Solve Codeforces 2256B: Domino Tiles

Solve a Codeforces problem using constraint simplification and chain decomposition.

Introduction

30 Second Summary

Sometimes you can solve a puzzle by trying every combination. But when the number of combinations exceeds the atoms in the universe, you need a smarter approach.

In this project, you will solve Codeforces problem 2256B "Domino Tiles" by building a complete solution pipeline. You will go from a correct-but-slow brute force to a stress-tested optimized solution that earns an Accepted verdict.

What You'll Build

You will paste your final solution into Codeforces, click submit, and watch every test case pass with an Accepted verdict.

By the end of this project, you'll have:

  • A brute force solver and an O(n) optimized solution you run side by side, watching the brute force choke on large inputs while the optimized version answers instantly.
  • A stress testing harness that fires 10,000 random inputs at both solvers and prints confirmation when they agree on every one.
  • A reusable Java CP template with fast I/O (BufferedReader + StringBuilder) you can copy into any future contest.
  • Secret Mission: Build a modular arithmetic toolkit with modPow, modInverse, and nCr that you will reuse in nearly every future counting problem.

Are there any prerequisites?

This project assumes you know basic Java syntax: loops, arrays, and conditionals. You do not need any prior competitive programming experience.

Before We Start

Before we dive in, let's lock in what you're about to build and why it matters for your competitive programming practice.

Set Up Your Java CP Template

Competitive programming problems process hundreds of thousands of inputs under tight time limits. Java's default Scanner class parses input with regex under the hood, making it too slow for these constraints.

In this step, you'll create a reusable Java template with BufferedReader for fast input and StringBuilder for batched output. This template handles the standard multi-test-case format that most Codeforces problems use.

In this step, get ready to:
  • Verify JDK 21 is installed and available in your terminal.
  • Create Main.java with a fast I/O template for competitive programming.
  • Compile and run the template with a test input to confirm it works.
Verify your Java installation

Before writing any code, you need to confirm that JDK 21 is available on your machine. The Codeforces judge runs Java 21, so your local environment should match.

  • In VS Code, open the integrated terminal by pressing Ctrl+` (macOS and Windows).
  • Check your Java version by running this command:
java -version

✔️ I see version 21 or higher

Your Java installation is ready. You should see output that includes openjdk version "21 or similar. Move on to the next substep.

ⓧ I see an older version

Your system has an older Java version installed. You need JDK 21 to match the Codeforces judge environment.

  • Download JDK 21 from the Oracle Java Downloads page.
  • Run the installer for your operating system and follow the prompts.
  • Close and reopen your VS Code terminal, then run java -version again to confirm it now shows version 21.

Why close and reopen the terminal?

Your terminal loads environment variables (like the path to Java) when it starts. After installing a new JDK, the old terminal session still points to the previous version. Reopening the terminal picks up the updated path.

ⓧ Command not found

Java is not installed on your machine yet.

  • Download JDK 21 from the Oracle Java Downloads page.
  • Select the installer for your operating system (macOS or Windows) and run it.
  • Close and reopen your VS Code terminal, then run java -version again.
  • You should now see output containing openjdk version "21 or similar.
Create Main.java with the fast I/O template

Most Codeforces problems follow the same input pattern: a number of test cases t, then t blocks of problem-specific input. This template handles that pattern with fast I/O built in.

  • In the VS Code Explorer sidebar, right-click inside your project folder and select New File.
  • Name the file Main.java and press Enter.
  • Paste the following template into Main.java:
import java.io.*;
import java.util.*;

public class Main {
    // Modular arithmetic constant required by the problem
    static final int MOD = 998244353;

    public static void main(String[] args) throws IOException {
        // Fast input using BufferedReader instead of Scanner
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        // Collect all output for a single print at the end
        StringBuilder sb = new StringBuilder();
        // Read the number of test cases
        int t = Integer.parseInt(br.readLine().trim());
        while (t-- > 0) {
            // Read string length and the domino string for each test case
            int n = Integer.parseInt(br.readLine().trim());
            String s = br.readLine().trim();
            sb.append(solve(n, s)).append('\n');
        }
        System.out.print(sb);
    }

    // Placeholder - we'll build the real logic in later steps
    static long solve(int n, String s) {
        return 0;
    }
}

What does this template do?

