refactor(codeforces): simplify logic and cleanup comments in p2189C2

Author: lzwjava

src/main/java/com/lzw/solutions/codeforces/p2189C2/Main.java

@@ -1,4 +1,3 @@
-// grok-4.1-thinking
 package com.lzw.solutions.codeforces.p2189C2;
 
 import java.io.*;
@@ -14,59 +13,51 @@ public static void main(String[] args) throws IOException {
         for (int test = 0; test < t; test++) {
             int n = Integer.parseInt(br.readLine());
 
-            // Check if n is power of 2
+            // If n is power of 2, impossible
             if ((n & (n - 1)) == 0) {
                 pw.println(-1);
                 continue;
             }
 
-            // TreeSet to keep available numbers in sorted order
             TreeSet<Integer> available = new TreeSet<>();
             for (int i = 1; i <= n; i++) {
                 available.add(i);
             }
 
             int[] p = new int[n + 1];
-
-            // Place n at position n
             p[n] = n;
             available.remove(n);
 
-            // From n-1 down to 1
+            boolean ok = true;
             for (int pos = n - 1; pos >= 1; pos--) {
-                boolean found = false;
-
-                // Try smallest available numbers first
                 Iterator<Integer> it = available.iterator();
+                boolean found = false;
                 while (it.hasNext()) {
                     int v = it.next();
                     int w = v ^ pos;
-
-                    // Check if w is invalid or already used
-                    if (w < 1 || w > n || !available.contains(w)) {
-                        // v is good to place at pos
+                    // Only accept if w is valid and already placed
+                    if (w >= 1 && w <= n && !available.contains(w)) {
                         p[pos] = v;
                         available.remove(v);
                         found = true;
                         break;
                     }
                 }
-
-                // In theory this should never happen for non-power-of-2 n
                 if (!found) {
-                    pw.println(-1);
+                    ok = false;
                     break;
                 }
             }
 
-            if (p[1] == 0) continue; // failed case (should not occur)
-
-            // Output the permutation
-            for (int i = 1; i <= n; i++) {
-                if (i > 1) pw.print(" ");
-                pw.print(p[i]);
+            if (!ok) {
+                pw.println(-1);
+            } else {
+                for (int i = 1; i <= n; i++) {
+                    if (i > 1) pw.print(" ");
+                    pw.print(p[i]);
+                }
+                pw.println();
             }
-            pw.println();
         }
 
         pw.flush();