Check if Every Row and Column Contains All Numbers - Problem

An n x n matrix is valid if every row and every column contains all the integers from 1 to n (inclusive).

Given an n x n integer matrix matrix, return true if the matrix is valid. Otherwise, return false.

Input & Output

Example 1 — Valid Matrix

$ Input: matrix = [[1,2,3],[2,3,1],[3,1,2]]

› Output: true

💡 Note: Each row contains [1,2,3] and each column contains [1,2,3] in some order. Row 1: [1,2,3]✓, Row 2: [2,3,1]✓, Row 3: [3,1,2]✓. Column 1: [1,2,3]✓, Column 2: [2,3,1]✓, Column 3: [3,1,2]✓

Example 2 — Invalid Matrix

$ Input: matrix = [[1,1,1],[1,2,3],[1,2,3]]

› Output: false

💡 Note: Row 1 contains [1,1,1] which is missing 2 and 3, and has duplicate 1s. This violates the requirement that each row must contain all numbers from 1 to n exactly once.

Example 3 — 2x2 Valid Matrix

$ Input: matrix = [[1,2],[2,1]]

› Output: true

💡 Note: For n=2, each row and column must contain [1,2]. Row 1: [1,2]✓, Row 2: [2,1]✓, Column 1: [1,2]✓, Column 2: [2,1]✓

Constraints

n == matrix.length == matrix[i].length
1 ≤ n ≤ 100
1 ≤ matrix[i][j] ≤ n

Visualization

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Asked in

a Amazon 15 G Google 12 M Microsoft 8 🍎 Apple 6

The key insight is that each row and column must be a perfect permutation of numbers 1 to n. Best approach uses boolean arrays for O(1) lookups to verify presence of all required numbers. Time: O(n²), Space: O(n)

Common Approaches

✓ Brute Force with Sorting

⏱️ Time: O(n³ log n) Space: O(n)

For each row and column, create a copy, sort it, and verify it contains exactly the numbers 1 through n in order. This approach is straightforward but inefficient due to multiple sorting operations.

Hash Set Validation

⏱️ Time: O(n²) Space: O(n)

For each row and column, use a hash set to track which numbers we've seen. After processing each row/column, verify the set contains exactly the numbers 1 through n.

Single Pass with Boolean Arrays

⏱️ Time: O(n²) Space: O(n²)

Instead of using hash sets, use boolean arrays of size n+1 to track which numbers have been seen. This eliminates the overhead of hash operations and provides constant-time lookups.

Brute Force with Sorting — Algorithm Steps

Create expected array [1,2,...,n]
For each row: copy, sort, compare with expected
For each column: extract, sort, compare with expected
Return true only if all match

Visualization

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Step-by-Step Walkthrough

Sort Each Row

Copy and sort each row, compare with [1,2,3]

Sort Each Column

Extract and sort each column, compare with [1,2,3]

Result

Return true if all match expected sequence

Code -

solution.c — C

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

int compare(const void *a, const void *b) {
    return (*(int*)a - *(int*)b);
}

bool solution(int** matrix, int n) {
    int* expected = malloc(n * sizeof(int));
    for (int i = 0; i < n; i++) {
        expected[i] = i + 1;
    }
    
    // Check all rows
    for (int i = 0; i < n; i++) {
        int* sortedRow = malloc(n * sizeof(int));
        memcpy(sortedRow, matrix[i], n * sizeof(int));
        qsort(sortedRow, n, sizeof(int), compare);
        
        for (int j = 0; j < n; j++) {
            if (sortedRow[j] != expected[j]) {
                free(sortedRow);
                free(expected);
                return false;
            }
        }
        free(sortedRow);
    }
    
    // Check all columns
    for (int col = 0; col < n; col++) {
        int* column = malloc(n * sizeof(int));
        for (int row = 0; row < n; row++) {
            column[row] = matrix[row][col];
        }
        qsort(column, n, sizeof(int), compare);
        
        for (int j = 0; j < n; j++) {
            if (column[j] != expected[j]) {
                free(column);
                free(expected);
                return false;
            }
        }
        free(column);
    }
    
    free(expected);
    return true;
}

int main() {
    static char buffer[10000];
    fgets(buffer, sizeof(buffer), stdin);
    
    // Remove newline if present
    int len = strlen(buffer);
    if (len > 0 && buffer[len-1] == '\n') {
        buffer[len-1] = '\0';
    }
    
    // Find the start of the matrix
    char* ptr = strchr(buffer, '[');
    if (!ptr) return 1;
    
    // Count the number of rows by counting '[' after the first one
    int n = 0;
    char* temp = ptr + 1; // Skip the outer '['
    while (*temp) {
        if (*temp == '[') n++;
        temp++;
    }
    
    // Allocate matrix
    int** matrix = malloc(n * sizeof(int*));
    for (int i = 0; i < n; i++) {
        matrix[i] = malloc(n * sizeof(int));
    }
    
    // Parse the matrix
    ptr++; // Skip outer '['
    int row = 0;
    
    while (*ptr && row < n) {
        // Skip to the start of a row
        while (*ptr && *ptr != '[') ptr++;
        if (!*ptr) break;
        ptr++; // Skip '['
        
        // Parse numbers in this row
        int col = 0;
        char* endptr;
        while (*ptr && *ptr != ']' && col < n) {
            // Skip whitespace and commas
            while (*ptr && (*ptr == ' ' || *ptr == ',' || *ptr == '\t')) ptr++;
            
            if (*ptr >= '0' && *ptr <= '9') {
                matrix[row][col] = strtol(ptr, &endptr, 10);
                ptr = endptr;
                col++;
            } else if (*ptr != ']') {
                ptr++;
            }
        }
        
        // Skip past the closing ']' of this row
        while (*ptr && *ptr != ']') ptr++;
        if (*ptr == ']') ptr++;
        
        row++;
    }
    
    bool result = solution(matrix, n);
    printf(result ? "true\n" : "false\n");
    
    // Clean up
    for (int i = 0; i < n; i++) {
        free(matrix[i]);
    }
    free(matrix);
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n³ log n)

n rows + n columns, each sorted in O(n log n), total O(n × n log n) = O(n² log n)

⚠ Quadratic Growth

Space Complexity

O(n)

Temporary arrays for sorting each row/column

⚠ Quadratic Space

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Check if Every Row and Column Contains All Numbers - Problem

Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force with Sorting — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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