BufferedReader reads input in large chunks instead of character-by-character like Scanner. This avoids the regex overhead that makes Scanner too slow for competitive programming.

StringBuilder collects all output into one string, then prints it in a single call at the end. Calling System.out.println() inside the loop would flush the output buffer on every iteration, adding unnecessary overhead.

The solve method is a placeholder that returns 0 for now. You'll replace it with the real logic in later steps.

  • Save Main.java.
  • Confirm the file exists by running this in your terminal:
ls

You should see Main.java listed in the output.

Don't see Main.java?

  • Make sure you saved the file in VS Code (Cmd+S on macOS, Ctrl+S on Windows).
  • Check that your terminal is in the same folder as your file. Run pwd (macOS) or cd (Windows) to see your current directory.

Help me find my Main.java file

✔️ Awesome, I've got everything!

Great. Make sure you've saved Main.java before moving on.

ⓧ I'd like to double check the full code

Here is the complete Main.java so far. Compare it against your file and fix any differences.

import java.io.*;
import java.util.*;

public class Main {
    static final int MOD = 998244353;

    public static void main(String[] args) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        StringBuilder sb = new StringBuilder();
        int t = Integer.parseInt(br.readLine().trim());
        while (t-- > 0) {
            int n = Integer.parseInt(br.readLine().trim());
            String s = br.readLine().trim();
            sb.append(solve(n, s)).append('\n');
        }
        System.out.print(sb);
    }

    static long solve(int n, String s) {
        return 0;
    }
}
Compile and test the template

The template is in place, but you need to confirm it actually compiles and handles input correctly.

  • Compile and run the program by running this command:
javac Main.java && java Main

What does this command do?

javac Main.java compiles your Java source code into bytecode. The && means the second command only runs if compilation succeeds. java Main then runs the compiled program.

The program is now waiting for your input. It expects lines typed directly into the terminal.

  • Type 1 and press Enter. This tells the program there is 1 test case.
  • Type 2 and press Enter. This is the string length n.
  • Type ?? and press Enter. This is the domino string s.

Before you look at the output, what do you expect the program to print? Remember, the solve method is still a placeholder.

The program outputs 0 and exits. That is the placeholder value from solve. The template is working correctly.

Seeing a compilation error?

  • Check that the class name is exactly Main (capital M) and matches the filename Main.java. Java requires the public class name to match the filename.
  • Make sure every opening brace { has a matching closing brace }.
  • Verify the import lines at the top are exactly import java.io.*; and import java.util.*;.

Help me fix my compilation error

Program seems to hang after running?

  • The program is waiting for your input. Type the three lines of test input described above.
  • If you need to stop the program at any point, press Ctrl+C in the terminal.

Your template is compiled and running cleanly. Next up, you'll build a brute force solver that replaces the placeholder and actually solves the problem for small inputs.

Build a Brute Force Solver

Your fast I/O template reads input and spits out a placeholder 0. The skeleton works, but it does not actually solve anything yet. Before jumping to a clever algorithm, you need to confirm you truly understand what the problem is asking.

The safest way to verify your understanding is to try every possibility. You will build a brute force solver that replaces each ? with both 0 and 1, checks every combination, and counts the valid ones. This approach is far too slow for the full constraints, but it gives you a correct reference you can test against later.

In this step, get ready to:
  • Write a brute force method that enumerates all possible replacements for question mark characters.
  • Write a validity check that ensures consecutive domino weights differ.
  • Test the brute force against all four example test cases and confront its exponential limitation.
Implement the brute force enumeration

Each ? in the string can become 0 or 1. If there are k question marks, there are 2^k total combinations to try. A bitmask integer from 0 to 2^k - 1 encodes every combination neatly.

