# Count of arrangements of RGB balls with no duplicates in a set

• Last Updated : 17 Jan, 2022

Given R red balls, G green balls, B blue balls. A set can have 1, 2, or 3 balls. Also, a set could have all balls of the same color or all balls of different colors. All other possible sets are not considered valid. The task is to calculate the minimum possible sets required to place all balls.

Examples:

Input: R = 4, G = 2, B = 4
Output: 4
Explanation: There can be 4 sets which satisfies all condition mentioned above {R, R, R}, {B, B, B}, {G, R}, {G, B}.

Input: R = 1, G = 7, B = 1
Output: 3
Explanation: There are 3 valid sets {R, G, B}, {G, G, G}, {G, G, G}

Approach: This problem is analysis-based. Follow the steps below to solve the given problem.

• The problem statement can be solved with careful case analysis.
• Without loss of generality assume that R<=G<=B.
• So there will be at least R sets formed. Subtract R from G and B. All the sets formed from this step will have different balls in them.
• The remaining balls will be 0, G – R, B – R. For the remaining balls form sets of the same balls.
• After the above step remaining balls will be 0, (G – R)%3, (B – R)%3.
• Now there can be 3 cases:
• 2nd term is 0 and 3rd term is zero. No extra set will be required.
• (2nd term is 1 and 3rd term is 1) or (2nd term is 0 and 3rd term is 2 or vice versa). 1 extra set will be required.
• In all other cases, 2 more sets will be required.
• Check all the above conditions carefully and print the answer.

Below is the implementation of the above approach:

## C++

 `// C++ Program to implement``// the above approach``#include ``using` `namespace` `std;` `// Function to calculate minimum sets``// required with given conditions``int` `minimumSetBalls(``int` `R, ``int` `G, ``int` `B)``{``    ``// Push values R, G, B``    ``// in a vector and sort them``    ``vector<``int``> balls;``    ``balls.push_back(R);``    ``balls.push_back(G);``    ``balls.push_back(B);` `    ``// Sort the vector``    ``sort(balls.begin(), balls.end());` `    ``// Store the answer``    ``int` `ans = 0;``    ``ans += balls[0];``    ``balls[1] -= balls[0];``    ``balls[2] -= balls[0];``    ``balls[0] = 0;` `    ``// Check all mentioned conditions``    ``ans += balls[1] / 3;``    ``balls[1] %= 3;``    ``ans += balls[2] / 3;``    ``balls[2] %= 3;` `    ``if` `(balls[1] == 0``        ``&& balls[2] == 0) {``        ``// No extra set required``    ``}``    ``else` `if` `(balls[1] == 1``             ``&& balls[2] == 1) {``        ``ans++;``        ``// 1 extra set is required``    ``}``    ``else` `if` `((balls[1] == 2``              ``&& balls[2] == 0)``             ``|| (balls[1] == 0``                 ``&& balls[2] == 2)) {``        ``ans++;``        ``// 1 extra set is required``    ``}``    ``else` `{``        ``// 2 extra sets will be required``        ``ans += 2;``    ``}``    ``return` `ans;``}` `// Driver Code``int` `main()``{``    ``int` `R = 4, G = 2, B = 4;``    ``cout << minimumSetBalls(R, G, B);``}`

## Java

 `// Java program for the above approach``import` `java.io.*;``import` `java.util.*;` `public` `class` `GFG{` `// Function to calculate minimum sets``// required with given conditions``static` `int` `minimumSetBalls(``int` `R, ``int` `G, ``int` `B)``{``  ` `    ``// Push values R, G, B``    ``// in a vector and sort them``    ``ArrayList balls = ``new` `ArrayList();``    ``balls.add(R);``    ``balls.add(G);``    ``balls.add(B);``  ` `    ``// Sort the vector``    ``Collections.sort(balls);   ;``  ` `    ``// Store the answer``    ``int` `ans = ``0``;``    ``ans += balls.get(``0``);``    ``balls.set(``1``,balls.get(``1``) - balls.get(``0``));``    ``balls.set(``2``,balls.get(``2``) - balls.get(``0``));``    ``balls.set(``0``,``0``);``  ` `    ``// Check all mentioned conditions``    ``ans += balls.get(``1``) / ``3``;``    ``balls.set(``1``, balls.get(``1``) % ``3``);``    ``ans += balls.get(``2``) / ``3``;``    ``balls.set(``2``,balls.get(``2``) % ``3``);``  ` `    ``if` `(balls.get(``1``) == ``0``        ``&& balls.get(``2``) == ``0``)``    ``{``      ` `        ``// No extra set required``    ``}``    ``else` `if` `(balls.get(``1``) == ``1``             ``&& balls.get(``2``) == ``1``) {``        ``ans++;``        ``// 1 extra set is required``    ``}``    ``else` `if` `((balls.get(``1``) == ``2``              ``&& balls.get(``2``) == ``0``)``             ``|| (balls.get(``1``) == ``0``                 ``&& balls.get(``2``) == ``2``)) {``        ``ans++;``        ``// 1 extra set is required``    ``}``    ``else` `{``        ``// 2 extra sets will be required``        ``ans += ``2``;``    ``}``    ``return` `ans;``}`  `// Driver Code``public` `static` `void` `main(String []args)``{``    ``int` `R = ``4``, G = ``2``, B = ``4``;``    ``System.out.println( minimumSetBalls(R, G, B));``}``}` `// This code is contributed by AnkThon`

