LeetCode #794 — MEDIUM

Valid Tic-Tac-Toe State

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

Problem Statement

Given a Tic-Tac-Toe board as a string array board, return true if and only if it is possible to reach this board position during the course of a valid tic-tac-toe game.

The board is a 3 x 3 array that consists of characters ' ', 'X', and 'O'. The ' ' character represents an empty square.

Here are the rules of Tic-Tac-Toe:

Players take turns placing characters into empty squares ' '.
The first player always places 'X' characters, while the second player always places 'O' characters.
'X' and 'O' characters are always placed into empty squares, never 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.
No more moves can be played if the game is over.

Example 1:

Input: board = ["O  ","   ","   "]
Output: false
Explanation: The first player always plays "X".

Example 2:

Input: board = ["XOX"," X ","   "]
Output: false
Explanation: Players take turns making moves.

Example 3:

Input: board = ["XOX","O O","XOX"]
Output: true

Constraints:

board.length == 3
board[i].length == 3
board[i][j] is either 'X', 'O', or ' '.

Step 01

Brute Force Baseline

Problem summary: Given a Tic-Tac-Toe board as a string array board, return true if and only if it is possible to reach this board position during the course of a valid tic-tac-toe game. The board is a 3 x 3 array that consists of characters ' ', 'X', and 'O'. The ' ' character represents an empty square. Here are the rules of Tic-Tac-Toe: Players take turns placing characters into empty squares ' '. The first player always places 'X' characters, while the second player always places 'O' characters. 'X' and 'O' characters are always placed into empty squares, never 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. No more moves can be played if the game is over.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array

Example 1

["O  ","   ","   "]

Example 2

["XOX"," X ","   "]

Example 3

["XOX","O O","XOX"]

Core Insight

What unlocks the optimal approach

No official hints in dataset. Start from constraints and look for a monotonic or reusable state.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #794: Valid Tic-Tac-Toe State
class Solution {
    private String[] board;

    public boolean validTicTacToe(String[] board) {
        this.board = board;
        int x = count('X'), o = count('O');
        if (x != o && x - 1 != o) {
            return false;
        }
        if (win('X') && x - 1 != o) {
            return false;
        }
        return !(win('O') && x != o);
    }

    private boolean win(char x) {
        for (int i = 0; i < 3; ++i) {
            if (board[i].charAt(0) == x && board[i].charAt(1) == x && board[i].charAt(2) == x) {
                return true;
            }
            if (board[0].charAt(i) == x && board[1].charAt(i) == x && board[2].charAt(i) == x) {
                return true;
            }
        }
        if (board[0].charAt(0) == x && board[1].charAt(1) == x && board[2].charAt(2) == x) {
            return true;
        }
        return board[0].charAt(2) == x && board[1].charAt(1) == x && board[2].charAt(0) == x;
    }

    private int count(char x) {
        int cnt = 0;
        for (var row : board) {
            for (var c : row.toCharArray()) {
                if (c == x) {
                    ++cnt;
                }
            }
        }
        return cnt;
    }
}

// Accepted solution for LeetCode #794: Valid Tic-Tac-Toe State
func validTicTacToe(board []string) bool {
	var x, o int
	for _, row := range board {
		for _, c := range row {
			if c == 'X' {
				x++
			} else if c == 'O' {
				o++
			}
		}
	}
	win := func(x byte) bool {
		for i := 0; i < 3; i++ {
			if board[i][0] == x && board[i][1] == x && board[i][2] == x {
				return true
			}
			if board[0][i] == x && board[1][i] == x && board[2][i] == x {
				return true
			}
		}
		if board[0][0] == x && board[1][1] == x && board[2][2] == x {
			return true
		}
		return board[0][2] == x && board[1][1] == x && board[2][0] == x
	}
	if x != o && x-1 != o {
		return false
	}
	if win('X') && x-1 != o {
		return false
	}
	return !(win('O') && x != o)
}

