Minimum number of moves after which there exists a 3X3 coloured square
Given an N * N board which is initially empty and a sequence of queries, each query consists of two integers X and Y where the cell (X, Y) is painted. The task is to print the number of the query after which there will be a 3 * 3 square in the board with all the cells painted.
If there is no such square after processing all of the queries then print -1.
Examples:
Input: N = 4, q = {{1, 1}, {1, 2}, {1, 3}, {2, 2}, {2, 3}, {1, 4}, {2, 4}, {3, 4}, {3, 2}, {3, 3}, {4, 1}}
Output: 10
After the 10th move, there exists a 3X3 square, from (1, 1) to (3, 3) (1-based indexing).Input: N = 3, q = {(1, 1), {1, 2}, {1, 3}}
Output: -1
Approach: An important observation here is that every time we colour a square, it can be a part of the required square in 9 different ways (any cell of the 3 * 3 square) . For every possibility, check whether the current cell is a part of any square where all the 9 cells are painted. If the condition is satisfied print the number of queries processed so far else print -1 after all the queries have been processed.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach #include <bits/stdc++.h> using namespace std; // Function that returns true // if the coordinate is inside the grid bool valid( int x, int y, int n) { if (x < 1 || y < 1 || x > n || y > n) return false ; return true ; } // Function that returns the count of squares // that are coloured in the 3X3 square // with (cx, cy) being the top left corner int count( int n, int cx, int cy, vector<vector< bool > >& board) { int ct = 0; // Iterate through 3 rows for ( int x = cx; x <= cx + 2; x++) // Iterate through 3 columns for ( int y = cy; y <= cy + 2; y++) // Check if the current square // is valid and coloured if (valid(x, y, n)) ct += board[x][y]; return ct; } // Function that returns the query // number after which the grid will // have a 3X3 coloured square int minimumMoveSquare( int n, int m, vector<pair< int , int > > moves) { int x, y; vector<vector< bool > > board(n + 1); // Initialize all squares to be uncoloured for ( int i = 1; i <= n; i++) board[i].resize(n + 1, false ); for ( int i = 0; i < moves.size(); i++) { x = moves[i].first; y = moves[i].second; // Mark the current square as coloured board[x][y] = true ; // Check if any of the 3X3 squares // which contains the current square // is fully coloured for ( int cx = x - 2; cx <= x; cx++) for ( int cy = y - 2; cy <= y; cy++) if (count(n, cx, cy, board) == 9) return i + 1; } return -1; } // Driver code int main() { int n = 3; vector<pair< int , int > > moves = { { 1, 1 }, { 1, 2 }, { 1, 3 } }; int m = moves.size(); cout << minimumMoveSquare(n, m, moves); return 0; } |
Python3
# Python3 implementation of the approach # Function that returns True # if the coordinate is inside the grid def valid(x, y, n): if (x < 1 or y < 1 or x > n or y > n): return False ; return True ; # Function that returns the count of squares # that are coloured in the 3X3 square # with (cx, cy) being the top left corner def count(n, cx, cy, board): ct = 0 ; # Iterate through 3 rows for x in range (cx, cx + 3 ): # Iterate through 3 columns for y in range (cy + 3 ): # Check if the current square # is valid and coloured if (valid(x, y, n)): ct + = board[x][y]; return ct; # Function that returns the query # number after which the grid will # have a 3X3 coloured square def minimumMoveSquare(n, m, moves): x = 0 y = 0 board = [[] for i in range (n + 1 )] # Initialize all squares to be uncoloured for i in range ( 1 , n + 1 ): board[i] = [ False for i in range (n + 1 )] for i in range ( len (moves)): x = moves[i][ 0 ]; y = moves[i][ 1 ]; # Mark the current square as coloured board[x][y] = True ; # Check if any of the 3X3 squares # which contains the current square # is fully coloured for cx in range (x - 2 , x + 1 ): for cy in range (y - 2 , y + 1 ): if (count(n, cx, cy, board) = = 9 ): return i + 1 ; return - 1 ; # Driver code if __name__ = = '__main__' : n = 3 ; moves = [[ 1 , 1 ],[ 1 , 2 ], [ 1 , 3 ]] m = len (moves) print (minimumMoveSquare(n, m, moves)) # This code is contributed by rutvik_56. |
Javascript
