Algorithm
Problem link- https://www.spoj.com/problems/TOE1/
TOE1 - Tic-Tac-Toe ( I )
Tic Tac Toe is a child's game played on a 3 by 3 grid. One player, X, starts by placing an X at an unoccupied grid position. Then the other player, O, places an O at an unoccupied grid position. Play alternates between X and O until the grid is filled or one player's symbols occupy an entire line (vertical, horizontal, or diagonal) in the grid.
We will denote the initial empty Tic Tac Toe grid with nine dots. Whenever X or O plays we fill in an X or an O in the appropriate position. The example below illustrates each grid configuration from the beginning to the end of a game in which X wins.
Your job is to read a grid and to determine whether or not it could possibly be part of a valid Tic Tac Toe game. That is, is there a series of plays that can yield this grid somewhere between the start and end of the game?
Input
The first line of input contains N, the number of test cases. 4N-1 lines follow, specifying N grid configurations separated by empty lines.
Output
For each case print "yes" or "no" on a line by itself, indicating whether or not the configuration could be part of a Tic Tac Toe game.
Example
Input: 2 X.O OO. XXX O.X XX. OOO Output: yes no
Code Examples
#1 Code Example with C++ Programming
Code -
C++ Programming
#include <iostream>
#include <cstdio>
#include <vector>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <sstream>
#include <map>
#include <list>
using namespace std;
string g[3];
bool checkrows(char winner){
for(int i=0;i<3;i++){
if(g[i][0]==winner&&g[i][1]==winner&&g[i][2]==winner){
return true;
}
}
return false;
}
bool checkcols(char winner){
for(int i=0;i<3;i++){
if(g[0][i]==winner&&g[1][i]==winner&&g[2][i]==winner){
return true;
}
}
return false;
}
bool checkdiagnols(char winner){
if(g[0][0]==winner&&g[1][1]==winner&&g[2][2]==winner){
return true;
}
if(g[0][2]==winner&&g[1][1]==winner&&g[2][0]==winner){
return true;
}
return false;
}
int main(){
int n;
scanf("%d",&n);
while(n--){
bool possible=true;
for(int i=0;i<3;i++){
cin>>g[i];
}
int xcount=0,ocount=0;
for(int i=0;i<3;i++){
for(int j=0;j<3;j++){
if(g[i][j]=='X'){
xcount++;
continue;
}
if(g[i][j]=='O'){
ocount++;
}
}
}
if(ocount>xcount){
possible=false;
}
if((xcount-ocount)>1){
possible=false;
}
//cout<<xcount<<" "<<ocount<<endl;
bool winnerx=(checkrows('X')||checkcols('X')||checkdiagnols('X'));
if(winnerx){
if((xcount-ocount)!=1){
possible=false;
}
}
bool winnero=(checkrows('O')||checkcols('O')||checkdiagnols('O'));
if(winnero){
if((xcount-ocount)!=0){
possible=false;
}
}
if(winnerx==true&&winnero==true){
possible=false;
}
if(possible){
printf("yes\n");
}
else{
printf("no\n");
}
if(n!=0){
printf("\n");
}
}
return 0;
}Input
X.O
OO.
XXX
O.X
XX.
OOO
Output
no
#2 Code Example with Java Programming
Code -
Java Programming
#include<stdio.h>
char str[10],str1[4],str2[4],str3[4];
int main()
{
int i,j,no_X,no_O,flag,win_X,win_O,t;
scanf("%i",&t);
while(t--)
{
no_X = no_O = flag = win_X = win_O = 0;
scanf("%s%s%s",str1,str2,str3);
str[0] = str1[0];
str[1] = str1[1];
str[2] = str1[2];
str[3] = str2[0];
str[4] = str2[1];
str[5] = str2[2];
str[6] = str3[0];
str[7] = str3[1];
str[8] = str3[2];
for(i=0; i<9; i++)
{
if(str[i] == 'X')
no_X++;
else if(str[i] == 'O')
no_O++;
}
if(str[4] != '.')
{
if(str[4] == 'O')
{
if((str[0] == 'O' && str[8] == 'O') || (str[1] == 'O' && str[7] == 'O') || (str[2] == 'O' && str[6] == 'O') || (str[3] == 'O' && str[5] == 'O'))
win_O = 1;
}
else
{
if((str[0] == 'X' && str[8] == 'X') || (str[1] == 'X' && str[7] == 'X') || (str[2] == 'X' && str[6] == 'X') || (str[3] == 'X' && str[5] == 'X'))
win_X = 1;
}
}
if(str[1] != '.')
{
if(str[0] == 'O')
{
if((str[3] == 'O' && str[6] == 'O') || (str[1] == 'O' && str[2] == 'O'))
win_O =1;
}
else
{
if((str[3] == 'X' && str[6] == 'X') || (str[1] == 'X' && str[2] == 'X'))
win_X =1;
}
}
if(str[8] != '.')
{
if(str[8] == 'O')
{
if((str[5] == 'O' && str[2] == 'O') || (str[7] == 'O' && str[6] == 'O'))
win_O = 1;
}
else
{
if((str[5] == 'X' && str[2] == 'X') || (str[7] == 'X' && str[6] == 'X'))
win_X = 1;
}
}
if(win_X == 1 && win_O == 1)
flag = 1;
else
{
if(win_X == 1 && win_O == 0 && no_X != no_O + 1)
flag = 1;
else if(win_X == 0 && win_O == 1 && no_X != no_O)
flag = 1;
else if(no_X != no_O+1 && no_X != no_O)
{
flag = 1;
}
}
if(flag)
printf("no\n");
else
printf("yes\n");
}
return 0;
}Input
X.O
OO.
XXX
O.X
XX.
OOO
Output
no
Demonstration
SPOJ Solution-TOE1 Tic-Tac-Toe ( I )-Solution in C, C++, Java, Python