CodingGyan

Toeplitz Matrix Check

easy Matrix (2D arrays)

Problem

A Toeplitz matrix has the same value on every diagonal that goes from the top-left to the bottom-right. Print true if the matrix is Toeplitz, else false.

Constraints

  • 1 <= r, c <= 500

Examples

Input

3 4
1 2 3 4
5 1 2 3
9 5 1 2

Output

true

Input

2 2
1 2
2 2

Output

false

Solutions

Best approachTime O(r * c)Space O(1)

If every cell equals its top-left neighbour, then every diagonal is constant, because each diagonal is a chain of such neighbours. This needs no list of diagonals.

Same code works in Java810172125Basic features only, so nothing changes between versions.
/*
 * Problem : Process a matrix.
 * Input   : r and c, then r rows of c integers.
 * Output  : true or false.
 * Approach: Compare with the top-left neighbour.
 * Time    : O(r * c)
 * Space   : O(1)
 * Java    : 8 or newer.
 */
import java.util.Scanner;

public class Main {
    // This method solves the task. It receives the input and returns the answer.
    static boolean solve(int[][] m) {
        // Step 1: A cell must be equal to its neighbour at the top-left. Cells in the first row or first column have no such neighbour.
        for (int i = 1; i < m.length; i++) {
            for (int j = 1; j < m[0].length; j++) {
                // Step 2: One difference on a diagonal is enough to decide.
                if (m[i][j] != m[i - 1][j - 1]) {
                    return false;
                }
            }
        }
        return true;
    }

    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        int rows = scanner.nextInt();
        int cols = scanner.nextInt();
        int[][] m = new int[rows][cols];
        for (int i = 0; i < rows; i++) {
            for (int j = 0; j < cols; j++) {
                m[i][j] = scanner.nextInt();
            }
        }
        System.out.println(solve(m));
    }
}
Hints (1)
  1. Which cells have no top-left neighbour?

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