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}.
Therefore, the answer is 4.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++ Program to implement // the above approach #include <bits/stdc++.h> 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 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<Integer> balls = new ArrayList<Integer>();
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 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# 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. |
<script> // JavaScript Program to implement
// the above approach
// Function to calculate minimum sets
// required with given conditions
function minimumSetBalls(R, G, B)
{
// Push values R, G, B
// in a vector and sort them
let balls = [];
balls.push(R);
balls.push(G);
balls.push(B);
// Sort the vector
balls.sort( function (a, b) { return a - b })
// Store the answer
let ans = 0;
ans += balls[0];
balls[1] -= balls[0];
balls[2] -= balls[0];
balls[0] = 0;
// Check all mentioned conditions
ans += Math.floor(balls[1] / 3);
balls[1] %= 3;
ans += Math.floor(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
let R = 4, G = 2, B = 4;
document.write(minimumSetBalls(R, G, B));
// This code is contributed by Potta Lokesh
</script>
|
4
Time Complexity: O(1)
Auxiliary Space: O(1)