Crying Colours CRYCOLR Solution Codechef

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Codechef Crying Colours CRYCOLR Solution

You have a total of 3N3N balls of colours RedRed, BlueBlue and GreenGreen. There are exactly NN balls of each colour.

These 3N3N balls are now distributed to 33 boxes such that each box contains exactly NN balls. You are given the contents of each box.

You would like the 1st1st box to contain all the red balls, the 2nd2nd box to contain all the green balls, and the 3rd3rd box to contain all the blue balls. Note that the given order of boxes is important here — it is not enough for each box to contain only balls of a single colour.

To achieve this, you can perform the following operation several (possibly, zero) times:

Pick any two distinct boxes, pick any one ball from each of these two boxes, and swap them.

Determine the minimum number of operations required to satisfy the given condition.

Input Format

The first line of input contains a single integer TT, denoting the number of test cases. The description of TT test cases follows.
Each test case consists of 44 lines of input.
The first line of each test case contains a single integer NN, denoting the number of balls of each colour.
The ii-th of the next three lines contains three space-separated integers Ri,GiRi,Gi and BiBi — the number of red, green, and blue balls in the ii-th box respectively.

Output Format

For each test case, output a single line containing one integer — the minimum number of operations required such that the given condition is satisfied.

Constraints

1≤T≤10001≤T≤1000
1≤N≤10001≤N≤1000
0≤Ri,Gi,Bi≤N0≤Ri,Gi,Bi≤N
R1+G1+B1=R2+G2+B2=R3+G3+B3=NR1+G1+B1=R2+G2+B2=R3+G3+B3=N
R1+R2+R3=G1+G2+G3=B1+B2+B3=NR1+R2+R3=G1+G2+G3=B1+B2+B3=N

Subtasks

Subtask #1 (100 points): Original constraints

Sample Input 1

Sample Output 1

0
3

Explanation

Test case 11: Initially,

The first box has 33 red balls and none of any other colours
The second box has 33 green balls and none of any other colours
The third box has 33 blue balls and none of any other colours

The condition is already satisfied, and so no moves are required.

Test case 22: One sequence of moves is as follows:

Swap a green ball from the first box with a red ball from the second box.
Swap a blue ball from the first box with a red ball from the third box.
Once again, swap a blue ball from the first box with a red ball from the third box.

Now the first box has only red balls, the second has only green balls, and the third has only blue ones — as required. It can be verified that no sequence of less than three moves achieves this result.

Click Below for solution

Solution

Program: Crying Colours CRYCOLR Solution in C++

#include <iostream>
#include <algorithm>
#include<cstdlib>
using namespace std;

int main() {
	short t,n;
	cin>>t;
	while(t--){
	    cin>>n;
	    short arr1[3],arr2[3],arr3[3];
	    for(int i=0;i<3;i++) cin>>arr1[i];
	    for(int i=0;i<3;i++) cin>>arr2[i];
	    for(int i=0;i<3;i++) cin>>arr3[i];
	    short result=min(arr1[1],arr2[0])+min(arr1[2],arr3[0])+min(arr2[2],arr3[1]);
	    result+=(2*abs(arr1[1]-arr2[0]));
	    cout<<result<<"\n";
	}
	return 0;
}

Program: Crying Colours CRYCOLR Solution in Java

import java.util.*;
import java.lang.*;
import java.io.*;
class Codechef
{
	public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int T = in.nextInt();
        while (T-- > 0) {
            int N = in.nextInt();
            int [][]arr= new int[3][3];
            for (int i=0;i<3;i++){
                for(int j=0;j<3;j++){
                    arr[i][j]=in.nextInt();
                }
            }
            int count=0;
            if(arr[0][0]==N && arr[1][1]==N && arr[2][2]==N){
                System.out.println(count);
                continue;
            }
            int a=arr[0][1]+arr[0][2]+arr[1][2];
            int b=arr[1][0]+arr[2][0]+arr[2][1];
            System.out.println(Math.max(a,b));
        }
    }
}

Program: Crying Colours CRYCOLR Solution in Python

for _ in range(int(input())):
    n=int(input())
    a=[]
    for i in range(3):
        b=list(map(int,input().split()))
        a.append(b)
    if(a[0][0]==n and a[1][1]==n and a[2][2]==n):
        print(0)
        continue
    c=a[1][0]+a[2][0]+a[2][1]
    b=a[0][1]+a[0][2]+a[1][2]
    print(max(c,b))