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Hackerrank Organizing Containers of Balls Solution

Hackerrank Organizing Containers of Balls Solution

Beeze Aal
Beeze Aal

.MathJax_SVG_Display {text-align: center; margin: 1em 0em; position: relative; display: block!important; text-indent: 0; max-width: none; max-height: none; min-width: 0; min-height: 0; width: 100%} .MathJax_SVG .MJX-monospace {font-family: monospace} .MathJax_SVG .MJX-sans-serif {font-family: sans-serif} .MathJax_SVG {display: inline; font-style: normal; font-weight: normal; line-height: normal; font-size: 100%; font-size-adjust: none; text-indent: 0; text-align: left; text-transform: none; letter-spacing: normal; word-spacing: normal; word-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0; min-height: 0; border: 0; padding: 0; margin: 0} .MathJax_SVG * {transition: none; -webkit-transition: none; -moz-transition: none; -ms-transition: none; -o-transition: none} .mjx-svg-href {fill: blue; stroke: blue} .MathJax_SVG_LineBox {display: table!important} .MathJax_SVG_LineBox span {display: table-cell!important; width: 10000em!important; min-width: 0; max-width: none; padding: 0; border: 0; margin: 0}

David has several containers, each with a number of balls in it.  He has just enough containers to sort each type of ball he has into its own container.  David wants to sort the balls using his sort method.

As an example, David has  containers and  different types of balls, both of which are numbered from  to . The distribution of ball types per container are described by an  matrix of integers, . For example, consider the following diagram for :

image

In a single operation, David can swap two balls located in different containers.

The diagram below depicts a single swap operation:

image

David wants to perform some number of swap operations such that:

  • Each container contains only balls of the same type.
  • No two balls of the same type are located in different containers.

You must perform  queries where each query is in the form of a matrix, . For each query, print Possible on a new line if David can satisfy the conditions above for the given matrix.  Otherwise, print Impossible.

Function Description

Complete the organizingContainers function in the editor below.  It should return a string, either Possible or Impossible.

organizingContainers has the following parameter(s):

  • containter: a two dimensional array of integers that represent the number of balls of each color in each container

Input Format.MathJax_SVG_Display {text-align: center; margin: 1em 0em; position: relative; display: block!important; text-indent: 0; max-width: none; max-height: none; min-width: 0; min-height: 0; width: 100%} .MathJax_SVG .MJX-monospace {font-family: monospace} .MathJax_SVG .MJX-sans-serif {font-family: sans-serif} .MathJax_SVG {display: inline; font-style: normal; font-weight: normal; line-height: normal; font-size: 100%; font-size-adjust: none; text-indent: 0; text-align: left; text-transform: none; letter-spacing: normal; word-spacing: normal; word-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0; min-height: 0; border: 0; padding: 0; margin: 0} .MathJax_SVG * {transition: none; -webkit-transition: none; -moz-transition: none; -ms-transition: none; -o-transition: none} .mjx-svg-href {fill: blue; stroke: blue} .MathJax_SVG_LineBox {display: table!important} .MathJax_SVG_LineBox span {display: table-cell!important; width: 10000em!important; min-width: 0; max-width: none; padding: 0; border: 0; margin: 0}

The first line contains an integer , the number of queries.

Each of the next  sets of lines is as follows:

  1. The first line contains an integer , the number of containers (rows) and ball types (columns).
  2. Each of the next  lines contains  space-separated integers describing row .

Constraints.MathJax_SVG_Display {text-align: center; margin: 1em 0em; position: relative; display: block!important; text-indent: 0; max-width: none; max-height: none; min-width: 0; min-height: 0; width: 100%} .MathJax_SVG .MJX-monospace {font-family: monospace} .MathJax_SVG .MJX-sans-serif {font-family: sans-serif} .MathJax_SVG {display: inline; font-style: normal; font-weight: normal; line-height: normal; font-size: 100%; font-size-adjust: none; text-indent: 0; text-align: left; text-transform: none; letter-spacing: normal; word-spacing: normal; word-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0; min-height: 0; border: 0; padding: 0; margin: 0} .MathJax_SVG * {transition: none; -webkit-transition: none; -moz-transition: none; -ms-transition: none; -o-transition: none} .mjx-svg-href {fill: blue; stroke: blue} .MathJax_SVG_LineBox {display: table!important} .MathJax_SVG_LineBox span {display: table-cell!important; width: 10000em!important; min-width: 0; max-width: none; padding: 0; border: 0; margin: 0}

Scoring

  • For ¬†of score, .
  • For ¬†of score, .

