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++ 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; } |
-1
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