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Problem solution · Java

Two Furthest Houses With Different Colors

Two Furthest Houses With Different Colors: a Java solution using binary search. Learn the idea, check the complexity, and read the full code, with credit to walkccc LeetCode Solutions.

Technique
Binary search
Source
walkccc LeetCode Solutions
Length
13 lines
Start with the idea.

Try the problem first. If you get stuck, read the approach below, then write your own solution. The full code is at the bottom.

Approach

Binary search

For Two Furthest Houses With Different Colors, the implementation exploits a monotonic condition to discard half of the remaining search space after every check.

  1. Identify the ordered answer range or sorted search domain.
  2. Write a predicate whose truth changes only once.
  3. Move the appropriate boundary after each midpoint check and return the final feasible position.

Code notes

  • 13 lines of Java from the credited upstream file 2078.java.
  • The implementation visibly relies on sequence storage.
  • 2 loop blocks detected.

Complexity

Multiply the logarithmic number of midpoint checks by the cost of one predicate evaluation.

Check the problem constraints before deciding whether this complexity will pass.

Source

Code and credit

This code comes from walkccc LeetCode Solutions by P.-Y. Chen (walkccc) and is used under the MIT licence.

Open problem ↗Original source ↗
Full codeTwo Furthest Houses With Different Colors · JavaJava
Use this to learn the idea, then write your own version.
class Solution {  public int maxDistance(int[] colors) {    final int n = colors.length;    int i = 0;     // the leftmost index, where colors[i] != colors[n - 1]    int j = n - 1; // the rightmost index, where colors[j] != colors[0]    while (colors[i] == colors[n - 1])      ++i;    while (colors[j] == colors[0])      --j;    return Math.max(n - 1 - i, j);  }}

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