easy Toeplitz Matrix

Problem Statement

Given an m x n matrix, determine if the given matrix is a Toeplitz matrix.

A Toeplitz matrix is one in which every diagonal from top-left to bottom-right has the same elements.

In other words, the matrix should have the property that each element is equal to the element diagonally down to its right.

Example 1:

• Input:

[[1,2,3],
 [4,1,2],
 [5,4,1]]

• Expected Output: true
• Justification: All diagonals have the same elements. Diagonals of the above matrix are [5], [4, 4], [1, 1, 1], [2, 2], and [3].

Example 2:

• Input:

[[1,2],
 [3,4]]

• Expected Output: false
• Justification: The diagonal are [3], [1, 4], and [2]. A diagonal [1, 2] doesn't have the same elements.

Example 3:

• Input:

[[7,7,7],
 [7,7,7],
 [7,7,7]]

• Expected Output: true
• Justification: Each diagonal, although containing the same repeated element (7), satisfies the condition of a Toeplitz matrix.

Solution

To solve this problem, we will iterate through the elements of the matrix, checking if each element (except those in the last row and last column) matches with its diagonal successor. This approach works because a Toeplitz matrix is defined by the consistency of elements along its diagonals. By verifying this condition, we can ensure whether the matrix adheres to the Toeplitz criteria.

Step-by-step Algorithm

• Iterate over each row of the matrix, except the last row.
  • For each row, iterate over each column, except the last column.
    • Check if the current element is equal to the element diagonally down to its right.
    • If any such pair of elements is not equal, return false immediately.
• If the entire matrix is iterated without finding any mismatch, return true.

Algorithm Walkthrough

Using Example 1: matrix =

[[1,2,3],
 [4,1,2],
 [5,4,1]]

• Start with the first element (1) at matrix[0][0]. It matches with its diagonal successor (1) at matrix[1][1]. Continue.
• Move to the second element (2) in the first row at matrix[0][1]. It matches with its diagonal successor (2) at matrix[1][2]. Continue.
• Skip the last element (3) in the first row as it has no diagonal successor.
• Move to the first element (4) in the second row. It matches with its diagonal successor (4). Continue.
• Next, Move to the second element (1) in the second row. It matches with its diagonal successor (1). Continue.
• Skip the last element (2) in the second row.
• Since we've reached the last row, we conclude that the matrix is a Toeplitz matrix.

Code

public class Solution {

  // Method to check if a matrix is a Toeplitz matrix
  public boolean isToeplitzMatrix(int[][] matrix) {
    for (int i = 0; i < matrix.length - 1; i++) {
      for (int j = 0; j < matrix[0].length - 1; j++) {
        // Check if current element is equal to its diagonal successor
        if (matrix[i][j] != matrix[i + 1][j + 1]) {
          return false;
        }
      }
    }
    return true;
  }

  // Main method to test the algorithm
  public static void main(String[] args) {
    Solution solution = new Solution();
    // Test case 1
    System.out.println(
      solution.isToeplitzMatrix(
        new int[][] { { 1, 2, 3 }, { 4, 1, 2 }, { 5, 4, 1 } }
      )
    ); // true
    // Test case 2
    System.out.println(
      solution.isToeplitzMatrix(new int[][] { { 1, 2 }, { 3, 4 } })
    ); // false
    // Test case 3
    System.out.println(
      solution.isToeplitzMatrix(
        new int[][] { { 7, 7, 7 }, { 7, 7, 7 }, { 7, 7, 7 } }
      )
    ); // true
  }
}

Complexity Analysis

• Time Complexity: , where m is the number of rows and n is the number of columns in the matrix. Each element is checked once.
• Space Complexity: , as no extra space is used besides a few variables for iteration.

🎯 STRICT STANDOUT — Solution Toeplitz Matrix

0. Why / judgment (K3)

Family: Matrix diagonal invariant

Local check is complete: if every consecutive pair on a diagonal matches, whole diagonal constant. Implementation double loop from (1,1). Avoid map-of-diagonals unless streaming constraints force hashing by i-j.

1. Pattern + recognition + when-NOT (K12)

Pattern / template: Nested loops; compare to upper-left; return false on first mismatch.

When-NOT: Don’t extract each diagonal into arrays (extra space, same O(mn)).

2. Complexity derivation (K11)

T=Θ(mn), S=Θ(1).

3. Edge hand-run (K13)

Hand-run 2×2 fail case [[1,2],[3,4]]: only check i=j=1: 4 vs 1 → false.
All equal values: all checks pass → true.
Single cell: true.

4. Interviewer follow-ups & drills

Q1. i-j key approach complexity?
Model answer: Still O(mn) time; O(m+n) space for first-seen map.

Q2. Follow-up: modify one cell to fix?
Model answer: Search mismatches — different problem.

Q3. Row-shift characterization?
Model answer: Each row is previous row shifted right by 1 with new first entry.