What is the time complexity of Find Winner on a Tic Tac Toe Game?

The optimal solution has a time complexity of O(1).

What data structures are used to solve Find Winner on a Tic Tac Toe Game?

This problem uses Array, Hash Table, Matrix, Simulation. Understanding these concepts is essential for solving this Easy problem efficiently.

What are the key insights for solving Find Winner on a Tic Tac Toe Game?

**Direct Game Simulation on a 2D Array:** The most intuitive and effective way to analyze a fixed-size board game like Tic-Tac-Toe is to explicitly build and maintain its state. Using a `3x3` list of lists (or a 2D array) directly mirrors the game board, allowing straightforward placement of 'X's and 'O's based on the input `moves` sequence. **Exhaustive Win Condition Checking:** For a 3x3 Tic-Tac-Toe board, there are only 8 possible winning lines (3 rows, 3 columns, 2 diagonals). The `check_winner` helper function systematically iterates through each of these lines for *both* players. This fixed, small number of checks makes determining a winner efficient and reliable, regardless of the sequence of moves, as it's a constant time operation for a 3x3 grid.

Which companies ask LeetCode 1275. Find Winner on a Tic Tac Toe Game in interviews?

This problem is frequently asked at Qualcomm, Tesla, Zoho.

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Find Winner on a Tic Tac Toe Game - LeetCode 1275 Solution

Find Winner on a Tic Tac Toe Game - Complete Solution Guide

Find Winner on a Tic Tac Toe Game is LeetCode problem 1275, a Easy level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

Tic-tac-toe is played by two players A and B on a 3 x 3 grid. The rules of Tic-Tac-Toe are: Players take turns placing characters into empty squares ' ' . The first player A always places 'X' characters, while the second player B always places 'O' characters. 'X' and 'O' characters are always placed into empty squares, never on filled ones. The game ends when there are three of the same (non-empty) character filling any row, column, or diagonal. The game also ends if all squares are non-empty. N

Detailed Explanation

We're tasked with analyzing a sequence of moves in a standard 3x3 Tic-Tac-Toe game and determining the final state: whether Player A (X) won, Player B (O) won, the game ended in a draw, or it's still pending a winner. The `moves` input is an array of `[row, col]` pairs, representing the chronological order of placements. Player A always goes first, placing 'X', then Player B places 'O', and so on. Importantly, we're given the *entire* sequence of moves for a completed or partially completed game, not a dynamic turn-by-turn input. We just need to simulate the given game to its conclusion.

Solution Approach

The provided solution takes a direct, simulation-based approach. It meticulously reconstructs the game board by iterating through the `moves` list. A 3x3 grid, initialized with empty strings, serves as our game board. It alternates placing 'X' for Player A and 'O' for Player B based on whose `turn` it is. After all moves from the input `moves` list have been applied to the `grid`, the solution then performs a comprehensive check for a winner.

Step-by-Step Algorithm

Step 1: Initialize a 3x3 grid (e.g., a 2D array) to represent the Tic-Tac-Toe board. Initially, all cells are empty.
Step 2: Iterate through the `moves` list. For each move `[row, col]`, place the appropriate player's mark ('X' or 'O') on the board.
Step 3: After each move (or after all moves), check for a win condition by examining rows, columns, and diagonals. A player wins if they have three of their marks in a row.
Step 4: If a win is detected, return the winner ('A' or 'B').
Step 5: If no win is detected and all squares are filled, return 'Draw'.
Step 6: If no win is detected and the board is not full, return 'Pending'.

Key Insights

**Direct Game Simulation on a 2D Array:** The most intuitive and effective way to analyze a fixed-size board game like Tic-Tac-Toe is to explicitly build and maintain its state. Using a `3x3` list of lists (or a 2D array) directly mirrors the game board, allowing straightforward placement of 'X's and 'O's based on the input `moves` sequence.
**Exhaustive Win Condition Checking:** For a 3x3 Tic-Tac-Toe board, there are only 8 possible winning lines (3 rows, 3 columns, 2 diagonals). The `check_winner` helper function systematically iterates through each of these lines for *both* players. This fixed, small number of checks makes determining a winner efficient and reliable, regardless of the sequence of moves, as it's a constant time operation for a 3x3 grid.
**Prioritized Outcome Determination:** The final result ("A", "B", "Draw", "Pending") must be determined in a specific order after checking for individual player wins. First, check for Player A's win, then Player B's win. If neither has won, then distinguish between a "Draw" (all 9 squares filled, i.e., `len(moves) == 9`) and "Pending" (fewer than 9 squares filled and no winner). This order of checking is crucial to correctly identify the earliest possible end-state.

Complexity Analysis

Time Complexity: O(1)

Space Complexity: O(1)

Topics

This problem involves: Array, Hash Table, Matrix, Simulation.

Companies

Asked at: Qualcomm, Tesla, Zoho.

