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++
// CPP code to find whether// house can be cleaned or not#include <bits/stdc++.h>using namespace std;// Matrix A stores the stringchar A[105][105];// ans stores the pair of indices// to be cleaned by the machinevector<pair<int, int> > ans;// Function for printing the// vector of pairvoid 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;}// Function performing calculationsint solve(int n){ // push every first cell in // each row containing '.' 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 total number of cells are // n then house can be cleaned if (ans.size() == n) { print(); return 0; } ans.clear(); // push every first cell in // each column containing '.' 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 total number of cells are // n then house can be cleaned if (ans.size() == n) { print(); return 0; } cout << "house cannot be cleaned." << endl;}// Driver functionint main(){ int n = 3; string s = ""; s += ".**"; s += ".**"; s += ".**"; int k = 0; // Loop to insert letters from // string to array 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 { // Matrix A stores the string static char[][] A = new char[105][105]; // ans stores the pair of indices // to be cleaned by the machine static ArrayList<Pair<Integer, Integer>> ans = new ArrayList<>(); // Function for printing the // ArrayList of pair 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()); } // Function performing calculations static int solve(int n) { // push every first cell in // each row containing '.' 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 total number of cells are // n then house can be cleaned if (ans.size() == n) { print(); return 0; } ans.clear(); // push every first cell in // each column containing '.' 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 total number of cells are // n then house can be cleaned if (ans.size() == n) { print(); return 0; } System.out.println("house cannot be cleaned."); return 0; } // Driver function public static void main(String[] args) { int n = 3; String s = ""; s += ".**"; s += ".**"; s += ".**"; int k = 0; // Loop to insert letters from // string to array 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
# Python3 code to find whether# house can be cleaned or not# Matrix A stores the stringA = [[0 for i in range(105)] for j in range(105)]# ans stores the pair of indices# to be cleaned by the machineans = []# Function for printing the# vector of pairdef printt(): print("Yes, the house can be cleaned.") for i in range(len(ans)): print(ans[i][0], ans[i][1]) # Function performing calculationsdef solve(n): global ans # push every first cell in # each row containing '.' for i in range(n): for j in range(n): if (A[i][j] == '.'): ans.append([i + 1, j + 1]) break # If total number of cells are # n then house can be cleaned if (len(ans) == n): printt() return 0 ans = [] # push every first cell in # each column containing '.' for i in range(n): for j in range(n): if (A[j][i] == '.'): ans.append([i + 1, j + 1]) break # If total number of cells are # n then house can be cleaned if (len(ans) == n): printt() return 0 print("house cannot be cleaned.")# Driver functionn = 3s = ""s += ".**"s += ".**"s += ".**"k = 0# Loop to insert letters from# string to arrayfor i in range(n): for j in range(n): A[i][j] = s[k] k += 1solve(n)# This code is contributed by shubhamsingh10 |
Javascript
// JS code to find whether// house can be cleaned or not// Matrix A stores the stringlet A = [];// ans stores the pair of indices// to be cleaned by the machinelet ans = [];// Function for printing the// vector of pairfunction 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 performing calculationsfunction solve(n) {// push every first cell in// each row containing '.'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 total number of cells are// n then house can be cleanedif (ans.length == n) {print();return 0;}ans = [];// push every first cell in// each column containing '.'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 total number of cells are// n then house can be cleanedif (ans.length == n) {print();return 0;}document.write("house cannot be cleaned.");}// Driver functionfunction main() {let n = 3;let s = "...**";let k = 0;// Loop to insert letters from// string to arrayfor (let i = 0; i < n; i++) {A.push([]);for (let j = 0; j < n; j++) A[i].push(s[k++]);}solve(n);}main(); |
C#
// C# code to find whether// house can be cleaned or notusing System;using System.Collections.Generic;public class Program{// Matrix A stores the stringstatic char[,] A = new char[105, 105];// ans stores the pair of indices// to be cleaned by the machinestatic List<Tuple<int, int>> ans = new List<Tuple<int, int>>();// Function for printing the// vector of pairstatic 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); }}// Function performing calculationsstatic int Solve(int n){ // push every first cell in // each row containing '.' 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 total number of cells are // n then house can be cleaned if (ans.Count == n) { Print(); return 0; } ans.Clear(); // push every first cell in // each column containing '.' 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 total number of cells are // n then house can be cleaned if (ans.Count == n) { Print(); return 0; } Console.WriteLine("house cannot be cleaned."); return -1;}//Driver codestatic 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);}} |
Output
Yes, the house can be cleaned. 1 1 2 1 3 1
Time Complexity: O(n2)
Auxiliary Space: O(n2)
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