Maximum number of 3-person teams formed from two groups
Given two integers N1 and N2 where, N1 is the number of people in group 1 and N2 is the number of people in group 2. The task is to count the maximum number of 3-person teams that can be formed when at least a single person is chosen from both the groups.
Examples:
Input: N1 = 2, N2 = 8
Output: 2
Team 1: 2 members from group 2 and 1 member from group 1
Update: N1 = 1, N2 = 6
Team 2: 2 members from group 2 and 1 member from group 1
Update: N1 = 0, N2 = 4
No further teams can be formed.
Input: N1 = 4, N2 = 5
Output: 3
Approach: Choose a single person from the team with less members and choose 2 persons from the team with more members (while possible) and update count = count + 1. Print the count in the end.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach #include <bits/stdc++.h> using namespace std; // Function to return the count // of maximum teams possible int maxTeams( int N1, int N2) { int count = 0; // While it is possible to form a team while (N1 > 0 && N2 > 0 && N1 + N2 >= 3) { // Choose 2 members from group 1 // and a single member from group 2 if (N1 > N2) { N1 -= 2; N2 -= 1; } // Choose 2 members from group 2 // and a single member from group 1 else { N1 -= 1; N2 -= 2; } // Update the count count++; } // Return the count return count; } // Driver code int main() { int N1 = 4, N2 = 5; cout << maxTeams(N1, N2); return 0; } |
Java
// Java implementation of the approach class GFG { // Function to return the count // of maximum teams possible static int maxTeams( int N1, int N2) { int count = 0 ; // While it is possible to form a team while (N1 > 0 && N2 > 0 && N1 + N2 >= 3 ) { // Choose 2 members from group 1 // and a single member from group 2 if (N1 > N2) { N1 -= 2 ; N2 -= 1 ; } // Choose 2 members from group 2 // and a single member from group 1 else { N1 -= 1 ; N2 -= 2 ; } // Update the count count++; } // Return the count return count; } // Driver code public static void main(String []args) { int N1 = 4 , N2 = 5 ; System.out.println(maxTeams(N1, N2)); } } // This code is contributed by ihritik |
Python3
# Python3 implementation of the approach # Function to return the count # of maximum teams possible def maxTeams(N1, N2): count = 0 # While it is possible to form a team while (N1 > 0 and N2 > 0 and N1 + N2 > = 3 ) : # Choose 2 members from group 1 # and a single member from group 2 if (N1 > N2): N1 - = 2 N2 - = 1 # Choose 2 members from group 2 # and a single member from group 1 else : N1 - = 1 N2 - = 2 # Update the count count = count + 1 # Return the count return count # Driver code N1 = 4 N2 = 5 print (maxTeams(N1, N2)) # This code is contributed by ihritik |
C#
// C# implementation of the approach using System; class GFG { // Function to return the count // of maximum teams possible static int maxTeams( int N1, int N2) { int count = 0; // While it is possible to form a team while (N1 > 0 && N2 > 0 && N1 + N2 >= 3) { // Choose 2 members from group 1 // and a single member from group 2 if (N1 > N2) { N1 -= 2; N2 -= 1; } // Choose 2 members from group 2 // and a single member from group 1 else { N1 -= 1; N2 -= 2; } // Update the count count++; } // Return the count return count; } // Driver code public static void Main() { int N1 = 4, N2 = 5; Console.WriteLine(maxTeams(N1, N2)); } } // This code is contributed by ihritik |
Javascript
<script> // Javascript implementation of the approach // Function to return the count // of maximum teams possible function maxTeams(N1, N2) { let count = 0; // While it is possible to form a team while (N1 > 0 && N2 > 0 && N1 + N2 >= 3) { // Choose 2 members from group 1 // and a single member from group 2 if (N1 > N2) { N1 -= 2; N2 -= 1; } // Choose 2 members from group 2 // and a single member from group 1 else { N1 -= 1; N2 -= 2; } // Update the count count++; } // Return the count return count; } let N1 = 4, N2 = 5; document.write(maxTeams(N1, N2)); // This code is contributed by divyeshrabadiya07. </script> |
PHP
<?php // PHP implementation of the approach // Function to return the count // of maximum teams possible function maxTeams( $N1 , $N2 ) { $count = 0 ; // While it is possible to form a team while ( $N1 > 0 && $N2 > 0 && $N1 + $N2 >= 3) { // Choose 2 members from group 1 // and a single member from group 2 if ( $N1 > $N2 ) { $N1 -= 2; $N2 -= 1; } // Choose 2 members from group 2 // and a single member from group 1 else { $N1 -= 1; $N2 -= 2; } // Update the count $count ++; } // Return the count return $count ; } // Driver code $N1 = 4 ; $N2 = 5 ; echo maxTeams( $N1 , $N2 ); // This code is contributed by Ryuga ?> |
3
Time Complexity: O(min(N1,N2))
Auxiliary Space: O(1)
New Approach:- Here, another approach we can calculate the maximum number of teams that can be formed by dividing the total number of people from both groups by 3, rounding down to the nearest integer. This is because each team consists of 3 people.