  • In Main.java, add the following bruteForce method below the solve method:
static long bruteForce(int n, String s) {
    // Convert to char array so we can swap in values for '?'
    char[] arr = s.toCharArray();
    // Collect the positions of all '?' characters
    List<Integer> unknowns = new ArrayList<>();
    for (int i = 0; i < n; i++) {
        if (arr[i] == '?') unknowns.add(i);
    }
    int k = unknowns.size();
    long count = 0;
    // Try every combination: mask from 0 to 2^k - 1
    for (int mask = 0; mask < (1 << k); mask++) {
        // Assign each '?' based on the corresponding bit in mask
        for (int j = 0; j < k; j++) {
            arr[unknowns.get(j)] = (char) ('0' + ((mask >> j) & 1));
        }
        if (isValid(arr, n)) count++;
    }
    return count % MOD;
}

What does this code do?

  • unknowns stores the index of every ? in the string. These are the only positions that need to be filled in.
  • mask is an integer whose binary bits represent one specific replacement. Bit j of mask decides whether the j-th unknown becomes 0 or 1.
  • For each mask, the inner loop writes the chosen values into arr, then calls isValid to check whether the completed string satisfies the domino constraint.
  • The result is returned modulo MOD (998244353) as the problem requires.
  • Save Main.java. The bruteForce method calls isValid, which you will add next.
Implement the validity check

The problem says two consecutive dominoes must have different weights. Domino i spans positions i and i+1, so its weight is s[i] + s[i+1]. The check compares every adjacent pair of dominoes across the entire string.

  • Still in Main.java, add the isValid method below bruteForce:
static boolean isValid(char[] arr, int n) {
    // Check every pair of consecutive dominoes
    for (int i = 0; i + 2 < n; i++) {
        // Domino i has weight arr[i] + arr[i+1]
        int w1 = (arr[i] - '0') + (arr[i + 1] - '0');
        // Domino i+1 has weight arr[i+1] + arr[i+2]
        int w2 = (arr[i + 1] - '0') + (arr[i + 2] - '0');
        if (w1 == w2) return false;
    }
    return true;
}

What does this code do?

  • The loop walks through every pair of consecutive dominoes. Domino i covers positions i and i+1. Domino i+1 covers positions i+1 and i+2.
  • w1 and w2 are the numeric weights of those two dominoes. If any pair shares the same weight, the string is invalid.
  • The condition i + 2 < n ensures the loop stops before reading past the end of the array.
  • Save Main.java.
  • Compile by running this command to confirm both methods are accepted:
javac Main.java
  • The command should complete with no output. No output from javac means success.

Seeing a compilation error?

  • Check that both bruteForce and isValid are inside the Main class braces, not outside them.
  • Confirm the method signatures match exactly. bruteForce returns long and isValid returns boolean.
  • Make sure you have both import java.io.*; and import java.util.*; at the top of the file. ArrayList and List come from java.util.

Why won't my Main.java compile after adding bruteForce and isValid?

Test against the examples and confront the obstacle

Both methods are in place, but main still calls the placeholder solve method. You need to swap it to call bruteForce instead, then test against the official examples.

  • In Main.java, find this line inside the while loop in main:
sb.append(solve(n, s)).append('\n');
  • Replace solve with bruteForce, so the line now looks like this:
sb.append(bruteForce(n, s)).append('\n');

Now create a test file with all four official examples so you can run them in one shot.

  • Create a file called input.txt in your project folder with the following content:
4
2
??
5
0?1??
5
0?0??
8
00110011

What are these test cases?

  • Test 1: n=2, s=??. Two unknown tiles. With only one domino, there is no adjacent pair to conflict, so every replacement is valid.
  • Test 2: n=5, s=0?1??. Three unknowns with some fixed characters constraining the choices.
  • Test 3: n=5, s=0?0??. Same length, different fixed characters. The answer turns out to be zero. No valid replacements exist.
  • Test 4: n=8, s=00110011. No unknowns at all. Either the string is valid as-is or it is not.
  • Save both Main.java and input.txt.

Before you run this, what output do you expect for each of the four test cases? Think about how many ways you can fill in the ? characters while keeping consecutive domino weights different.

  • Compile and run the brute force against your test file:
javac Main.java && java Main < input.txt

You should see four lines of output:

4
2
0
1

If your output matches, your brute force correctly understands the problem. Every combination was tried, every invalid one was filtered out, and the counts are right.

Output doesn't match?

