Sudoku Validator - Problem

A Sudoku puzzle is a 9×9 grid filled with digits 1-9. A valid Sudoku has the following properties:

Each row contains the digits 1-9 without repetition
Each column contains the digits 1-9 without repetition
Each of the nine 3×3 sub-boxes contains the digits 1-9 without repetition

Given a completed 9×9 Sudoku board, determine if it is valid.

Note: The input will always be a complete 9×9 grid with digits 1-9 only (no empty cells).

Input & Output

Example 1 — Valid Sudoku

$ Input: board = [[5,3,4,6,7,8,9,1,2],[6,7,2,1,9,5,3,4,8],[1,9,8,3,4,2,5,6,7],[8,5,9,7,6,1,4,2,3],[4,2,6,8,5,3,7,9,1],[7,1,3,9,2,4,8,5,6],[9,6,1,5,3,7,2,8,4],[2,8,7,4,1,9,6,3,5],[3,4,5,2,8,6,1,7,9]]

› Output: true

💡 Note: This is a valid completed Sudoku. Each row, column, and 3×3 box contains all digits 1-9 exactly once.

Example 2 — Invalid Row

$ Input: board = [[5,3,5,6,7,8,9,1,2],[6,7,2,1,9,5,3,4,8],[1,9,8,3,4,2,5,6,7],[8,5,9,7,6,1,4,2,3],[4,2,6,8,5,3,7,9,1],[7,1,3,9,2,4,8,5,6],[9,6,1,5,3,7,2,8,4],[2,8,7,4,1,9,6,3,5],[3,4,5,2,8,6,1,7,9]]

› Output: false

💡 Note: The first row has two 5s at positions 0 and 2, making it invalid.

Example 3 — Invalid Box

$ Input: board = [[5,3,4,6,7,8,9,1,2],[6,7,2,1,9,5,3,4,8],[1,9,5,3,4,2,5,6,7],[8,5,9,7,6,1,4,2,3],[4,2,6,8,5,3,7,9,1],[7,1,3,9,2,4,8,5,6],[9,6,1,5,3,7,2,8,4],[2,8,7,4,1,9,6,3,5],[3,4,5,2,8,6,1,7,9]]

› Output: false

💡 Note: The top-left 3×3 box contains two 5s, violating the Sudoku rules.

Constraints

board.length == 9
board[i].length == 9
1 ≤ board[i][j] ≤ 9

Visualization

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Asked in

🍎 Apple 15 Ⓜ Microsoft 12 a Amazon 8

The key insight is using hash sets to track numbers we've seen in each row, column, and 3×3 box as we iterate through the board. The optimal approach validates the entire Sudoku in a single pass with O(1) time and space complexity.

Common Approaches

✓ Brute Force Validation

⏱️ Time: O(1) Space: O(1)

For each of the 27 groups (9 rows + 9 columns + 9 boxes), collect all 9 numbers and check for duplicates by comparing every pair. This requires multiple passes through the board.

Single Pass with Sets

⏱️ Time: O(1) Space: O(1)

Iterate through the board once, using sets to track numbers seen in each row, column, and 3×3 box. Detect duplicates instantly when adding to sets.

Brute Force Validation — Algorithm Steps

Check all 9 rows for duplicates using nested loops
Check all 9 columns for duplicates using nested loops
Check all 9 sub-boxes for duplicates using nested loops

Visualization

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Step-by-Step Walkthrough

Check Rows

For each row, compare every pair of elements

Check Columns

For each column, compare every pair of elements

Check Boxes

For each 3×3 box, extract numbers and compare pairs

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <stdbool.h>
#include <string.h>

bool solution(int board[9][9]) {
    // Check rows
    for (int r = 0; r < 9; r++) {
        for (int i = 0; i < 9; i++) {
            for (int j = i + 1; j < 9; j++) {
                if (board[r][i] == board[r][j]) {
                    return false;
                }
            }
        }
    }
    
    // Check columns
    for (int col = 0; col < 9; col++) {
        for (int i = 0; i < 9; i++) {
            for (int j = i + 1; j < 9; j++) {
                if (board[i][col] == board[j][col]) {
                    return false;
                }
            }
        }
    }
    
    // Check 3x3 boxes
    for (int boxRow = 0; boxRow < 3; boxRow++) {
        for (int boxCol = 0; boxCol < 3; boxCol++) {
            int nums[9];
            int idx = 0;
            for (int i = 0; i < 3; i++) {
                for (int j = 0; j < 3; j++) {
                    nums[idx++] = board[boxRow * 3 + i][boxCol * 3 + j];
                }
            }
            // Check for duplicates in this box
            for (int i = 0; i < 9; i++) {
                for (int j = i + 1; j < 9; j++) {
                    if (nums[i] == nums[j]) {
                        return false;
                    }
                }
            }
        }
    }
    
    return true;
}

int main() {
    int board[9][9];
    char line[1000];
    fgets(line, sizeof(line), stdin);
    
    // Parse the 2D array manually
    char *ptr = line + 1; // Skip first '['
    for (int i = 0; i < 9; i++) {
        ptr++; // Skip '['
        for (int j = 0; j < 9; j++) {
            board[i][j] = strtol(ptr, &ptr, 10);
            if (j < 8) ptr++; // Skip comma
        }
        ptr++; // Skip ']'
        if (i < 8) ptr++; // Skip comma
    }
    
    bool result = solution(board);
    printf(result ? "true\n" : "false\n");
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(1)

Fixed 9×9 board means constant operations (27 groups × 81 comparisons = 2187 operations)

✓ Linear Growth

Space Complexity

O(1)

Only uses temporary arrays of size 9 for each group check

✓ Linear Space

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Sudoku Validator - Problem

Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force Validation — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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