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# Place N boys and M girls in different rows such that count of persons placed in each row is maximized

Given two integers N and M representing the number of boys and girls, the task is to arrange them in number of different rows of same size such that each row contains the maximum number of students possible and each row should contain either boys or girls.
Note: No row can contain both boys and girls.

Example:

Input: N = 4, M = 2
Output: 2
Explanation:
The following order of arrangement satisfies the given conditions:
1st Row: B1, B2
2nd Row: B3, B4
3rd Row: G1, G2
Clearly, every row has either boys or girls.

Input: N = 6, M = 6
Output: 6
Explanation:
The following order of arrangement satisfies the given conditions:
1st Row: B1, B2, B3, B4, B5, B6
2nd Row: G1, G2, G3, G4, G5, G6

Approach: Follow the steps below to solve the problem

• Since each row can contain either boys or girls and size of all rows must be same, the most optimal arrangement possible is by placing greater common divisor of (N, M) number of elements in each row.
• Therefore, print GCD(N, M) as the required answer.

Below is the implementation of the above approach:

## C++14

 `// C++ Program to implement``// the above approach` `#include ``using` `namespace` `std;` `// Function to calculate``// GCD of two numbers``int` `gcd(``int` `a, ``int` `b)``{``    ``if` `(b == 0)``        ``return` `a;` `    ``return` `gcd(b, a % b);``}` `// Function to count maximum persons``// that can be placed in a row``int` `maximumRowValue(``int` `n, ``int` `m) { ``return` `gcd(n, m); }` `// Driver Code``int` `main()``{``    ``// Input``    ``int` `N = 4;``    ``int` `M = 2;` `    ``// Function to count maximum``    ``// persons that can be placed in a row``    ``cout << maximumRowValue(N, M);``}`

## Java

 `// Java Program to implement``// the above approach``import` `java.util.*;``class` `GFG``{` `  ``// Function to calculate``  ``// GCD of two numbers``  ``static` `int` `gcd(``int` `a, ``int` `b)``  ``{``    ``if` `(b == ``0``)``      ``return` `a;` `    ``return` `gcd(b, a % b);``  ``}` `  ``// Function to count maximum persons``  ``// that can be placed in a row``  ``static` `int` `maximumRowValue(``int` `n, ``int` `m) {``    ``return` `gcd(n, m);``  ``}` `  ``// Driver Code``  ``public` `static` `void` `main(String args[])``  ``{``    ``// Input``    ``int` `N = ``4``;``    ``int` `M = ``2``;` `    ``// Function to count maximum``    ``// persons that can be placed in a row``    ``System.out.print(maximumRowValue(N, M));``  ``}``}` `// This code is contributed by SURENDRA_GANGWAR.`

## Python3

 `# Python3 Program to implement``# the above approach` `# Function to calculate``# GCD of two numbers``def` `gcd(a, b):``    ``if` `(b ``=``=` `0``):``        ``return` `a` `    ``return` `gcd(b, a ``%` `b)` `# Function to count maximum persons``# that can be placed in a row``def` `maximumRowValue(n, m):``    ``return` `gcd(n, m)` `# Driver Code``if` `__name__ ``=``=` `'__main__'``:``  ` `    ``# Input``    ``N ``=` `4``    ``M ``=` `2` `    ``# Function to count maximum``    ``# persons that can be placed in a row``    ``print` `(maximumRowValue(N, M))` `# This code is contributed by mohit kumar 29.`

## C#

 `// C# Program to implement``// the above approach``using` `System;``class` `GFG``{` `  ``// Function to calculate``  ``// GCD of two numbers``  ``static` `int` `gcd(``int` `a, ``int` `b)``  ``{``    ``if` `(b == 0)``      ``return` `a;` `    ``return` `gcd(b, a % b);``  ``}` `  ``// Function to count maximum persons``  ``// that can be placed in a row``  ``static` `int` `maximumRowValue(``int` `n, ``int` `m) { ``return` `gcd(n, m); }` `  ``// Driver Code``  ``public` `static` `void` `Main(String[] args)``  ``{``    ` `    ``// Input``    ``int` `N = 4;``    ``int` `M = 2;` `    ``// Function to count maximum``    ``// persons that can be placed in a row``    ``Console.WriteLine(maximumRowValue(N, M));``  ``}``}` `// This code is contributed by code_hunt.`

## Javascript

 ``

Output:

`2`

Time Complexity: O(log N)
Auxiliary Space: O(1)

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