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

• Last Updated : 07 Jun, 2021

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 ``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;``}`

## 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

 ``
Output:
`-1`

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