<script> // Javascript implementation of the approach // Function that returns true // if the coordinate is inside the grid function valid(x, y, n) { if (x < 1 || y < 1 || x > n || y > n) return false ; return true ; } // Function that returns the count of squares // that are coloured in the 3X3 square // with (cx, cy) being the top left corner function count(n, cx, cy, board) { var ct = 0; // Iterate through 3 rows for ( var x = cx; x <= cx + 2; x++) // Iterate through 3 columns for ( var y = cy; y <= cy + 2; y++) // Check if the current square // is valid and coloured if (valid(x, y, n)) ct += board[x][y]; return ct; } // Function that returns the query // number after which the grid will // have a 3X3 coloured square function minimumMoveSquare(n, m, moves) { var x, y; var board = Array.from(Array(n+1), ()=>Array(n+1).fill( false )); for ( var i = 0; i < moves.length; i++) { x = moves[i][0]; y = moves[i][1]; // Mark the current square as coloured board[x][y] = true ; // Check if any of the 3X3 squares // which contains the current square // is fully coloured for ( var cx = x - 2; cx <= x; cx++) for ( var cy = y - 2; cy <= y; cy++) if (count(n, cx, cy, board) == 9) return i + 1; } return -1; } // Driver code var n = 3; var moves = [ [ 1, 1 ], [ 1, 2 ], [ 1, 3 ] ]; var m = moves.length; document.write( minimumMoveSquare(n, m, moves)); // This code is contributed by famously. </script> |
Java
import java.util.*; public class Main { // Function that returns True if the coordinate is inside the grid public static boolean valid( int x, int y, int n) { if (x < 1 || y < 1 || x > n || y > n) { return false ; } return true ; } // Function that returns the count of squares // that are coloured in the 3X3 square // with (cx, cy) being the top left corner public static int count( int n, int cx, int cy, boolean [][] board) { int ct = 0 ; // Iterate through 3 rows for ( int x = cx; x < cx + 3 ; x++) { // Iterate through 3 columns for ( int y = cy; y < cy + 3 ; y++) { // Check if the current square is valid and coloured if (valid(x, y, n)) { if (board[x][y]) { ct++; } } } } return ct; } // Function that returns the query number after which the grid will // have a 3X3 coloured square public static int minimumMoveSquare( int n, int m, int [][] moves) { int x = 0 , y = 0 ; boolean [][] board = new boolean [n + 1 ][n + 1 ]; // Initialize all squares to be uncoloured for ( int i = 1 ; i <= n; i++) { Arrays.fill(board[i], false ); } for ( int i = 0 ; i < m; i++) { x = moves[i][ 0 ]; y = moves[i][ 1 ]; // Mark the current square as coloured board[x][y] = true ; // Check if any of the 3X3 squares which contains the current square is fully coloured for ( int cx = x - 2 ; cx <= x; cx++) { for ( int cy = y - 2 ; cy <= y; cy++) { if (count(n, cx, cy, board) == 9 ) { return i + 1 ; } } } } return - 1 ; } // Driver code public static void main(String[] args) { int n = 3 ; int [][] moves = { { 1 , 1 }, { 1 , 2 }, { 1 , 3 } }; int m = moves.length; System.out.println(minimumMoveSquare(n, m, moves)); } } // This code is contributed by Prince Kumar |
C#
// C# porgrm for the above approach using System; public class Program { // Function that returns True if the coordinate is inside the grid public static bool Valid( int x, int y, int n) { if (x < 1 || y < 1 || x > n || y > n) { return false ; } return true ; } // Function that returns the count of squares // that are coloured in the 3X3 square // with (cx, cy) being the top left corner public static int Count( int n, int cx, int cy, bool [][] board) { int ct = 0; // Iterate through 3 rows for ( int x = cx; x < cx + 3; x++) { // Iterate through 3 columns for ( int y = cy; y < cy + 3; y++) { // Check if the current square is valid and coloured if (Valid(x, y, n)) { if (board[x][y]) { ct++; } } } } return ct; } // Function that returns the query number after which the grid will // have a 3X3 coloured square public static int MinimumMoveSquare( int n, int m, int [][] moves) { int x = 0, y = 0; bool [][] board = new bool [n + 1][]; // Initialize all squares to be uncoloured for ( int i = 1; i <= n; i++) { board[i] = new bool [n + 1]; for ( int j = 1; j <= n; j++) { board[i][j] = false ; } } for ( int i = 0; i < m; i++) { x = moves[i][0]; y = moves[i][1]; // Mark the current square as coloured board[x][y] = true ; // Check if any of the 3X3 squares which contains the current square is fully coloured for ( int cx = x - 2; cx <= x; cx++) { for ( int cy = y - 2; cy <= y; cy++) { if (Count(n, cx, cy, board) == 9) { return i + 1; } } } } return -1; } // Driver code public static void Main() { int n = 3; int [][] moves = new int [3][]; moves[0] = new int [] { 1, 1 }; moves[1] = new int [] { 1, 2 }; moves[2] = new int [] { 1, 3 }; int m = moves.Length; Console.WriteLine(MinimumMoveSquare(n, m, moves)); } } // This code is contributed by adityasharmadev01 |
-1
Time Complexity: O(m)
Auxiliary Space: O(1)