  • If you see four 0s, you probably forgot to swap solve for bruteForce in the main method.
  • Check that input.txt starts with 4 on the first line (the number of test cases), not the first test case directly.
  • Verify the isValid loop condition is i + 2 < n, not i + 1 < n. You need three consecutive characters to form two overlapping dominoes.

Why is my brute force output wrong for the domino tiles examples?

Why can't you just submit this?

The brute force works perfectly for these small examples, but think about what happens when n = 200,000 and every character is ?. That means k = 200,000 unknowns and 2^200,000 combinations to try.

That number is larger than the number of atoms in the observable universe. No computer will finish that enumeration before the 1-second time limit. You need a fundamentally different approach for the real constraints.

✔️ Awesome, I've got everything!

Great. Double check you have saved Main.java.

ⓧ I'd like to double check the full code

Your Main.java should contain these pieces in order. Scroll up and confirm each one is present.

  • The import statements and MOD constant at the top.
  • The main method with the fast I/O loop, now calling bruteForce(n, s) instead of solve(n, s).
  • The original solve method (still returning 0). You will replace this with the real solution later.
  • The bruteForce method with bitmask enumeration.
  • The isValid method checking consecutive domino weights.

Your brute force proves you understand the problem, and it will serve as a ground-truth reference for stress testing. Next up, you will discover an algebraic shortcut hidden inside the constraint that collapses the problem from exponential to linear time.

Discover the Constraint and Build the Efficient Solution

Your brute force solver nails every example test case. But it has an exponential ceiling. For a string of 200,000 question marks, it would need to check 2^200000 possibilities.

The validity constraint looks rigid, but it hides an algebraic shortcut that collapses the entire problem from exponential to linear time. In this step, you will simplify the constraint, decompose the problem into two independent chains, and build an O(n) solution.

In this step, get ready to:
  • Simplify the domino-weight constraint to a single algebraic rule.
  • Implement an O(n) solution using chain decomposition.
  • Verify the efficient solution matches the brute force on all examples.
Simplify the constraint algebraically

Look back at the isValid method you wrote. The condition it checks for each pair of consecutive dominoes is:

// Domino i has weight s[i] + s[i+1]
// Domino i+1 has weight s[i+1] + s[i+2]
// Validity requires: s[i] + s[i+1] != s[i+1] + s[i+2]

Notice that s[i+1] appears on both sides of the inequality. Cancel it from both sides. The entire constraint simplifies to s[i] != s[i+2].

Every character must differ from the character two positions away. That single rule replaces all the domino-weight arithmetic.

What does s[i] != s[i+2] mean?

If every character must differ from the one two steps ahead, the string splits into two independent subsequences.

  • Even-indexed positions (0, 2, 4, ...) form one chain that must strictly alternate between 0 and 1.
  • Odd-indexed positions (1, 3, 5, ...) form a second independent chain that must also strictly alternate.
  • Each chain has at most 2 valid patterns: start with 0 or start with 1.
  • The total answer is the product of valid patterns per chain. The maximum possible answer for any input is 4 (2 options per chain, 2 chains).

Why chain decomposition instead of DP?

A full DP tracking the last two character values would also run in O(n), but it is unnecessary once you see the decomposition. The chain approach is simpler to implement and teaches the transferable skill of constraint simplification.

Implement the chain-counting method

The core of the efficient solution is a helper that checks how many valid alternating patterns exist for one chain. You need a method that tries both starting values (0 and 1) and counts how many are consistent with the fixed characters.

  • In Main.java, add the following countChain method below the solve method:
static long countChain(String s, int start, int n) {
    int count = 0;
    // Try both possible starting values (0 and 1)
    for (int startVal = 0; startVal <= 1; startVal++) {
        boolean valid = true;
        int expected = startVal;
        // Walk the chain, checking every other position
        for (int i = start; i < n; i += 2) {
            char c = s.charAt(i);
            // Fixed characters must match the expected alternating value
            if (c != '?' && (c - '0') != expected) {
                valid = false;
                break;
            }
            // Flip expected for the next position in the chain
            expected = 1 - expected;
        }
        if (valid) count++;
    }
    return count;
}

What does this code do?

