Cleaning the room - GeeksforGeeks

19 Mar, 2020
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++
Python3
// 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;}
# 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 += 1  solve(n)  # This code is contributed by shubhamsingh10
Yes, the house can be cleaned.
1 1
2 1
3 1

SHUBHAMSINGH10
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