Check if a variable equal X or Y - ruby

I'm stuck with something very basic in Ruby. Maybe I did not understand how the operator || works?
a_to_i = [a = 12, b = 0, c = 0, d = 0, e = 0, f = 0, g = 0, h = 0, i = 0]
puts a
puts a == (3 || 12)
I'm waiting the output : true
But it returns false instead.
Do you see where am i doing wrong ? Thank you very much in advance.


3 || 12 returns 3, as it is a truthy value.
You need to ask two questions here:
puts (a == 3 || a == 12)
Alternatively, if the list of options is slightly longer:
puts [3, 12].include? a


The result of the statment (3 || 12) is 3, always resulting in a == 3. You'll need to check both values separately.
puts((a == 3) || (a == 12))

Related

Comparing two arrays with each other

I am trying to solve some problems in ruby get a hold.
I am trying compare two arrays with each other.
Array1 = [1,0,1,0,1,1]
Array2= [0,0,1,0,1,0]
I am getting this input from the user. Then I am comparing the votes. If both persons have upvoted 1 at same index, I am trying increment an empty array by 1.
def count_votes
bob_votes = gets.chomp
alice_votes = gets.chomp
bvotes = bob_votes.split('')
avotes = alice_votes.split('')
common_upvotes = []
bvotes.each.with_index(0) do |el, i|
if bvotes[i] == 1
common_upvotes << 1
end
end
I actually want to compare avotes with bvotes and then increment empty array by 1. I need a little help

You can use Array#zip and Enumerable#count:
array1 = [1,0,1,0,1,1]
array2= [0,0,1,0,1,0]
array1.zip(array2)
#⇒ [[1, 0], [0, 0], [1, 1], [0, 0], [1, 1], [1, 0]]
array1.zip(array2).count { |v1, v2| v1 == v2 && v1 == 1 }
#⇒ 2
or (credits to #engineersmnky):
array1.zip(array2).count { |v1, v2| v1 & v2 == 1 }
or even better (credits to #Stefan):
array1.zip(array2).count { |values| values.all?(1) }
or
array1.
zip(array2).
reject { |v1, v2| v1 == 0 || v2 == 0 }.
count
#⇒ 2
Sidenote: capitalized Array1 declares a constant. To declare a variable, use array1 instead.

The number of indices i for which Array1[i] == 1 && Array2[i] == 1 is
Array1.each_index.count { |i| Array1[i] == 1 && Array2[i] == 1 }
#=> 2
The array of indices i for which Array1[i] == 1 && Array2[i] == 1 is
Array1.each_index.select { |i| Array1[i] == 1 && Array2[i] == 1 }
#=> [2, 4]
The number of indices i for which Array1[i] == Array2[i] is
Array1.each_index.count { |i| Array1[i] == Array2[i] }
#=> 4

If you want to build a new array tracking the index where there is a match of upvotes:
a1 = [1,0,1,0,1,1]
a2= [0,0,1,0,1,0]
p [a1, a2].transpose.map {|x| x.reduce(:&)}
#=> [0, 0, 1, 0, 1, 0]
For just counting, this is another way:
a1 = [1,0,1,0,1,1]
a2= [0,0,1,0,1,0]
votes = 0
a1.each_with_index do |a1, idx|
votes +=1 if (a1 + a2[idx]) == 2
end
p votes #=> 2
In one line:
a1.each_with_index { |a1, idx| votes += 1 if (a1 + a2[idx]) == 2 }

How does recursion work in ruby?

I have this method which switches the number digit 5 with 7.
def switch_digit(num)
if num <= 0
return 0
end
digit = num % 10
if (digit == 5)
digit = 7
end
return switch_digit(num/10) * 10 + digit
end
switch_digit(5952)
Can someone explain why once the method hits the base case it doesn't return 0?
How does this recursive method actually work? Does it append the returned digit with the next digit?

I added a little change to your code, to be aware it's working.
Finally I also expected the value of the method was 0, but it is not.
The end is reached, but the returned value is not 0. Why?
def switch_digit(num, array)
if num <= 0
array << num
p array
puts "The end"
return 0
end
digit = num % 10
array << [digit, num]
if (digit == 5)
digit = 7
end
return p switch_digit(num/10, array) * 10 + digit
end
p "returned value = " + switch_digit(123456789, Array.new).to_s
Which outputs:
#=> [[9, 123456789], [8, 12345678], [7, 1234567], [6, 123456], [5, 12345], [4, 1234], [3, 123], [2, 12], [1, 1], 0]
#=> The end
#=> 1
#=> 12
#=> 123
#=> 1234
#=> 12347
#=> 123476
#=> 1234767
#=> 12347678
#=> 123476789
#=> "returned value = 123476789"

