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Minimum number of moves after which there exists a 3X3 coloured square

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  • Last Updated : 15 Sep, 2022
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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>

Output

-1

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