How do you solve Check if Every Row and Column Contains All Numbers on LeetCode?

Check if Every Row and Column Contains All Numbers (LeetCode #2133) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Every row and column must contain unique integers from 1 to n.

What is the time complexity of Check if Every Row and Column Contains All Numbers?

The optimal solution for Check if Every Row and Column Contains All Numbers runs in O(n) time and uses O(n) space. The time complexity is O(n) because we only need to iterate through the matrix once, and the space complexity is O(n) due to the row and column count arrays.

What is the brute force solution for Check if Every Row and Column Contains All Numbers?

The brute force approach for Check if Every Row and Column Contains All Numbers is Brute Force, with O(n²) time complexity and O(1) space complexity. We can check each row and each column individually to see if they contain all numbers from 1 to n. If any row or column is missing a number, we can immediately return false. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Check if Every Row and Column Contains All Numbers?

Check if Every Row and Column Contains All Numbers uses the following concepts: Array, Hash Table, Matrix. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Check if Every Row and Column Contains All Numbers?

Explain your thought process clearly. Consider edge cases like minimum and maximum n. Discuss time and space complexity during your solution.

What are common mistakes when solving Check if Every Row and Column Contains All Numbers?

Not checking both rows and columns. Assuming the input is always valid without validation.

#2133

Check if Every Row and Column Contains All Numbers

Easy

ArrayHash TableMatrixHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

Instead of using sets for each row and column, we can use a single array to track occurrences of numbers. This reduces the overhead of creating sets and allows us to check validity more efficiently.

⚙️

Algorithm

4 steps

1Step 1: Create two arrays of size n initialized to zero, one for rows and one for columns.
2Step 2: Iterate through each element in the matrix, incrementing the corresponding row and column counts.
3Step 3: After filling the counts, check if each row and column has exactly one occurrence of each number from 1 to n.
4Step 4: If any row or column fails this check, return false; otherwise, return true.

solution.py12 lines

1def checkMatrix(matrix):
2    n = len(matrix)
3    row_count = [0] * n
4    col_count = [0] * n
5    for i in range(n):
6        for j in range(n):
7            num = matrix[i][j] - 1
8            row_count[i] += 1
9            col_count[j] += 1
10            if row_count[i] > n or col_count[j] > n:
11                return False
12    return True if all(count == n for count in row_count) and all(count == n for count in col_count) else False

ℹ

Complexity note: The time complexity is O(n) because we only need to iterate through the matrix once, and the space complexity is O(n) due to the row and column count arrays.

1Every row and column must contain unique integers from 1 to n.
2Using sets can simplify checking for duplicates.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.