  • start picks which chain to check. Pass 0 for even-indexed positions, 1 for odd-indexed positions.
  • The outer loop tries both starting values (0 and 1). The inner loop walks every other position (i += 2), comparing each fixed character against what the alternating pattern expects.
  • A ? character matches any expected value, so the loop skips right past it.
  • The method returns 0, 1, or 2, representing how many of the two possible alternating patterns are consistent with the chain's fixed characters.
  • Save Main.java. This method becomes useful in the next substep when you wire it into solve.

Seeing a compilation error?

  • Make sure the method is inside the Main class braces, not outside them.
  • Check that the loop uses i += 2 (stepping by 2), not i++.
  • Confirm the return type is long, matching what solve expects.

Why won't my countChain method compile in my Main class?

Wire the solution and verify against all examples

Two changes connect your new countChain method to the rest of the program.

  • In Main.java, find the solve method with the placeholder. It currently looks like this:
static long solve(int n, String s) {
    return 0;
}
  • Replace the entire body so solve now reads:
static long solve(int n, String s) {
    // Count valid patterns for even-indexed and odd-indexed chains
    long evenCount = countChain(s, 0, n);
    long oddCount = countChain(s, 1, n);
    return (evenCount * oddCount) % MOD;
}

What does this code do?

countChain(s, 0, n) counts valid patterns for even-indexed positions. countChain(s, 1, n) does the same for odd-indexed positions. Since the two chains are independent, the total number of valid strings is their product, taken mod 998244353.

  • Now scroll up to main and find this line:
sb.append(bruteForce(n, s)).append('\n');
  • Replace bruteForce(n, s) with solve(n, s), so the line looks like this:
sb.append(solve(n, s)).append('\n');
  • Save Main.java.

Before you run this, what do you expect the output to be? The brute force gave you 4, 2, 0, 1. Will the efficient solution agree?

  • Compile and run your program by running this command:
javac Main.java && java Main
  • Paste the following example input, then press Ctrl+D (macOS/Linux) or Ctrl+Z then Enter (Windows) to send EOF:
4
2
??
5
0?1??
5
0?0??
8
00110011

The output on four lines is 4, 2, 0, 1. The results match the brute force exactly, and the solution returns instantly regardless of input size.

Output doesn't match?

  • Make sure you updated the call in main from bruteForce(n, s) to solve(n, s). If main still calls bruteForce, you will get correct output but from the wrong method.
  • If the output is all zeros, confirm you replaced return 0 inside solve with the three-line body that calls countChain.
  • Check that countChain passes 0 for the even chain and 1 for the odd chain as the start parameter.

Why is my efficient solution returning wrong results for the example test cases?

✔️ Awesome, I've got everything!

Great. Double check you have saved Main.java.

ⓧ I'd like to double check the full code

Your Main.java should now have these sections in order. Scroll up through your file and confirm each one is present.

  • Imports (java.io.* and java.util.*) and the MOD constant.
  • main method with fast I/O, now calling solve(n, s) (not bruteForce).
  • solve method calling countChain for even and odd chains, returning their product mod MOD.
  • countChain method that tries both starting values and walks every other position.
  • bruteForce and isValid methods from the previous step (still present for stress testing later).

Your efficient solution matches the brute force on every example. But examples are weak tests. Next up, you will pit the brute force against the optimized solution on thousands of random inputs and then submit for an Accepted verdict on Codeforces.

Stress Test and Submit

Your efficient solution handles the four example test cases correctly and runs in O(n) time. But examples are hand-picked by problem setters to illustrate the problem, not to catch bugs.

In this step, you'll build a stress test that compares your brute force against your optimized solution on 10,000 random inputs. Once every input agrees, you'll submit to Codeforces and earn your Accepted verdict.

In this step, get ready to:
  • Write a stress test that generates random inputs and compares brute force against optimized output.
  • Run the stress test and confirm all 10,000 cases match.
  • Submit your solution to Codeforces and earn an Accepted verdict.
Write a random test generator and stress test loop

The four examples cover basic scenarios, but they miss edge cases like all-question-mark strings of varying lengths, or strings where one chain is fully fixed and the other is all unknowns. A stress test generates thousands of random inputs and verifies that your brute force and optimized solution always agree.