The base case returns 0, but the overall result is determined by the return switch_digit(num/10) * 10 + digit
Follow the code through with a smaller example e.g. switch_digit(15):
num <= 0 # no
digit = num % 10 # 5
digit == 5 # yep, so swap it for a 7
return switch_digit(num/10) * 10 + 7
num/10 is 1 so what does the recursive switch_digit(1) evaluate to?
num <= 0 # no
digit = num % 10 # 1
digit == 1 # so leave it unchanged
return switch_digit(num/10) * 10 + 1
num/10 is 0 so now we hit the base case
switch_digit(15) == switch_digit(1) * 10 + 7
switch_digit(1) == switch_digit(0) * 10 + 1
switch_digit(0) == 0 # the base case
working back up, plugging in values from lower down results:
switch_digit(1) == 0 * 10 + 1 == 1
switch_digit(15) == 1 * 10 + 7 == 17
I'd also add that there's nothing specific to Ruby about how recursion is handled here. Any other descriptions of recursion, or classic example such as a recursive factorial function should help you get a better understanding.

What is the Grooviest way to implement an incrementing integer list?

I need a data structure which is a list of integers but each time an integer is added to it the value it stores is the sum of the values it contains plus the value that is added.
For example:
def incrementingList = []
incrementingList.add(5) // now it has 5
incrementingList.add(3) // now it has 5,8
incrementingList.add(2) // now it has 5,8,10
Is there a groovy way to implement this so it can be ready to use as in the example?
UPDATE
What if it is possible for this list to contain 0s and if the last element is a 0 then it should increment by the last non 0 element?

You can use metaprogramming to define custom method:
List.metaClass.plusAdd = { e ->
delegate.isEmpty() ? delegate.add(e) : delegate.add(delegate.last() + e)
}
def l = []
l.plusAdd(5)
l.plusAdd(3)
l.plusAdd(2)
assert l == [5, 8, 10]
EDIT
Update for adding last non-zero element:
List.metaClass.plusAdd = { e ->
if(delegate.isEmpty()) {
delegate << e
} else {
def nonZeros = delegate.findAll { it > 0 }
delegate << (nonZeros ? nonZeros.last() + e : e)
}
}
def l = []
l.plusAdd(5)
l.plusAdd(3)
l.plusAdd(2)
assert l == [5, 8, 10]
l = [5, 0]
l.plusAdd(5)
assert l == [5, 0, 10]
l = [1,0]
l.plusAdd(5)
assert l == [1, 0 ,6]

this should be simple and working
incrementingList.add(incrementingList.last() + 5)
check if not empty:
incrementingList.add((incrementingList.size() > 0 ) ? incrementingList.last() + 5 : 5)
,
​def incrementingList = []
incrementingList.add((incrementingList.size() > 0 ) ? incrementingList.last() + 5 : 5)
incrementingList.add((incrementingList.size() > 0 ) ? incrementingList.last() + 5 : 5)
incrementingList.add((incrementingList.size() > 0 ) ? incrementingList.last() + 5 : 5)
println incrementingList​
output:
[5, 10, 15]

Slightly different approach:
def addToList(incList, val) {
if (incList.size() > 0) {
incList << incList.last() + val
} else {
incList << val
}
return incList
}
def incList = []
addToList(incList, 3)
addToList(incList, 2)
addToList(incList, 5)
println incList
output:
[3, 5, 10]
Updated:
def addToList(incList, val) {
if (incList.size() > 0) {
if (val != 0) {
incList << val + incList.last()
} else {
def addVal = incList.size() > 1 ? (incList[-1] - incList[-2]) : incList[-1]
incList << addVal + incList.last()
}
} else {
incList << val
}
}
def incList = []
addToList(incList, 5)
addToList(incList, 3)
addToList(incList, 0)
addToList(incList, 5)
addToList(incList, 0)
println incList
output:
[5, 8, 11, 16, 21]

Suggestion, based on the update request (as I understand it)...
[Edit: the 'empty' case is covered by the 'no non-zero' case, so it can be simplified]
List.metaClass.addAsSum = { e ->
def nonZero = delegate.reverse().find { it != 0 }
if (nonZero != null) {
delegate.add(nonZero + e)
} else {
delegate.add(e)
}
}
Test runs:
def list
// test 0
list = []
list.addAsSum(5)
assert list == [5]
// test 1
list = []
list.addAsSum(5)
list.addAsSum(3)
list.addAsSum(2)
assert list == [5, 8, 10]
// test 2
list = []
list.addAsSum(5)
list.addAsSum(3)
list.addAsSum(0)
list.addAsSum(2)
assert list == [5, 8, 8, 10]
// test 3
list = []
list.addAsSum(0)
list.addAsSum(0)
list.addAsSum(0)
list.addAsSum(2)
list.addAsSum(4)
assert list == [0, 0, 0, 2, 6]

Adding a requirement that on a list filled only with zeros it should simply add the new element, this is my suggestion:
List.metaClass.fill = { n ->
delegate ? delegate.add((delegate.reverse().find { it > 0 } ?: 0) + n) : delegate.add(n)
delegate
}
assert [].fill(1).fill(2).fill(3) == [1, 3, 6]
assert [5, 0].fill(5) == [5, 0, 10]
assert [0, 0].fill(5).fill(5) == [0, 0, 5, 10]
note that I return delegate in the fill() such that you can call it in sequence.