Output Format.MathJax_SVG_Display {text-align: center; margin: 1em 0em; position: relative; display: block!important; text-indent: 0; max-width: none; max-height: none; min-width: 0; min-height: 0; width: 100%} .MathJax_SVG .MJX-monospace {font-family: monospace} .MathJax_SVG .MJX-sans-serif {font-family: sans-serif} .MathJax_SVG {display: inline; font-style: normal; font-weight: normal; line-height: normal; font-size: 100%; font-size-adjust: none; text-indent: 0; text-align: left; text-transform: none; letter-spacing: normal; word-spacing: normal; word-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0; min-height: 0; border: 0; padding: 0; margin: 0} .MathJax_SVG * {transition: none; -webkit-transition: none; -moz-transition: none; -ms-transition: none; -o-transition: none} .mjx-svg-href {fill: blue; stroke: blue} .MathJax_SVG_LineBox {display: table!important} .MathJax_SVG_LineBox span {display: table-cell!important; width: 10000em!important; min-width: 0; max-width: none; padding: 0; border: 0; margin: 0}

For each query, print Possible on a new line if David can satisfy the conditions above for the given matrix.  Otherwise, print Impossible.

Sample Input 0.MathJax_SVG_Display {text-align: center; margin: 1em 0em; position: relative; display: block!important; text-indent: 0; max-width: none; max-height: none; min-width: 0; min-height: 0; width: 100%} .MathJax_SVG .MJX-monospace {font-family: monospace} .MathJax_SVG .MJX-sans-serif {font-family: sans-serif} .MathJax_SVG {display: inline; font-style: normal; font-weight: normal; line-height: normal; font-size: 100%; font-size-adjust: none; text-indent: 0; text-align: left; text-transform: none; letter-spacing: normal; word-spacing: normal; word-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0; min-height: 0; border: 0; padding: 0; margin: 0} .MathJax_SVG * {transition: none; -webkit-transition: none; -moz-transition: none; -ms-transition: none; -o-transition: none} .mjx-svg-href {fill: blue; stroke: blue} .MathJax_SVG_LineBox {display: table!important} .MathJax_SVG_LineBox span {display: table-cell!important; width: 10000em!important; min-width: 0; max-width: none; padding: 0; border: 0; margin: 0} 221 11 120 21 1

Sample Output 0.MathJax_SVG_Display {text-align: center; margin: 1em 0em; position: relative; display: block!important; text-indent: 0; max-width: none; max-height: none; min-width: 0; min-height: 0; width: 100%} .MathJax_SVG .MJX-monospace {font-family: monospace} .MathJax_SVG .MJX-sans-serif {font-family: sans-serif} .MathJax_SVG {display: inline; font-style: normal; font-weight: normal; line-height: normal; font-size: 100%; font-size-adjust: none; text-indent: 0; text-align: left; text-transform: none; letter-spacing: normal; word-spacing: normal; word-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0; min-height: 0; border: 0; padding: 0; margin: 0} .MathJax_SVG * {transition: none; -webkit-transition: none; -moz-transition: none; -ms-transition: none; -o-transition: none} .mjx-svg-href {fill: blue; stroke: blue} .MathJax_SVG_LineBox {display: table!important} .MathJax_SVG_LineBox span {display: table-cell!important; width: 10000em!important; min-width: 0; max-width: none; padding: 0; border: 0; margin: 0} PossibleImpossible

Explanation 0.MathJax_SVG_Display {text-align: center; margin: 1em 0em; position: relative; display: block!important; text-indent: 0; max-width: none; max-height: none; min-width: 0; min-height: 0; width: 100%} .MathJax_SVG .MJX-monospace {font-family: monospace} .MathJax_SVG .MJX-sans-serif {font-family: sans-serif} .MathJax_SVG {display: inline; font-style: normal; font-weight: normal; line-height: normal; font-size: 100%; font-size-adjust: none; text-indent: 0; text-align: left; text-transform: none; letter-spacing: normal; word-spacing: normal; word-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0; min-height: 0; border: 0; padding: 0; margin: 0} .MathJax_SVG * {transition: none; -webkit-transition: none; -moz-transition: none; -ms-transition: none; -o-transition: none} .mjx-svg-href {fill: blue; stroke: blue} .MathJax_SVG_LineBox {display: table!important} .MathJax_SVG_LineBox span {display: table-cell!important; width: 10000em!important; min-width: 0; max-width: none; padding: 0; border: 0; margin: 0}

We perform the following  queries:

  1. The diagram below depicts one possible way to satisfy David's requirements for the first query:
image


Thus, we print Possible on a new line.