  • In Main.java, scroll to the bottom of the class (just before the closing } of the class) and add the following stressTest method:
    // Generate random test cases and compare brute force vs optimized
    static void stressTest() {
        Random rng = new Random(42);
        for (int iter = 0; iter < 10000; iter++) {
            // Random string length between 2 and 12
            int n = 2 + rng.nextInt(11);
            StringBuilder sb = new StringBuilder();
            for (int i = 0; i < n; i++) {
                // Each character is randomly '0', '1', or '?'
                int r = rng.nextInt(3);
                sb.append(r == 0 ? '0' : r == 1 ? '1' : '?');
            }
            String s = sb.toString();
            long bf = bruteForce(n, s);
            long opt = solve(n, s);
            // Stop immediately if outputs disagree
            if (bf != opt) {
                System.out.println("MISMATCH: n=" + n + " s=" + s + " bf=" + bf + " opt=" + opt);
                return;
            }
        }
        System.out.println("All 10000 tests passed!");
    }

What does this code do?

The method seeds a Random with 42 so the test is reproducible. On each of 10,000 iterations, it builds a string of length 2 to 12 using random characters from 0, 1, and ?.

It runs both bruteForce and solve on the same input and compares their results. If any mismatch is found, it prints the failing input and stops immediately. This small n keeps brute force feasible while generating enough variety to catch subtle bugs.

  • Save Main.java.
Run the stress test

To run the stress test, you need to temporarily bypass the normal input loop in main. You'll add two lines that call stressTest() and return before the input loop runs.

  • In Main.java, find the main method. Locate this first line inside the method body:
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
  • Add these two lines directly above it, so the stress test runs and exits before the input loop:
        stressTest();
        return;
  • Save Main.java.

Before you run this, what do you expect to see? Will any of the 10,000 random inputs produce a mismatch between your brute force and optimized solution?

  • Compile and run the stress test by running this command:
javac Main.java && java Main

You should see this output:

All 10000 tests passed!

Why does this matter?

If even one of those 10,000 random tests had produced a different answer between brute force and optimized, you would have found a bug. This is the gold standard for competitive programming: a stress test catches edge cases that hand-picked examples miss.

Seeing a MISMATCH?

If the output shows a MISMATCH line, read the printed n and s values carefully. Run both methods on that specific input by hand to find where your countChain logic diverges from the brute force. Check that expected = 1 - expected flips correctly and that ? characters are being skipped properly.

If you see a compile error, make sure the stressTest method is inside the Main class (before its closing brace) and that you added stressTest(); return; inside the main method body.

help me debug my stress test

Submit to Codeforces

With 10,000 random tests passing, your solution is ready for the real judge. Time to remove the temporary stress test call, restore normal input, and submit.

  • In Main.java, find the two lines you just added at the top of main:
        stressTest();
        return;
  • Delete both lines so main starts with the BufferedReader line again, like this:
    public static void main(String[] args) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
  • Save Main.java.
  • Verify the restored code still works by compiling and running with the example input:
javac Main.java && java Main
  • Type this input (press Enter after each line):
4
2
??
5
0?1??
5
0?0??
8
00110011
  • After the last line, press Ctrl+Z then Enter (Windows) or Ctrl+D (macOS/Linux) to send EOF.

You should see the familiar output: 4, 2, 0, 1 on four lines.

Now it's time to submit. You'll need a Codeforces account for this.

  • Open your browser and go to:
https://codeforces.com/contest/2256/problem/B
  • If you don't have a Codeforces account, click Register in the top right corner and create a free account. Then sign in.
  • On the problem page, look for the submission area on the right side. Select Java 21 (64-bit) from the language dropdown.
  • Copy your submission code from Main.java. You need the imports, the class declaration, MOD, main, solve, and countChain. Leave out bruteForce, isValid, and stressTest since the judge doesn't need them.
  • Paste the code into the text area on the submission page.
  • Click Submit.

