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:
CPP
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std;// Function that returns true// if the coordinate is inside the gridbool 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 cornerint 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 squareint 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 codeint 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 griddef 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 cornerdef 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 squaredef 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 codeif __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 gridfunction 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 cornerfunction 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 squarefunction 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 codevar 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> |
-1
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