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Problem solution · Java

Beautiful Arrangement

Beautiful Arrangement: a Java solution using depth-first search. Learn the idea, check the complexity, and read the full code, with credit to walkccc LeetCode Solutions.

Technique
Depth-first search
Source
walkccc LeetCode Solutions
Length
30 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

Depth-first search

For Beautiful Arrangement, the implementation follows one branch at a time, making it suitable for components, trees, backtracking, or dependency exploration.

  1. Define the state carried into one recursive or stack frame.
  2. Mark or choose the current state before exploring children.
  3. Combine child results or undo the choice when the branch finishes.

Code notes

  • 30 lines of Java from the credited upstream file 526.java.
  • The implementation visibly relies on hash lookup, ordered lookup.
  • 1 loop block detected, together with recursive traversal.

Complexity

Count unique states for graph traversal; for backtracking, count the branching factor and maximum depth.

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 codeBeautiful Arrangement · JavaJava
Use this to learn the idea, then write your own version.
class Solution {  public int countArrangement(int n) {    final String filled = "x".repeat(n + 1);    StringBuilder sb = new StringBuilder(filled);    Map<String, Integer> mem = new HashMap<>();     return dfs(n, 1, sb, mem);  }   private int dfs(int n, int num, StringBuilder sb, Map<String, Integer> mem) {    if (num == n + 1)      return 1;    final String filled = sb.toString();    if (mem.containsKey(filled))      return mem.get(filled);     int count = 0;     for (int i = 1; i <= n; ++i)      if (sb.charAt(i) == 'x' && (num % i == 0 || i % num == 0)) {        sb.setCharAt(i, 'o');        count += dfs(n, num + 1, sb, mem);        sb.setCharAt(i, 'x');      }     mem.put(filled, count);    return count;  }}

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