## Python3

 `# Python3 Program to implement the above approach` `# Function to calculate minimum sets``# required with given conditions``def` `minimumSetBalls(R, G, B) :` `    ``# Push values R, G, B``    ``# in a vector and sort them``    ``balls ``=` `[];``    ``balls.append(R);``    ``balls.append(G);``    ``balls.append(B);` `    ``# Sort the vector``    ``balls.sort();` `    ``# Store the answer``    ``ans ``=` `0``;``    ``ans ``+``=` `balls[``0``];``    ``balls[``1``] ``-``=` `balls[``0``];``    ``balls[``2``] ``-``=` `balls[``0``];``    ``balls[``0``] ``=` `0``;` `    ``# Check all mentioned conditions``    ``ans ``+``=` `balls[``1``] ``/``/` `3``;``    ``balls[``1``] ``%``=` `3``;``    ``ans ``+``=` `balls[``2``] ``/``/` `3``;``    ``balls[``2``] ``%``=` `3``;` `    ``if` `(balls[``1``] ``=``=` `0` `and` `balls[``2``] ``=``=` `0``) :``        ``pass``    ` `    ``elif` `(balls[``1``] ``=``=` `1` `and` `balls[``2``] ``=``=` `1``) :``        ``ans ``+``=` `1``;``        ``# 1 extra set is required` `    ``elif` `((balls[``1``] ``=``=` `2` `and` `balls[``2``] ``=``=` `0``) ``or` `(balls[``1``] ``=``=` `0` `and` `balls[``2``] ``=``=` `2``)) :``        ``ans ``+``=` `1``;``        ``# 1 extra set is required``    ``else` `:``        ``# 2 extra sets will be required``        ``ans ``+``=` `2``;` `    ``return` `ans;` `# Driver Code``if` `__name__ ``=``=` `"__main__"` `:` `    ``R ``=` `4``; G ``=` `2``; B ``=` `4``;``    ``print``(minimumSetBalls(R, G, B));``    ` `    ``# This code is contributed by AnkThon`

## C#

 `// C# program for the above approach``using` `System;``using` `System.Collections.Generic;` `class` `GFG{` `// Function to calculate minimum sets``// required with given conditions``static` `int` `minimumSetBalls(``int` `R, ``int` `G, ``int` `B)``{``  ` `    ``// Push values R, G, B``    ``// in a vector and sort them``    ``List<``int``> balls = ``new` `List<``int``>();``    ``balls.Add(R);``    ``balls.Add(G);``    ``balls.Add(B);``  ` `    ``// Sort the vector``    ``balls.Sort();``  ` `    ``// Store the answer``    ``int` `ans = 0;``    ``ans += balls[0];``    ``balls[1] -= balls[0];``    ``balls[2] -= balls[0];``    ``balls[0] = 0;``  ` `    ``// Check all mentioned conditions``    ``ans += balls[1] / 3;``    ``balls[1] %= 3;``    ``ans += balls[2] / 3;``    ``balls[2] %= 3;``  ` `    ``if` `(balls[1] == 0``        ``&& balls[2] == 0)``    ``{``      ` `        ``// No extra set required``    ``}``    ``else` `if` `(balls[1] == 1``             ``&& balls[2] == 1) {``        ``ans++;``        ``// 1 extra set is required``    ``}``    ``else` `if` `((balls[1] == 2``              ``&& balls[2] == 0)``             ``|| (balls[1] == 0``                 ``&& balls[2] == 2)) {``        ``ans++;``        ``// 1 extra set is required``    ``}``    ``else` `{``        ``// 2 extra sets will be required``        ``ans += 2;``    ``}``    ``return` `ans;``}`  `// Driver Code``public` `static` `void` `Main()``{``    ``int` `R = 4, G = 2, B = 4;``    ``Console.WriteLine( minimumSetBalls(R, G, B));``}``}` `// This code is contributed by sanjoy_62.`

## Javascript

 ``
Output:
`4`

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

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