3. i = 0, j = 3 -> colors[0] != colors[3] (max_dist = max(0, 3 - 0) = 3) 4. i = 0, j = 4 -> colors[0] == colors[4] (skip) 5. i = 0, j = 5 -> colors[0] == colors[5] (skip) 6. i = 0, j = 6 -> colors[0] == colors[6] (skip) Final max_dist = 3.

#2078

Two Furthest Houses With Different Colors

Easy

Array↗Greedy↗

LeetCode ↗

Approaches

💡

Intuition

Time O(n²)Space O(1)

The brute-force approach checks every possible pair of houses to find the maximum distance between two houses of different colors. This is straightforward but inefficient for larger inputs.

⚙️

Algorithm

3 steps

1Step 1: Initialize a variable to keep track of the maximum distance found.
2Step 2: Use two nested loops to iterate through all pairs of houses.
3Step 3: For each pair, check if the colors are different. If they are, calculate the distance and update the maximum distance if this distance is greater.

solution.py8 lines

1def maxDistance(colors):
2    max_dist = 0
3    n = len(colors)
4    for i in range(n):
5        for j in range(i + 1, n):
6            if colors[i] != colors[j]:
7                max_dist = max(max_dist, j - i)
8    return max_dist

ℹ

Complexity note: The time complexity is O(n²) because we are using two nested loops to check every pair of houses. The space complexity is O(1) as we are using a constant amount of space.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.