Tic Tac Toe Codechef Solution

Tic Tac Toe Solution Problem Code: TICTACTO

Two drunken players Alice and Bob are playing a modified version of Tic-Tac-Toe.

Initially, there is a N×MN×M empty grid.

Alice and Bob take alternate turns starting with Alice. In her turn, Alice chooses an empty cell on the board and puts a "A" on the chosen cell. In Bob’s turn, he chooses an empty cell and puts a "B" on the chosen cell.

The player who first gets any K×KK×K subgrid completely filled with his/her initial wins the game. By a K×KK×K subgrid, we mean the intersection of KK consecutive rows and KK consecutive columns. The players do not stop making turns after one of them wins and they play all N⋅MN⋅M turns until the grid is filled.

You are given the sequence of moves played by each player (The cell selected by the player in his/her turn). You have to output the winner of the game or report that there is no winner.


  • The first line contains an integer TT, the number of test cases. Then the test cases follow.
  • The first line of each test case contains three integers NN, MM, KK.
  • Each of the next N⋅MN⋅M lines contains two integers XX, YY (the row and column of the cell chosen by the player with the current turn)


For each test case, output "Alice" if Alice wins the game or "Bob" if Bob wins the game or "Draw" if no one wins.


  • 1≤T≤1051≤T≤105
  • 1≤N,M≤10001≤N,M≤1000
  • 1≤K≤min(N,M)1≤K≤min(N,M)
  • 1≤X≤N1≤X≤N
  • 1≤Y≤M1≤Y≤M
  • The sum of N⋅MN⋅M over all test cases does not exceed 106106.


Subtask 1 (10 points): 1≤N≤10, 1≤M≤101≤N≤10, 1≤M≤10

Subtask 2 (20 points): 1≤N≤50, 1≤M≤501≤N≤50, 1≤M≤50

Subtask 3 (70 points): Original Constraints

Sample Input

3 4 2
1 1
3 1
1 2
2 3
2 1
2 4
3 2
3 4
1 3
3 3
2 2
1 4

Sample Output



After 1010 turns of the game the state of the grid is:


At this moment, Bob has achieved a subgrid of 2×22×2 with his initials while Alice has not achieved it yet so Bob is the winner.


#include <bits/stdc++.h>
using namespace std;
struct play{
    int pos;
    char per;
    int val = 0;
int main() {
	int t,n,m,k;
	    play arr[n][m];
	    int x,y;
	    for(int i=0;i<n*m;i++){
	        arr[x-1][y-1].per = i%2==0 ? 'A':'B';
	        arr[x-1][y-1].pos = i;
	    int found = n*m;
	    string winner = "Draw";
	    for(int i=0;i<n;i++){
	        for(int j=0;j<m;j++){
	            char ch = arr[i][j].per;
	            if(i==0 or j==0){
	                arr[i][j].val = 1;
	            }else if(ch == arr[i-1][j-1].per and ch == arr[i-1][j].per and ch == arr[i][j-1].per){
	                arr[i][j].val = 1 + min({arr[i-1][j-1].val,arr[i-1][j].val,arr[i][j-1].val});
	                arr[i][j].val = 1;
	       for(int i=k-1;i<n;i++){
	        for(int j=k-1;j<m;j++){
	            if(arr[i][j].val >= k){
	                char win = arr[i][j].per;
	                int latest = 0;
	                for(int I = i-k+1;I<=i;I++){
	                    for(int J =j-k+1;J<=j;J++){
	                        latest = max(latest,arr[I][J].pos);
	                found = min(found,latest);
	                if(found == latest){
	                    winner = (win=='B' ? "Bob":"Alice");
	       if(found != m*n){
	return 0;

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