Given a room with square grids having ‘*’ and ‘.’ representing untidy and normal cells respectively. You need to find whether room can be cleaned or not. There is a machine which helps you in this task, but it is capable of cleaning only normal cell. Untidy cells cannot be cleaned with machine, until you have cleaned the normal cell in its row or column. Now, see to it whether room can be cleaned or not. The input is as follows : First line contains the size of the room. The next n lines contains description for each row where the row[i][j] is ‘‘ if it is more untidy than others else it is ‘‘ if it is normal cell. Examples:
Input : 3
.**
.**
.**
Output :Yes, the room can be cleaned.
1 1
2 1
3 1
Input :4
****
..*.
..*.
..*.
Output : house cannot be cleaned.
Approach : The minimum number of cells can be n. It is the only answer possible as it need to have an element of type ‘‘ in every different row and column. If particular column and a given row contain ‘‘ in all the cells then, it is known that the house cannot be cleaned. Traverse every row and find the ‘‘ that can be used for the machine. Use this step two times, check every column for every row and then check for every row for every column. Then check if any of the two gives answer as n. If yes then house can be cleaned otherwise not. This approach will give us the minimum answer required. In the first example the machine will clean cell (1, 1), (2, 1), (3, 1) in order to clean the entire room. In the second example every cell in the row has ‘‘ and every cell in column contains ‘‘, therefore the house cannot be cleaned. row cannot be cleaned in any way.
C++
#include <bits/stdc++.h>
using namespace std;
char A[105][105];
vector<pair< int , int > > ans;
void print()
{
cout << "Yes, the house can be"
<< " cleaned." << endl;
for ( int i = 0; i < ans.size(); i++)
cout << ans[i].first << " "
<< ans[i].second << endl;
}
int solve( int n)
{
for ( int i = 0; i < n; i++) {
for ( int j = 0; j < n; j++) {
if (A[i][j] == '.' ) {
ans.push_back(make_pair(i + 1, j + 1));
break ;
}
}
}
if (ans.size() == n) {
print();
return 0;
}
ans.clear();
for ( int i = 0; i < n; i++) {
for ( int j = 0; j < n; j++) {
if (A[j][i] == '.' ) {
ans.push_back(make_pair(i + 1, j + 1));
break ;
}
}
}
if (ans.size() == n) {
print();
return 0;
}
cout << "house cannot be cleaned."
<< endl;
}
int main()
{
int n = 3;
string s = "" ;
s += ".**" ;
s += ".**" ;
s += ".**" ;
int k = 0;
for ( int i = 0; i < n; i++) {
for ( int j = 0; j < n; j++)
A[i][j] = s[k++];
}
solve(n);
return 0;
}
|
Java
import java.util.*;
class Main {
static char [][] A = new char [ 105 ][ 105 ];
static ArrayList<Pair<Integer, Integer>> ans = new ArrayList<>();
static void print() {
System.out.println( "Yes, the house can be cleaned." );
for (Pair<Integer, Integer> p : ans)
System.out.println(p.getKey() + " " + p.getValue());
}
static int solve( int n) {
for ( int i = 0 ; i < n; i++) {
for ( int j = 0 ; j < n; j++) {
if (A[i][j] == '.' ) {
ans.add( new Pair<>(i + 1 , j + 1 ));
break ;
}
}
}
if (ans.size() == n) {
print();
return 0 ;
}
ans.clear();
for ( int i = 0 ; i < n; i++) {
for ( int j = 0 ; j < n; j++) {
if (A[j][i] == '.' ) {
ans.add( new Pair<>(i + 1 , j + 1 ));
