# 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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