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

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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++ implementation of the approach    #include 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 >& 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 > moves) {     int x, y;     vector > 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 > moves = { { 1, 1 },                                       { 1, 2 },                                       { 1, 3 } };     int m = moves.size();        cout << minimumMoveSquare(n, m, moves);        return 0; }

Output:

-1

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