break ;
}
}
}
if (ans.size() == n) {
print();
return 0 ;
}
System.out.println( "house cannot be cleaned." );
return 0 ;
}
public static void main(String[] args) {
int n = 3 ;
String s = "" ;
s += ".**" ;
s += ".**" ;
s += ".**" ;
int k = 0 ;
for ( int i = 0 ; i < n; i++) {
for ( int j = 0 ; j < n; j++)
A[i][j] = s.charAt(k++);
}
solve(n);
}
}
class Pair<K, V> {
private final K key;
private final V value;
public Pair(K key, V value) {
this .key = key;
this .value = value;
}
public K getKey() {
return key;
}
public V getValue() {
return value;
}
}
|
Python3
A = [[ 0 for i in range ( 105 )] for j in range ( 105 )]
ans = []
def printt():
print ( "Yes, the house can be cleaned." )
for i in range ( len (ans)):
print (ans[i][ 0 ], ans[i][ 1 ])
def solve(n):
global ans
for i in range (n):
for j in range (n):
if (A[i][j] = = '.' ):
ans.append([i + 1 , j + 1 ])
break
if ( len (ans) = = n):
printt()
return 0
ans = []
for i in range (n):
for j in range (n):
if (A[j][i] = = '.' ):
ans.append([i + 1 , j + 1 ])
break
if ( len (ans) = = n):
printt()
return 0
print ( "house cannot be cleaned." )
n = 3
s = ""
s + = ".**"
s + = ".**"
s + = ".**"
k = 0
for i in range (n):
for j in range (n):
A[i][j] = s[k]
k + = 1
solve(n)
|
Javascript
let A = [];
let ans = [];
function print() {
console.log( "Yes, the house can be cleaned." + "<br>" );
for (let i = 0; i < ans.length; i++)
console.log(ans[i][0] + " " + ans[i][1] + "<br>" );
}
function solve(n) {
for (let i = 0; i < n; i++) {
for (let j = 0; j < n; j++) {
if (A[i][j] == "." ) {
ans.push([i + 1, j + 1]);
break ;
}
}
}
if (ans.length == n) {
print();
return 0;
}
ans = [];
for (let i = 0; i < n; i++) {
for (let j = 0; j < n; j++) {
if (A[j][i] == "." ) {
ans.push([i + 1, j + 1]);
break ;
}
}
}
if (ans.length == n) {
print();
return 0;
}
document.write( "house cannot be cleaned." );
}
function main() {
let n = 3;
let s = "...**" ;
let k = 0;
for (let i = 0; i < n; i++) {
A.push([]);
for (let j = 0; j < n; j++) A[i].push(s[k++]);
}
solve(n);
}
main();
|
C#
using System;
using System.Collections.Generic;
public class Program
{
static char [,] A = new char [105, 105];
static List<Tuple< int , int >> ans = new List<Tuple< int , int >>();
static void Print()
{
Console.WriteLine( "Yes, the house can be cleaned." );
for ( int i = 0; i < ans.Count; i++)
{
Console.WriteLine(ans[i].Item1 + " " + ans[i].Item2);
}
}
static int Solve( int n)
{
for ( int i = 0; i < n; i++)
{
for ( int j = 0; j < n; j++)
{
if (A[i, j] == '.' )
{
ans.Add(Tuple.Create(i + 1, j + 1));
break ;
}
}
}
if (ans.Count == n)
{
Print();
return 0;
}
ans.Clear();
for ( int i = 0; i < n; i++)
{
for ( int j = 0; j < n; j++)
{
if (A[j, i] == '.' )
{
ans.Add(Tuple.Create(i + 1, j + 1));
break ;
}
}
}
if (ans.Count == n)
{
Print();
return 0;
}
Console.WriteLine( "house cannot be cleaned." );
return -1;
}
static void Main()
{
int n = 3;
string s = "" ;
s += ".**" ;
s += ".**" ;
s += ".**" ;
int k = 0;
for ( int i = 0; i < n; i++)
{
for ( int j = 0; j < n; j++)
{
A[i, j] = s[k++];
}
}
Solve(n);
}
}
|
OutputYes, the house can be cleaned.
1 1
2 1
3 1
Time Complexity: O(n2)
Auxiliary Space: O(n2)
Last Updated :
05 Apr, 2023
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