  1. The diagram below depicts the matrix for the second query:
image


No matter how many times we swap balls of type  and  between the two containers, we'll never end up with one container only containing type  and the other container only containing type . Thus, we print Impossible on a new line.

Sample Input 1.MathJax_SVG_Display {text-align: center; margin: 1em 0em; position: relative; display: block!important; text-indent: 0; max-width: none; max-height: none; min-width: 0; min-height: 0; width: 100%} .MathJax_SVG .MJX-monospace {font-family: monospace} .MathJax_SVG .MJX-sans-serif {font-family: sans-serif} .MathJax_SVG {display: inline; font-style: normal; font-weight: normal; line-height: normal; font-size: 100%; font-size-adjust: none; text-indent: 0; text-align: left; text-transform: none; letter-spacing: normal; word-spacing: normal; word-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0; min-height: 0; border: 0; padding: 0; margin: 0} .MathJax_SVG * {transition: none; -webkit-transition: none; -moz-transition: none; -ms-transition: none; -o-transition: none} .mjx-svg-href {fill: blue; stroke: blue} .MathJax_SVG_LineBox {display: table!important} .MathJax_SVG_LineBox span {display: table-cell!important; width: 10000em!important; min-width: 0; max-width: none; padding: 0; border: 0; margin: 0} 231 3 12 1 23 3 330 2 11 1 12 0 0

Sample Output 1.MathJax_SVG_Display {text-align: center; margin: 1em 0em; position: relative; display: block!important; text-indent: 0; max-width: none; max-height: none; min-width: 0; min-height: 0; width: 100%} .MathJax_SVG .MJX-monospace {font-family: monospace} .MathJax_SVG .MJX-sans-serif {font-family: sans-serif} .MathJax_SVG {display: inline; font-style: normal; font-weight: normal; line-height: normal; font-size: 100%; font-size-adjust: none; text-indent: 0; text-align: left; text-transform: none; letter-spacing: normal; word-spacing: normal; word-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0; min-height: 0; border: 0; padding: 0; margin: 0} .MathJax_SVG * {transition: none; -webkit-transition: none; -moz-transition: none; -ms-transition: none; -o-transition: none} .mjx-svg-href {fill: blue; stroke: blue} .MathJax_SVG_LineBox {display: table!important} .MathJax_SVG_LineBox span {display: table-cell!important; width: 10000em!important; min-width: 0; max-width: none; padding: 0; border: 0; margin: 0} ImpossiblePossible

Solution in java8

Approach 1.

import java.util.*;
class BallsAndContainers
{
    public static void main(String[] args) 
    {
        Scanner sc=new Scanner(System.in);
        int i,j,k,l,m,n,t,p,q;
        t=sc.nextInt();
        for(l=1;l<=t;l++)
        {
            n=sc.nextInt();
            int containers[]=new int[n];
            int balltypes[]=new int[n];
            for(i=0;i<n;i++)
            {
                for(j=0;j<n;j++)
                {
                    m=sc.nextInt();
                    containers[i]+=m;
                    balltypes[j]+=m;
                }
            }
            String ans="Possible";
            for(i=0;i<n;i++)
            {
                j=0;
                for(j=i;j<n;j++)
                {
                    if(containers[i]==balltypes[j])
                    {
                        int temp=balltypes[j];
                        balltypes[j]=balltypes[i];
                        balltypes[i]=temp;
                        break;
                    }
                }
                if(j==n)
                {
                    ans="Impossible";
                    break;
                }
            }
            System.out.println(ans);
        }
    }
}

Approach 2.

