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

• Difficulty Level : Easy
• Last Updated : 21 May, 2021

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).

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.#includeusing namespace std;#define R 3#define C 4 // returning the greatest common divisor of two numberint 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 Programint 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 = 3C = 4 # returning the greatest common# divisor of two numberdef 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 Codeif __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.

Javascript



Output:

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

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