However, we need to make sure that at least one person is chosen from both groups. So, if the number of people in one of the groups is less than 3, we cannot form a team and the answer would be 0. Otherwise, the answer would be the maximum number of teams that can be formed as calculated earlier.
Algorithm:-
- Calculate the total number of people by adding N1 and N2.
- Check if N1, N2, or the total number of people is less than 3. If so, it is impossible to form a team and the function returns 0.
- Calculate the maximum number of teams possible by dividing the total number of people by 3 (since each team must have 3 members).
- Return the maximum number of teams as the output of the function.
- In the driver code, the function is called with the values N1=4 and N2=5, and the result is printed.
Here’s the implementation of the above approach:
C++
#include <iostream> using namespace std; // Function to return the count of maximum teams possible int maxTeams( int N1, int N2) { int total_people = N1 + N2; if (N1 < 1 || N2 < 1 || total_people < 3) { // If either N1 or N2 is less than 1, or the total number of people // is less than 3, it's not possible to form a team. return 0; } // Calculate the maximum number of teams that can be formed by dividing // the total number of people by 3 (each team requires 3 people). int max_teams = total_people / 3; return max_teams; } // Driver code int main() { int N1 = 4; int N2 = 5; cout << maxTeams(N1, N2) << endl; return 0; } |
Java
public class Main { // Function to return the count of maximum teams // possible static int maxTeams( int N1, int N2) { int totalPeople = N1 + N2; if (N1 < 1 || N2 < 1 || totalPeople < 3 ) { // If either N1 or N2 is less than 1, or the // total number of people is less than 3, it's // not possible to form a team. return 0 ; } // Calculate the maximum number of teams that can be // formed by dividing the total number of people by // 3 (each team requires 3 people). int maxTeams = totalPeople / 3 ; return maxTeams; } public static void main(String[] args) { int N1 = 4 ; int N2 = 5 ; System.out.println(maxTeams(N1, N2)); } } |
Python3
# Python3 implementation of the approach # Function to return the count of maximum teams possible def maxTeams(N1, N2): total_people = N1 + N2 if N1 < 1 or N2 < 1 or total_people < 3 : # Cannot form a team with less than 3 people or # if at least one group has no people return 0 max_teams = total_people / / 3 return max_teams # Driver code N1 = 4 N2 = 5 print (maxTeams(N1, N2)) |
C#
using System; public class GFG { // Function to return the count of maximum teams possible public static int MaxTeams( int N1, int N2) { int totalPeople = N1 + N2; if (N1 < 1 || N2 < 1 || totalPeople < 3) { // If either N1 or N2 is less than 1, or the total number of people // is less than 3, it's not possible to form a team. return 0; } // Calculate the maximum number of teams that can be formed by dividing // the total number of people by 3 (each team requires 3 people). int maxTeams = totalPeople / 3; return maxTeams; } // Driver code public static void Main() { int N1 = 4; int N2 = 5; Console.WriteLine(MaxTeams(N1, N2)); } } |
Javascript
// Function to return the count of maximum teams possible function maxTeams(N1, N2) { const totalPeople = N1 + N2; // Check if either N1 or N2 is less than 1, or the total number of people // is less than 3, it's not possible to form a team. if (N1 < 1 || N2 < 1 || totalPeople < 3) { return 0; } // Calculate the maximum number of teams that can be formed by dividing // the total number of people by 3 (each team requires 3 people). const maxTeamsCount = Math.floor(totalPeople / 3); return maxTeamsCount; } // Driver code const N1 = 4; const N2 = 5; console.log(maxTeams(N1, N2)); // Output: 3 |
Output:-
3
Time Complexity:- The time complexity of this code is O(1)
Auxiliary Space:- The auxiliary space complexity of this code is also O(1)