Submitting the full file works too

If trimming feels error-prone, you can paste the entire Main.java contents including bruteForce, isValid, and stressTest. The judge ignores unused methods. The trimmed version is just cleaner.

Before the verdict loads, what do you think the result will be? You've tested against all four examples and 10,000 random inputs. Is there any reason the judge would disagree?

The Codeforces verdict shows Accepted with all test cases passed in well under the 1-second time limit.

Not seeing Accepted?

If you see a Compilation Error, double-check that you included the import java.io.*; and import java.util.*; lines at the top. Also make sure the class is named Main and all braces are balanced.

If you see a Wrong Answer, verify that your solve and countChain methods match the code from the previous step exactly. The stress test passing locally and the judge disagreeing usually means something was lost during copy-paste.

help me debug my Codeforces submission

✔️ Awesome, I've got everything!

Great. Make sure you've saved Main.java with the main method restored to normal input mode (no stressTest() call).

ⓧ I'd like to double check the full code

Your Main.java should now contain these methods, in order:

  • main with the normal BufferedReader input loop (no temporary stressTest() call).
  • solve calling countChain for even and odd positions.
  • countChain trying both starting values and returning the count of valid patterns.
  • bruteForce with bitmask enumeration over unknown positions.
  • isValid checking that consecutive domino weights differ.
  • stressTest generating 10,000 random inputs and comparing both methods.

Scroll up and confirm each method is present in your file. The file should have the MOD constant at the top of the class, followed by these six methods.

Secret mission

Build a Modular Arithmetic Toolkit

The domino tiles problem only ever produces answers up to 4, so modular arithmetic was technically a no-op. But nearly every future counting problem will produce numbers far too large for any integer type. Build a reusable modPow, modInverse, and nCr toolkit that you will copy-paste into dozens of future contest solutions.

Clean Up Your Resources

Clean Up Your Resources

This project runs entirely locally and uses only free resources. There are no ongoing costs or cloud services to worry about.

Resources you used:

  • Your project folder on the Desktop containing Main.java (solution, brute force, stress test harness, and modular arithmetic toolkit).
  • A free Codeforces account with your Accepted submission on problem 2256B.

Decide whether to keep your resources as they are, archive them for later, or delete them entirely.

Keep everything running

No action needed. Choose this if you plan to reuse your code in future competitive programming contests.

  • Your Main.java contains a complete fast I/O template, brute force solver, stress test harness, and modular arithmetic toolkit. All of these are ready to copy into future contest solutions.
  • Your Codeforces account is free and stays active indefinitely.

Pause - I'll come back to this later

Move your project folder into a dedicated archive so your Desktop stays clean. You can come back to the code anytime.

  • Create a solved-problems folder on your Desktop (if you do not already have one) and move your project folder into it.
  • Your Codeforces account is free and requires no action to pause.

Delete - I don't want to use this again

Remove the project folder entirely. Your Codeforces account is free, so there is no cost reason to delete it.

  • Delete your project folder from the Desktop. On macOS/Linux, open a terminal and run:
cd ~/Desktop
ls
  • Find your project folder name in the output, then run:
rm -rf your-project-folder-name
  • On Windows, open PowerShell and run:
cd ~\Desktop
ls
  • Find your project folder name in the output, then run:
Remove-Item -Recurse -Force your-project-folder-name
  • Run ls again to confirm the folder is gone.

Nice Work!

Nice Work!

Nice work! You just solved Codeforces 2256B from scratch, going from a brute force that chokes on real constraints all the way to an Accepted verdict on the judge.

What you learned:

  • Built a reusable Java CP template with fast I/O (BufferedReader + StringBuilder) and discovered that the domino-weight constraint simplifies to s[i] != s[i+2], collapsing an exponential brute force into an O(n) solution using two independent alternating chains.
  • Wrote a stress testing harness that compared brute force against optimized output on 10,000+ random inputs, giving you confidence to submit without guessing.
  • Submitted to Codeforces and earned an Accepted verdict on problem 2256B, proving the solution handles all edge cases within the 1-second time limit.
  • Secret Mission: Built a modular arithmetic toolkit with modPow, modInverse, factorial precomputation, and nCr mod p, ready for any future counting problem.

Ready to quiz yourself?