How do you solve Toeplitz Matrix on LeetCode?

Toeplitz Matrix (LeetCode #766) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: Each diagonal in a Toeplitz matrix has the same elements.

What is the brute force solution for Toeplitz Matrix?

The brute force approach for Toeplitz Matrix is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach checks each diagonal in the matrix one by one, ensuring that all elements in that diagonal are the same. This is straightforward but can be inefficient for larger matrices. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Toeplitz Matrix?

Toeplitz Matrix uses the following concepts: Array, Matrix. The recommended patterns to study are: Array, Matrix.

What are the interview tips for Toeplitz Matrix?

Always clarify the constraints and edge cases before diving into the solution. Think about how you can reduce the number of checks needed. Practice explaining your thought process as you code.

What are common mistakes when solving Toeplitz Matrix?

Not considering the edges of the matrix properly. Confusing the diagonal checks with row or column checks.

#766

Toeplitz Matrix

Easy

ArrayMatrixArrayMatrix

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

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Intuition

Time O(n)Space O(1)

The optimal solution leverages the observation that if each element in the matrix is equal to its top-left neighbor, then the matrix is Toeplitz. This reduces unnecessary checks and improves efficiency.

⚙️

Algorithm

3 steps

1Step 1: Iterate through the matrix starting from the second row and second column.
2Step 2: For each element, check if it is equal to the element diagonally above-left.
3Step 3: If any element is not equal, return false. If all checks pass, return true.

solution.py9 lines

1# Full working Python code
2
3def isToeplitzMatrix(matrix):
4    rows, cols = len(matrix), len(matrix[0])
5    for i in range(1, rows):
6        for j in range(1, cols):
7            if matrix[i][j] != matrix[i-1][j-1]:
8                return False
9    return True

ℹ

Complexity note: The time complexity is O(n) because we only check each element once (excluding the first row and column). The space complexity is O(1) since we do not use any additional space.

1Each diagonal in a Toeplitz matrix has the same elements.
2We can check the condition by comparing each element with its top-left neighbor.

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