Given a matrix of **n** rows and **m** columns. The task is to replace each matrix element with Greatest Common Divisor of its row or column, whichever is maximum. That is, for each element (i, j) replace it from GCD of i’th row or GCD of j’th row, whichever is greater.

Examples:

Input :mat[3][4] = {1, 2, 3, 3, 4, 5, 6, 6 7, 8, 9, 9}Output :1 1 3 3 1 1 3 3 1 1 3 3 For index (0,2), GCD of row 0 is 1, GCD of row 2 is 3. So replace index (0,2) with 3 (3>1).

The idea is to us concept discussed here LCM of an array to find the GCD of row and column.

Using the **brute force**, we can traverse element of matrix, find the GCD of row and column corresponding to the element and replace it with maximum of both.

An **Efficient method** is to make two arrays of size n and m for row and column respectively. And store the GCD of each row and each column. An Array of size n will contain GCD of each row and array of size m will contain the GCD of each column. And replace each element with maximum of its corresponding row GCD or column GCD.

Below is C++ implementation of this approach:

// C++ program to replace each each element with // maximum of GCD of row or column. #include<bits/stdc++.h> using namespace std; #define R 3 #define C 4 // returning the greatest common divisor of two number int gcd(int a, int b) { if (b == 0) return a; return gcd(b, a%b); } // Finding GCD of each row and column and replacing // with each element with maximum of GCD of row or // column. void replacematrix(int mat[R][C], int n, int m) { int rgcd[R] = { 0 }, cgcd[C] = { 0 }; // Calculating GCD of each row and each column in // O(mn) and store in arrays. for (int i = 0; i < n; i++) { for (int j = 0; j < m; j++) { rgcd[i] = gcd(rgcd[i], mat[i][j]); cgcd[j] = gcd(cgcd[j], mat[i][j]); } } // Replacing matrix element for (int i = 0; i < n; i++) for (int j = 0; j < m; j++) mat[i][j] = max(rgcd[i], cgcd[j]); } // Driven Program int main() { int m[R][C] = { 1, 2, 3, 3, 4, 5, 6, 6, 7, 8, 9, 9, }; replacematrix(m, R, C); for (int i = 0; i < R; i++) { for (int j = 0; j < C; j++) cout << m[i][j] << " "; cout<<endl; } return 0; }

Output:

1 1 3 3 1 1 3 3 1 1 3 3

**Time Complexity : **O(mn).

**Axillary Space : **O(m + n).

This article is contributed by **Anuj Chauahan(anuj0503)**. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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