import java.io.*;
import java.math.*;
import java.security.*;
import java.text.*;
import java.util.*;
import java.util.concurrent.*;
import java.util.regex.*;

public class Solution {

    // Complete the organizingContainers function below.
    static String organizingContainers(int[][] container,int n) {
    int type[]=new int[n];
        int capa[]=new int[n];
        for(int i=0;i<n;i++)
        {
            for(int j=0;j<n;j++)
            {
                type[j]+=container[i][j];
                capa[i]+=container[i][j];
            }
        }
        Arrays.sort(type);
        Arrays.sort(capa);
        for(int i=0;i<type.length;i++)
        {
            if(type[i]!=capa[i])
                return "Impossible";
        }
        return "Possible";
    }

    private static final Scanner scanner = new Scanner(System.in);

    public static void main(String[] args) throws IOException {
        BufferedWriter bufferedWriter = new BufferedWriter(new FileWriter(System.getenv("OUTPUT_PATH")));

        int q = scanner.nextInt();
        scanner.skip("(\r\n|[\n\r\u2028\u2029\u0085])?");

        for (int qItr = 0; qItr < q; qItr++) {
            int n = scanner.nextInt();
            scanner.skip("(\r\n|[\n\r\u2028\u2029\u0085])?");

            int[][] container = new int[n][n];

            for (int i = 0; i < n; i++) {
                String[] containerRowItems = scanner.nextLine().split(" ");
                scanner.skip("(\r\n|[\n\r\u2028\u2029\u0085])?");

                for (int j = 0; j < n; j++) {
                    int containerItem = Integer.parseInt(containerRowItems[j]);
                    container[i][j] = containerItem;
                }
            }

            String result = organizingContainers(container,n);

            bufferedWriter.write(result);
            bufferedWriter.newLine();
        }

        bufferedWriter.close();

        scanner.close();
    }
}

Approach 3.

import java.io.*;
import java.math.*;
import java.security.*;
import java.text.*;
import java.util.*;
import java.util.concurrent.*;
import java.util.regex.*;

public class Solution {

    // Complete the organizingContainers function below.
    static String organizingContainers(int[][] container) {

        int c[] = new int[container.length];
        int t[] = new int[container.length];

        for (int i = 0; i < container.length; i++) {
            for (int j = 0; j < container[0].length; j++) {

                c[i] += container[i][j];
                t[j] += container[i][j];
            }
        }

        Arrays.sort(c);
        Arrays.sort(t);

        if (Arrays.equals(c, t)) {
            return "Possible";
        } else {
            return "Impossible";
        }
    }

    private static final Scanner scanner = new Scanner(System.in);

    public static void main(String[] args) throws IOException {
        BufferedWriter bufferedWriter = new BufferedWriter(new FileWriter(System.getenv("OUTPUT_PATH")));

        int q = scanner.nextInt();
        scanner.skip("(\r\n|[\n\r\u2028\u2029\u0085])?");

        for (int qItr = 0; qItr < q; qItr++) {
            int n = scanner.nextInt();
            scanner.skip("(\r\n|[\n\r\u2028\u2029\u0085])?");

            int[][] container = new int[n][n];

            for (int i = 0; i < n; i++) {
                String[] containerRowItems = scanner.nextLine().split(" ");
                scanner.skip("(\r\n|[\n\r\u2028\u2029\u0085])?");

                for (int j = 0; j < n; j++) {
                    int containerItem = Integer.parseInt(containerRowItems[j]);
                    container[i][j] = containerItem;
                }
            }

            String result = organizingContainers(container);

            bufferedWriter.write(result);
            bufferedWriter.newLine();
        }

        bufferedWriter.close();

        scanner.close();
    }
}

Solution in python3

Approach 1.

#!/bin/python3

import math
import os
import random
import re
import sys

# Complete the organizingContainers function below.
def organizingContainers(container):
    caps = []
    for i in range(len(container)):
        caps.append( sum(container[i]) )
    #print(caps)

    typenums = []
    for i in range(len(container)):
        n = 0
        for j in range(len(container)):
            n += container[j][i]
        typenums.append(n)
    
    if sorted(caps) == sorted(typenums):
        return 'Possible'
    else:
        return 'Impossible'

if __name__ == '__main__':
    fptr = open(os.environ['OUTPUT_PATH'], 'w')

    q = int(input())

    for q_itr in range(q):
        n = int(input())

        container = []

        for _ in range(n):
            container.append(list(map(int, input().rstrip().split())))

        result = organizingContainers(container)

        fptr.write(result + '\n')

    fptr.close()

Approach 2.

