# Replace every matrix element with maximum of GCD of row or column

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 = {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).
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 the implementation of this approach:

## C++

 `// C++ program to replace each each element with ` `// maximum of GCD of row or column. ` `#include ` `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<

## Java

 `// Java program to replace each each element with ` `// maximum of GCD of row or column. ` `import` `java .io.*; ` ` `  `class` `GFG ` `{ ` `      ``static` `int` `R = ``3``; ` `      ``static` `int` `C = ``4``; ` ` `  `      ``// returning the greatest common  ` `      ``// divisor of two number ` `      ``static` `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. ` `static` `void` `replacematrix(``int` `[][]mat, ``int` `n, ``int` `m) ` `{ ` `    ``int` `[]rgcd = ``new` `int``[R] ; ` `    ``int` `[]cgcd = ``new` `int``[C]; ` ` `  `    ``// 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] = Math.max(rgcd[i], cgcd[j]); ` `} ` ` `  `// Driver program ` `    ``static` `public` `void` `main (String[] args){ ` `    ``int` `[][]m = ` `    ``{ ` `        ``{``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++) ` `        ``System.out.print(m[i][j] + ``" "``); ` `        ``System.out.println(); ` `    ``} ` `    ``} ` `} ` ` `  `//This code is contributed by vt_m. `

## Python3

# Python3 program to replace each each element
# with maximum of GCD of row or column.

R = 3
C = 4

# returning the greatest common
# divisor of two number
def gcd(a, 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.
def replacematrix(mat, n, m):

rgcd =  * R
cgcd =  * C

# Calculating GCD of each row and each
# column in O(mn) and store in arrays.
for i in range (n):
for j in range (m):

rgcd[i] = gcd(rgcd[i], mat[i][j])
cgcd[j] = gcd(cgcd[j], mat[i][j])

# Replacing matrix element
for i in range (n):
for j in range (m):
mat[i][j] = max(rgcd[i], cgcd[j])

# Driver Code
if __name__ == “__main__”:

m = [[1, 2, 3, 3],
[4, 5, 6, 6],
[7, 8, 9, 9]]

replacematrix(m, R, C)

for i in range(R):
for j in range (C):
print ( m[i][j], end = ” “)
print ()

# This code is contributed by ita_c

## C#

 `// C# program to replace each each element with ` `// maximum of GCD of row or column. ` `using` `System; ` ` `  `class` `GFG ` `{ ` `      ``static` `int` `R = 3; ` `      ``static` `int` `C = 4; ` `   `  `      ``// returning the greatest common  ` `      ``// divisor of two number ` `      ``static` `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. ` `static` `void` `replacematrix(``int` `[,]mat, ``int` `n, ``int` `m) ` `{ ` `    ``int` `[]rgcd = ``new` `int``[R] ; ` `    ``int` `[]cgcd = ``new` `int``[C]; ` ` `  `    ``// 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] = Math.Max(rgcd[i], cgcd[j]); ` `} ` ` `  `// Driver program ` `    ``static` `public` `void` `Main (){ ` `    ``int` `[,]m = ` `    ``{ ` `        ``{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++) ` `        ``Console.Write(m[i,j] + ``" "``); ` `        ``Console.WriteLine(); ` `    ``} ` `    ``} ` `} ` ` `  `//This code is contributed by vt_m. `

Output:

```1 1 3 3
1 1 3 3
1 1 3 3
```

Time Complexity : O(mn).
Axillary Space : O(m + n).

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

This article is attributed to GeeksforGeeks.org

## tags:

Matrix GCD-LCM Matrix

code

load comments