Using `map` to solve 'FizzBuzz'

I tired to use map to solve fizzbuzz.
def fizzbuzz(n)
array =(1..n).to_a
array.map{|x|
if x % 3 == 0 && x % 5 == 0
x = 'FizzBuzz'
elsif x % 3 == 0
x = 'Fizz'
elsif x % 5 == 0
x = 'Buzz'
end
}
array
end
Somehow, it doesn't work. Do you know what's wrong?

Method map does not change the original array. Use the bang version map! instead.

Using map! as suggested by #tmc and some other changes try:
def fizzbuzz(n)
array =(1..n).to_a
array.map!{|x|
if x % 3 == 0 && x % 5 == 0
x = 'FizzBuzz'
elsif x % 3 == 0
x = 'Fizz'
elsif x % 5 == 0
x = 'Buzz'
else
x = x
end
}
p array
end
fizzbuzz(10) #=> [1, 2, "Fizz", 4, "Buzz", "Fizz", 7, 8, "Fizz", "Buzz"]
As you can see I've added a call to the method fizzbuzz with an argument of 10 which you can change. And I've used p to inspect the array as well as a final else statement.

All Possible Tic Tac Toe Winning Combinations

I had an interview were I was asked a seemingly simple algorithm question: "Write an algorithm to return me all possible winning combinations for tic tac toe." I still can't figure out an efficient way to handle this. Is there a standard algorithm or common that should be applied to similar questions like this that I'm not aware of?

This is one of those problems that's actually simple enough for brute force and, while you could use combinatorics, graph theory, or many other complex tools to solve it, I'd actually be impressed by applicants that recognise the fact there's an easier way (at least for this problem).
There are only 39, or 19,683 possible combinations of placing x, o or <blank> in the grid, and not all of those are valid.
First, a valid game position is one where the difference between x and o counts is no more than one, since they have to alternate moves.
In addition, it's impossible to have a state where both sides have three in a row, so they can be discounted as well. If both have three in a row, then one of them would have won in the previous move.
There's actually another limitation in that it's impossible for one side to have won in two different ways without a common cell (again, they would have won in a previous move), meaning that:
XXX
OOO
XXX
cannot be achieved, while:
XXX
OOX
OOX
can be. But we can actually ignore that since there's no way to win two ways without a common cell without having already violated the "maximum difference of one" rule, since you need six cells for that, with the opponent only having three.
So I would simply use brute force and, for each position where the difference is zero or one between the counts, check the eight winning possibilities for both sides. Assuming only one of them has a win, that's a legal, winning game.
Below is a proof of concept in Python, but first the output of time when run on the process sending output to /dev/null to show how fast it is:
real 0m0.169s
user 0m0.109s
sys 0m0.030s
The code:
def won(c, n):
if c[0] == n and c[1] == n and c[2] == n: return 1
if c[3] == n and c[4] == n and c[5] == n: return 1
if c[6] == n and c[7] == n and c[8] == n: return 1
if c[0] == n and c[3] == n and c[6] == n: return 1
if c[1] == n and c[4] == n and c[7] == n: return 1
if c[2] == n and c[5] == n and c[8] == n: return 1
if c[0] == n and c[4] == n and c[8] == n: return 1
if c[2] == n and c[4] == n and c[6] == n: return 1
return 0
pc = [' ', 'x', 'o']
c = [0] * 9
for c[0] in range (3):
for c[1] in range (3):
for c[2] in range (3):
for c[3] in range (3):
for c[4] in range (3):
for c[5] in range (3):
for c[6] in range (3):
for c[7] in range (3):
for c[8] in range (3):
countx = sum([1 for x in c if x == 1])
county = sum([1 for x in c if x == 2])
if abs(countx-county) < 2:
if won(c,1) + won(c,2) == 1:
print " %s | %s | %s" % (pc[c[0]],pc[c[1]],pc[c[2]])
print "---+---+---"
print " %s | %s | %s" % (pc[c[3]],pc[c[4]],pc[c[5]])
print "---+---+---"
print " %s | %s | %s" % (pc[c[6]],pc[c[7]],pc[c[8]])
print
As one commenter has pointed out, there is one more restriction. The winner for a given board cannot have less cells than the loser since that means the loser just moved, despite the fact the winner had already won on the last move.
I won't change the code to take that into account but it would be a simple matter of checking who has the most cells (the last person that moved) and ensuring the winning line belonged to them.

