Skip to content
Related Articles

Related Articles

Improve Article

Cleaning the room

  • Difficulty Level : Easy
  • Last Updated : 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 n 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 1^{st} row has ‘*‘ and every cell in 3^{rd} column contains ‘*‘, therefore the house cannot be cleaned. 1^{st} 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 string
char A[105][105];
  
// ans stores the pair of indices
// to be cleaned by the machine
vector<pair<int, int> > ans;
  
// Function for printing the
// vector of pair
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;
}
  
// Function performing calculations
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.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 function
int 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




# Python3 code to find whether
# house can be cleaned or not
  
# Matrix A stores the string
A = [[0 for i in range(105)] for j in range(105)]
  
# ans stores the pair of indices
# to be cleaned by the machine
ans = []
  
# Function for printing the
# vector of pair
def printt():
      
    print("Yes, the house can be cleaned.")
    for i in range(len(ans)):
        print(ans[i][0], ans[i][1])
          
# Function performing calculations
def 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 function
n = 3
s = ""
s += ".**"
s += ".**"
s += ".**"
k = 0
  
# Loop to insert letters from
# string to array
for i in range(n):
    for j in range(n):
        A[i][j] = s[k]
        k += 1
  
solve(n)
  
# This code is contributed by shubhamsingh10


Output:
Yes, the house can be cleaned.
1 1
2 1
3 1

Attention reader! Don’t stop learning now. Get hold of all the important Comcompetitivepetitve Programming concepts with the Competitive Programming Live  course.




My Personal Notes arrow_drop_up
Recommended Articles
Page :