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Javascript Program to Check if all rows of a matrix are circular rotations of each other

Last Updated : 23 Sep, 2024
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Given a matrix of n*n size, the task is to find whether all rows are circular rotations of each other or not. 

Examples: 

Input: mat[][] = 1, 2, 3
3, 1, 2
2, 3, 1
Output: Yes
All rows are rotated permutation
of each other.

Input: mat[3][3] = 1, 2, 3
3, 2, 1
1, 3, 2
Output: No
Explanation : As 3, 2, 1 is not a rotated or
circular permutation of 1, 2, 3

The idea is based on the below article. 
A Program to check if strings are rotations of each other or not

Steps :  

  1. Create a string of first-row elements and concatenate the string with itself so that string search operations can be efficiently performed. Let this string be str_cat.
  2. Traverse all remaining rows. For every row being traversed, create a string str_curr of current row elements. If str_curr is not a substring of str_cat, return false.
  3. Return true.

Below is the implementation of the above steps. 

JavaScript
// Javascript program to check if all rows of a matrix 
// are rotations of each other 

// Returns true if all rows of mat[0..n-1][0..n-1] 
// are rotations of each other. 
function isPermutedMatrix(mat, n) {

    // Creating a string that contains
    // elements of first row. 
    let str_cat = "";
    for (let i = 0; i < n; i++) {
        str_cat = str_cat + "-" +
            (mat[0][i]).toString();
    }

    // Concatenating the string with itself 
    // so that substring search operations  
    // can be performed on this 
    str_cat = str_cat + str_cat;

    // Start traversing remaining rows 
    for (let i = 1; i < n; i++) {

        // Store the matrix into vector in the form 
        // of strings 
        let curr_str = "";
        for (let j = 0; j < n; j++) {
            curr_str = curr_str + "-" +
                (mat[i][j]).toString();
        }

        // Check if the current string is present in 
        // the concatenated string or not 
        if (str_cat.includes(curr_str)) {
            return true;
        }
    }
    return false;
}

// Drivers code 
let n = 4;
let mat = [[1, 2, 3, 4],
[4, 1, 2, 3],
[3, 4, 1, 2],
[2, 3, 4, 1]];

if (isPermutedMatrix(mat, n))
    console.log("Yes")
else
    console.log("No")

Output
Yes

Complexity Analysis:

  • Time Complexity: O(n^3) 
  • Auxiliary Space: O(n)

Please refer complete article on Check if all rows of a matrix are circular rotations of each other for more details!


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