#!/bin/python3

import os

# Complete the organizingContainers function below.
def organizingContainers(container):
     # print(container)
    colsums=[]
    rowsums=[]
    for i in range(len(container)):
        row=0
        col=0
        for j in range(len(container)):
            row+=container[i][j]
            col+=container[j][i]
        rowsums.append(row)
        colsums.append(col)
    
    rowsums.sort()
    colsums.sort()
    print(rowsums,colsums)
    for i in range(len(rowsums)):
        if rowsums[i]!=colsums[i]:
            return "Impossible"
    return "Possible"

if __name__ == '__main__':
    fptr = open(os.environ['OUTPUT_PATH'], 'w')

    q = int(input())

    for q_itr in range(q):
        n = int(input())

        container = []

        for _ in range(n):
            container.append(list(map(int, input().rstrip().split())))

        result = organizingContainers(container)

        fptr.write(result + '\n')

    fptr.close()

Approach 3.

#!/bin/python3

import math
import os
import random
import re
import sys

# Complete the organizingContainers function below.
def organizingContainers(container):
    # print(container)
    colsums=[]
    rowsums=[]
    for i in range(len(container)):
        row=0
        col=0
        for j in range(len(container)):
            row+=container[i][j]
            col+=container[j][i]
        rowsums.append(row)
        colsums.append(col)
    
    rowsums.sort()
    colsums.sort()
    print(rowsums,colsums)
    for i in range(len(rowsums)):
        if rowsums[i]!=colsums[i]:
            return "Impossible"
    return "Possible"

if __name__ == '__main__':
    fptr = open(os.environ['OUTPUT_PATH'], 'w')

    q = int(input())

    for q_itr in range(q):
        n = int(input())

        container = []

        for _ in range(n):
            container.append(list(map(int, input().rstrip().split())))

        result = organizingContainers(container)

        fptr.write(result + '\n')

    fptr.close()

Solution in cpp

Approach 1.

#include <bits/stdc++.h>

using namespace std;

// Complete the organizingContainers function below.
int main() 
{
    int t;
    cin>>t;
    while(t--)
    {
        int n;
        cin>>n;
        vector<vector<int>> M(n);
        for (int i = 0; i < n; i++) 
        {
            M[i].resize(n);

            for (int j = 0; j < n; j++)
            {
                cin >> M[i][j];
            }
        }
        vector<int> col(n,0),row(n,0);
        for(int i=0;i<n;i++)
        {
            for(int j=0;j<n;j++)
            {
                col[i]+=M[i][j];
                row[i]+=M[j][i];
            }
        }
        sort(col.begin(),col.end());
        sort(row.begin(),row.end());

        if(col==row)
            cout<<"Possible"<<endl;
        else 
            cout<<"Impossible"<<endl;
    }
        return 0;
    
}
    

Approach 2.

#include <bits/stdc++.h>
using namespace std;

int main()
{   int q;
    cin >> q;
    while (q--) 
    {
        int n,j=0,i=0,flag=1;
        cin >> n;
        int container[n][n];
        long long type[n] ={} , containercnt [n] ={};
        for ( i = 0; i < n; i++) 
        {   
            for ( j = 0; j < n; j++)
            {
                cin >> container[i][j];
                containercnt[i]+=container[i][j];
            }         
        }
        for ( i = 0; i < n; i++) 
        for ( j = 0; j < n; j++)
        type[i]+=container[j][i];
        sort (type,type+n);
        sort (containercnt,containercnt+n);
        for ( i = 0; i < n; i++) 
        {  
           if (type[i]!=containercnt[i])   
           {
            flag=0;
            break;
           }
        }
        if (flag ==1) cout<<"Possible\n";
        else cout<<"Impossible\n";
    }
    return 0;
}

Approach 3.

#include <stdio.h>
#include <vector>
#include <iterator>
#include <string>
#include <string.h>
#include <algorithm>
#include <math.h>
#include <map>
#include <iostream>
#include <set>

using namespace std;


int main() {
    
    long long q, n, c[200], k[200];
    cin >> q;
    for (int t = 0; t < q; t++) {
        memset(c, 0, sizeof(c));
        memset(k, 0, sizeof(k));
        cin >> n;
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                long long tmp;
                cin >> tmp;
                k[i] += tmp;
                c[j] += tmp;
            }
        }
        sort(c, c + n);
        sort(k, k + n);
        for (int i = 0; i < n; i++) {
            if (k[i] != c[i]) {
                cout << "Impossible" << endl;
                break;
            }
            if (i == n - 1) {
                cout << "Possible" << endl;
            }
        }
    }
    return 0;
}