Check whether a Matrix is a Latin Square or not

Given a square matrix of size N x N, the task is to check if it is Latin square or not.

A square matrix is a Latin Square if each cell of the matrix contains one of N different values (in the range [1, N]), and no value is repeated within a row or a column.

Examples:

Input: 1 2 3 4
       2 1 4 3
       3 4 1 2
       4 3 2 1
Output: YES

Input: 2 2 2 2
       2 3 2 3
       2 2 2 3
       2 2 2 2
Output: NO

Naive Approach:

  1. For every element, we first check whether the given element is already present in the given row and given column by iterating over all the elements of the given row and given column.
  2. If not, then check whether the value is less than or equal to N, if yes, move for the next element.
  3. If any of the above points is false, then the matrix is not a Latin square.

Efficient Approach: Here is the more efficient approach using Set data structure in C++:



  1. Define sets for each row and each column and create two arrays of sets, one for all the rows and the other for columns.
  2. Iterate over all the element and insert the value of the given element in corresponding row set and in the corresponding column set.
  3. Also, check whether the given value is less than N or not. If not, Print “NO” and return.
  4. Now, Iterate over all row sets and column sets and check if the size of set is less than N or not.
  5. If Yes, Print “YES”. Otherwise Print “NO”.

Below is the implementation of the above approach.

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// C++ program to check if given matrix
// is natural latin square or not
  
#include <bits/stdc++.h>
using namespace std;
  
void CheckLatinSquare(int mat[4][4])
{
    // Size of square matrix is NxN
    int N = sizeof(mat[0]) / sizeof(mat[0][0]);
  
    // Vector of N sets corresponding
    // to each row.
    vector<set<int> > rows(N);
  
    // Vector of N sets corresponding
    // to each column.
    vector<set<int> > cols(N);
  
    // Number of invalid elements
    int invalid = 0;
  
    for (int i = 0; i < N; i++) {
        for (int j = 0; j < N; j++) {
            rows[i].insert(mat[i][j]);
            cols[j].insert(mat[i][j]);
  
            if (mat[i][j] > N || mat[i][j] <= 0) {
                invalid++;
            }
        }
    }
    // Number of rows with
    // repeatative elements.
    int numrows = 0;
  
    // Number of columns with
    // repeatative elements.
    int numcols = 0;
  
    // Checking size of every row
    // and column
    for (int i = 0; i < N; i++) {
        if (rows[i].size() != N) {
            numrows++;
        }
        if (cols[i].size() != N) {
            numcols++;
        }
    }
  
    if (numcols == 0 && numrows == 0
        && invalid == 0)
        cout << "YES" << endl;
    else
        cout << "NO" << endl;
  
    return;
}
  
// Driver code
int main()
{
  
    int Matrix[4][4] = { { 1, 2, 3, 4 },
                         { 2, 1, 4, 3 },
                         { 3, 4, 1, 2 },
                         { 4, 3, 2, 1 } };
  
    // Funtion call
    CheckLatinSquare(Matrix);
  
    return 0;
}

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Output:

YES

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