# Sum of GCDs of each row of the given matrix

• Last Updated : 01 Aug, 2022

Given a matrix mat[][] of size N * N, the task is to find the sum of GCDs of all the rows of the given matrix.

Examples:

Input: arr[][] = {
{2, 4, 6, 8},
{3, 6, 9, 12},
{4, 8, 12, 16},
{5, 10, 15, 20}};
Output: 14
gcd(2, 4, 6, 8) = 2
gcd(3, 6, 9, 12) = 3
gcd(4, 8, 12, 16) = 4
gcd(5, 10, 15, 20) = 5
2 + 3 + 4 + 5 = 14

Input: arr[][] = {
{1, 1, 1, 1},
{2, 2, 2, 2},
{3, 3, 3, 3},
{4, 4, 4, 4}};
Output: 10

Approach: Find the GCD of each row of the matrix and add it to the total sum. The total sum will be the sum of GCD of all the rows.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach``#include ``using` `namespace` `std;` `// Function to return gcd of a and b``int` `gcd(``int` `a, ``int` `b)``{``    ``if` `(a == 0)``        ``return` `b;``    ``return` `gcd(b % a, a);``}` `// Function to return the sum of gcd``// of each row of the given matrix``int` `sumGcd(``int` `arr[][4], ``int` `n)``{` `    ``// To store the required sum``    ``int` `sum = 0;` `    ``for` `(``int` `i = 0; i < n; i++) {` `        ``// To store the gcd of the current row``        ``int` `gcdRow = arr[i][0];``        ``for` `(``int` `j = 1; j < n; j++)``            ``gcdRow = gcd(arr[i][j], gcdRow);` `        ``// Add gcd of the current row to the sum``        ``sum += gcdRow;``    ``}` `    ``// Return the required sum``    ``return` `sum;``}` `// Driver code``int` `main()``{``    ``int` `arr[][4] = { { 2, 4, 6, 8 },``                     ``{ 3, 6, 9, 12 },``                     ``{ 4, 8, 12, 16 },``                     ``{ 5, 10, 15, 20 } };``    ``int` `size = ``sizeof``(arr) / ``sizeof``(arr[0]);` `    ``cout << sumGcd(arr, size);` `    ``return` `0;``}`

## Java

 `// Java implementation of the approach` `class` `GFG``{` `    ``// Function to return gcd of a and b``    ``static` `int` `gcd(``int` `a, ``int` `b)``    ``{``        ``if` `(a == ``0``)``            ``return` `b;``        ``return` `gcd(b % a, a);``    ``}``    ` `    ``// Function to return the sum of gcd``    ``// of each row of the given matrix``    ``static` `int` `sumGcd(``int` `arr[][], ``int` `n)``    ``{``    ` `        ``// To store the required sum``        ``int` `sum = ``0``;``    ` `        ``for` `(``int` `i = ``0``; i < n; i++)``        ``{``    ` `            ``// To store the gcd of the current row``            ``int` `gcdRow = arr[i][``0``];``            ``for` `(``int` `j = ``1``; j < n; j++)``                ``gcdRow = gcd(arr[i][j], gcdRow);``    ` `            ``// Add gcd of the current row to the sum``            ``sum += gcdRow;``        ``}``    ` `        ``// Return the required sum``        ``return` `sum;``    ``}``    ` `    ``// Driver code``    ``public` `static` `void` `main (String[] args)``    ``{``        ``int` `arr[][] = { { ``2``, ``4``, ``6``, ``8` `},``                        ``{ ``3``, ``6``, ``9``, ``12` `},``                        ``{ ``4``, ``8``, ``12``, ``16` `},``                        ``{ ``5``, ``10``, ``15``, ``20` `} };``                        ` `        ``int` `size = arr.length ;``    ` `        ``System.out.println(sumGcd(arr, size));``    ` `    ``}``}` `// This code is contributed by AnkitRai01`

## Python3

 `# Python3 implementation of the approach` `# Function to return gcd of a and b``def` `gcd(a, b):``    ``if` `(a ``=``=` `0``):``        ``return` `b``    ``return` `gcd(b ``%` `a, a)`  `# Function to return the Sum of gcd``# of each row of the given matrix``def` `SumGcd(arr, n):` `    ``# To store the required Sum``    ``Sum` `=` `0` `    ``for` `i ``in` `range``(n):` `        ``# To store the gcd of the current row``        ``gcdRow ``=` `arr[i][``0``]` `        ``for` `j ``in` `range``(``1``,n):``            ``gcdRow ``=` `gcd(arr[i][j], gcdRow)` `        ``# Add gcd of the current row to the Sum``        ``Sum` `+``=` `gcdRow` `    ``# Return the required Sum``    ``return` `Sum`  `# Driver code` `arr``=` `[ [ ``2``, ``4``, ``6``, ``8` `],``    ``[ ``3``, ``6``, ``9``, ``12` `],``    ``[ ``4``, ``8``, ``12``, ``16` `],``    ``[ ``5``, ``10``, ``15``, ``20` `] ]` `size ``=` `len``(arr)` `print``(SumGcd(arr, size))` `# This code is contributed by mohit kumar 29`

## C#

 `// C# implementation of the approach``using` `System;``class` `GFG``{` `    ``// Function to return gcd of a and b``    ``static` `int` `gcd(``int` `a, ``int` `b)``    ``{``        ``if` `(a == 0)``            ``return` `b;``        ``return` `gcd(b % a, a);``    ``}``    ` `    ``// Function to return the sum of gcd``    ``// of each row of the given matrix``    ``static` `int` `sumGcd(``int` `[ , ]arr, ``int` `n)``    ``{``    ` `        ``// To store the required sum``        ``int` `sum = 0;``    ` `        ``for` `(``int` `i = 0; i < n; i++)``        ``{``    ` `            ``// To store the gcd of the current row``            ``int` `gcdRow = arr[i, 0];``            ``for` `(``int` `j = 1; j < n; j++)``                ``gcdRow = gcd(arr[i, j], gcdRow);``    ` `            ``// Add gcd of the current row to the sum``            ``sum += gcdRow;``        ``}``    ` `        ``// Return the required sum``        ``return` `sum;``    ``}``    ` `    ``// Driver code``    ``public` `static` `void` `Main ()``    ``{``        ``int` `[ , ] arr = ``new` `int``[ 4, 4 ]{{ 2, 4, 6, 8 },``                                        ``{ 3, 6, 9, 12 },``                                        ``{ 4, 8, 12, 16 },``                                        ``{ 5, 10, 15, 20 }};``                        ` `        ``int` `size = 4;``        ``Console.WriteLine(sumGcd(arr, size));``    ``}``}` `// This code is contributed by ihritik`

## PHP

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## Javascript

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Output

`14`

Time Complexity: O(n2 log(min(a,b))
Auxiliary Space: O(1)

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