Find Winner on a Tic Tac Toe Game | LeetCode 1275

Easy

Array Hash Table Matrix Simulation

Time: O(1)Space: O(1)

Solution Code

Available in:

class Solution {
    public String tictactoe(int[][] moves) {
        char[][] board = new char[3][3];
        for (int i = 0; i < moves.length; i++) {
            int row = moves[i][0];
            int col = moves[i][1];
            board[row][col] = (i % 2 == 0) ? 'X' : 'O';
        }

        for (int i = 0; i < 3; i++) {
            if (board[i][0] != '\0' && board[i][0] == board[i][1] && board[i][1] == board[i][2]) {
                return (board[i][0] == 'X') ? "A" : "B";
            }
            if (board[0][i] != '\0' && board[0][i] == board[1][i] && board[1][i] == board[2][i]) {
                return (board[0][i] == 'X') ? "A" : "B";
            }
        }

        if (board[0][0] != '\0' && board[0][0] == board[1][1] && board[1][1] == board[2][2]) {
            return (board[0][0] == 'X') ? "A" : "B";
        }
        if (board[0][2] != '\0' && board[0][2] == board[1][1] && board[1][1] == board[2][0]) {
            return (board[0][2] == 'X') ? "A" : "B";
        }

        if (moves.length == 9) {
            return "Draw";
        } else {
            return "Pending";
        }
    }
}

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Problem Info

Description

Problem 1275: Find Winner on a Tic Tac Toe Game - Detailed explanation and solution approaches.

Topics

Array Hash Table Matrix Simulation

Companies

QualcommTeslaZoho

Stats

Difficulty:Easy

Languages:4

Companies:3

Problem #:1275

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

6 steps

1
Step 1: Initialize a 3x3 grid (e.g., a 2D array) to represent the Tic-Tac-Toe board. Initially, all cells are empty.
2
Step 2: Iterate through the `moves` list. For each move `[row, col]`, place the appropriate player's mark ('X' or 'O') on the board.
3
Step 3: After each move (or after all moves), check for a win condition by examining rows, columns, and diagonals. A player wins if they have three of their marks in a row.
4
Step 4: If a win is detected, return the winner ('A' or 'B').
5
Step 5: If no win is detected and all squares are filled, return 'Draw'.
6
Step 6: If no win is detected and the board is not full, return 'Pending'.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(1)

⚡ Constant time - Super fast! Same speed regardless of input size.

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The time complexity is constant because the board size is fixed (3x3). The number of operations for checking the win condition and filling the board is always bounded by a constant. Similarly, the space complexity is constant because only a 3x3 board and a few variables are used, regardless of the input size. The input `moves` list has a maximum length of 9, which is a constant.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Bit manipulation

Represent the board as a single integer using bit manipulation. Each bit represents a cell on the board. Win conditions can be checked using bitwise operations.

Time: O(1)Space: O(1)

Trade-offs: Might be slightly faster for win condition checks but less readable. The constant-time performance remains the same.

✨

Key Insights

1**Direct Game Simulation on a 2D Array:** The most intuitive and effective way to analyze a fixed-size board game like Tic-Tac-Toe is to explicitly build and maintain its state. Using a `3x3` list of lists (or a 2D array) directly mirrors the game board, allowing straightforward placement of 'X's and 'O's based on the input `moves` sequence.
2**Exhaustive Win Condition Checking:** For a 3x3 Tic-Tac-Toe board, there are only 8 possible winning lines (3 rows, 3 columns, 2 diagonals). The `check_winner` helper function systematically iterates through each of these lines for *both* players. This fixed, small number of checks makes determining a winner efficient and reliable, regardless of the sequence of moves, as it's a constant time operation for a 3x3 grid.
3**Prioritized Outcome Determination:** The final result ("A", "B", "Draw", "Pending") must be determined in a specific order after checking for individual player wins. First, check for Player A's win, then Player B's win. If neither has won, then distinguish between a "Draw" (all 9 squares filled, i.e., `len(moves) == 9`) and "Pending" (fewer than 9 squares filled and no winner). This order of checking is crucial to correctly identify the earliest possible end-state.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Hash Table

A data structure that maps keys to values for quick lookup. Like a dictionary!

Edge Cases

Edge case 1: Empty moves list: The solution correctly handles an empty `moves` list, returning 'Pending'.
Edge case 2: Game ends in a draw: The solution correctly identifies draws when all squares are filled without a winner.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Incorrect win condition check: Failing to consider all diagonal possibilities or handling boundary conditions correctly.

✗Common mistake 2: Incorrect turn management: Mixing up the players' turns and assigning 'X' and 'O' incorrectly.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Game development: This algorithm can be directly applied to create the logic for a Tic-Tac-Toe game in any programming environment.

🏦Real-world scenario 2: Simple game AI: The win-checking part of this algorithm can be adapted to create a very basic AI opponent for a Tic-Tac-Toe game.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

Even for easy problems, discuss time and space complexity. Interviewers want to see you think analytically.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

👨‍💻 From Experience

This is one of the first problems I solved when learning Array. It's a great foundation for more complex problems.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: June 16, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 13

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJul 6

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJun 29

❓

Frequently Asked Questions

When should I use in-place modification vs creating a new array?

Use in-place modification when space complexity matters (O(1) space requirement) and the original array can be modified. Create a new array when you need to preserve the original data or when the problem involves significant restructuring that would complicate in-place logic.

How do I choose between a hash map and a hash set?

Use a hash set when you only need to track presence/absence of elements (like detecting duplicates). Use a hash map when you need to associate values with keys (like counting frequencies or storing indices). Hash sets use less memory since they only store keys.

What should I learn from easy problems?

Easy problems introduce core patterns that appear in harder problems. Master basic operations (iteration, conditionals, simple data structures), recognize common patterns (counting, searching, basic transformations), and practice explaining your thought process clearly.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Find Winner on a Tic Tac Toe Game"?

For this easy-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether hash table techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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