# Accepted solution for LeetCode #794: Valid Tic-Tac-Toe State
class Solution:
    def validTicTacToe(self, board: List[str]) -> bool:
        def win(x):
            for i in range(3):
                if all(board[i][j] == x for j in range(3)):
                    return True
                if all(board[j][i] == x for j in range(3)):
                    return True
            if all(board[i][i] == x for i in range(3)):
                return True
            return all(board[i][2 - i] == x for i in range(3))

        x = sum(board[i][j] == 'X' for i in range(3) for j in range(3))
        o = sum(board[i][j] == 'O' for i in range(3) for j in range(3))
        if x != o and x - 1 != o:
            return False
        if win('X') and x - 1 != o:
            return False
        return not (win('O') and x != o)

// Accepted solution for LeetCode #794: Valid Tic-Tac-Toe State
struct Solution;

impl Solution {
    fn valid_tic_tac_toe(board: Vec<String>) -> bool {
        let mut a = vec![vec![]; 3];
        let mut x = 0;
        let mut o = 0;
        for (i, row) in board.iter().enumerate() {
            for c in row.chars() {
                match c {
                    'X' => {
                        x += 1;
                    }
                    'O' => {
                        o += 1;
                    }
                    _ => {}
                }
                a[i].push(c);
            }
        }
        let win_x = Self::win(&a, 'X');
        let win_o = Self::win(&a, 'O');
        x == o + 1 && win_x >= 0 && win_o == 0 || x == o && win_x == 0 && win_o <= 1
    }

    fn win(board: &[Vec<char>], c: char) -> i32 {
        let mut rows = vec![0; 3];
        let mut cols = vec![0; 3];
        let mut diagonals = vec![0; 2];
        for i in 0..3 {
            for j in 0..3 {
                let v = if board[i][j] == c { 1 } else { 0 };
                rows[i] += v;
                cols[j] += v;
                if i == j {
                    diagonals[0] += v;
                }
                if i + j == 2 {
                    diagonals[1] += v;
                }
            }
        }
        let mut sum = 0;
        if rows.iter().any(|&row| row == 3) {
            sum += 1;
        }
        if cols.iter().any(|&col| col == 3) {
            sum += 1;
        }
        if diagonals.iter().any(|&diagonal| diagonal == 3) {
            sum += 1;
        }
        sum
    }
}

#[test]
fn test() {
    let board = vec_string!["O  ", "   ", "   "];
    let res = false;
    assert_eq!(Solution::valid_tic_tac_toe(board), res);
    let board = vec_string!["XOX", " X ", "   "];
    let res = false;
    assert_eq!(Solution::valid_tic_tac_toe(board), res);
    let board = vec_string!["XXX", "   ", "OOO"];
    let res = false;
    assert_eq!(Solution::valid_tic_tac_toe(board), res);
    let board = vec_string!["XOX", "O O", "XOX"];
    let res = true;
    assert_eq!(Solution::valid_tic_tac_toe(board), res);
    let board = vec_string!["XXX", "OOX", "OOX"];
    let res = true;
    assert_eq!(Solution::valid_tic_tac_toe(board), res);
}

// Accepted solution for LeetCode #794: Valid Tic-Tac-Toe State
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #794: Valid Tic-Tac-Toe State
// class Solution {
//     private String[] board;
// 
//     public boolean validTicTacToe(String[] board) {
//         this.board = board;
//         int x = count('X'), o = count('O');
//         if (x != o && x - 1 != o) {
//             return false;
//         }
//         if (win('X') && x - 1 != o) {
//             return false;
//         }
//         return !(win('O') && x != o);
//     }
// 
//     private boolean win(char x) {
//         for (int i = 0; i < 3; ++i) {
//             if (board[i].charAt(0) == x && board[i].charAt(1) == x && board[i].charAt(2) == x) {
//                 return true;
//             }
//             if (board[0].charAt(i) == x && board[1].charAt(i) == x && board[2].charAt(i) == x) {
//                 return true;
//             }
//         }
//         if (board[0].charAt(0) == x && board[1].charAt(1) == x && board[2].charAt(2) == x) {
//             return true;
//         }
//         return board[0].charAt(2) == x && board[1].charAt(1) == x && board[2].charAt(0) == x;
//     }
// 
//     private int count(char x) {
//         int cnt = 0;
//         for (var row : board) {
//             for (var c : row.toCharArray()) {
//                 if (c == x) {
//                     ++cnt;
//                 }
//             }
//         }
//         return cnt;
//     }
// }

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(1)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.