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:

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// 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;
}

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Output:

-1


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