Another way could be to start with each of the eight winning positions,
xxx ---
--- xxx
--- --- ... etc.,
and recursively fill in all legal combinations (start with inserting 2 o's, then add an x for each o ; avoid o winning positions):
xxx xxx xxx
oo- oox oox
--- o-- oox ... etc.,

Today I had an interview with Apple and I had the same question. I couldn't think well at that moment. Later one on, before going to a meeting I wrote the function for the combinations in 15 minutes, and when I came back from the meeting I wrote the validation function again in 15 minutes. I get nervous at interviews, Apple not trusts my resume, they only trust what they see in the interview, I don't blame them, many companies are the same, I just say that something in this hiring process doesn't look quite smart.
Anyways, here is my solution in Swift 4, there are 8 lines of code for the combinations function and 17 lines of code to check a valid board.
Cheers!!!
// Not used yet: 0
// Used with x : 1
// Used with 0 : 2
// 8 lines code to get the next combination
func increment ( _ list: inout [Int], _ base: Int ) -> Bool {
for digit in 0..<list.count {
list[digit] += 1
if list[digit] < base { return true }
list[digit] = 0
}
return false
}
let incrementTicTacToe = { increment(&$0, 3) }
let win0_ = [0,1,2] // [1,1,1,0,0,0,0,0,0]
let win1_ = [3,4,5] // [0,0,0,1,1,1,0,0,0]
let win2_ = [6,7,8] // [0,0,0,0,0,0,1,1,1]
let win_0 = [0,3,6] // [1,0,0,1,0,0,1,0,0]
let win_1 = [1,4,7] // [0,1,0,0,1,0,0,1,0]
let win_2 = [2,5,8] // [0,0,1,0,0,1,0,0,1]
let win00 = [0,4,8] // [1,0,0,0,1,0,0,0,1]
let win11 = [2,4,6] // [0,0,1,0,1,0,1,0,0]
let winList = [ win0_, win1_, win2_, win_0, win_1, win_2, win00, win11]
// 16 lines to check a valid board, wihtout countin lines of comment.
func winCombination (_ tictactoe: [Int]) -> Bool {
var count = 0
for win in winList {
if tictactoe[win[0]] == tictactoe[win[1]],
tictactoe[win[1]] == tictactoe[win[2]],
tictactoe[win[2]] != 0 {
// If the combination exist increment count by 1.
count += 1
}
if count == 2 {
return false
}
}
var indexes = Array(repeating:0, count:3)
for num in tictactoe { indexes[num] += 1 }
// '0' and 'X' must be used the same times or with a diference of one.
// Must one and only one valid combination
return abs(indexes[1] - indexes[2]) <= 1 && count == 1
}
// Test
var listToIncrement = Array(repeating:0, count:9)
var combinationsCount = 1
var winCount = 0
while incrementTicTacToe(&listToIncrement) {
if winCombination(listToIncrement) == true {
winCount += 1
}
combinationsCount += 1
}
print("There is \(combinationsCount) combinations including possible and impossible ones.")
print("There is \(winCount) combinations for wining positions.")
/*
There are 19683 combinations including possible and impossible ones.
There are 2032 combinations for winning positions.
*/
listToIncrement = Array(repeating:0, count:9)
var listOfIncremented = ""
for _ in 0..<1000 { // Win combinations for the first 1000 combinations
_ = incrementTicTacToe(&listToIncrement)
if winCombination(listToIncrement) == true {
listOfIncremented += ", \(listToIncrement)"
}
}
print("List of combinations: \(listOfIncremented)")
/*
List of combinations: , [2, 2, 2, 1, 1, 0, 0, 0, 0], [1, 1, 1, 2, 2, 0, 0, 0, 0],
[2, 2, 2, 1, 0, 1, 0, 0, 0], [2, 2, 2, 0, 1, 1, 0, 0, 0], [2, 2, 0, 1, 1, 1, 0, 0, 0],
[2, 0, 2, 1, 1, 1, 0, 0, 0], [0, 2, 2, 1, 1, 1, 0, 0, 0], [1, 1, 1, 2, 0, 2, 0, 0, 0],
[1, 1, 1, 0, 2, 2, 0, 0, 0], [1, 1, 0, 2, 2, 2, 0, 0, 0], [1, 0, 1, 2, 2, 2, 0, 0, 0],
[0, 1, 1, 2, 2, 2, 0, 0, 0], [1, 2, 2, 1, 0, 0, 1, 0, 0], [2, 2, 2, 1, 0, 0, 1, 0, 0],
[2, 2, 1, 0, 1, 0, 1, 0, 0], [2, 2, 2, 0, 1, 0, 1, 0, 0], [2, 2, 2, 1, 1, 0, 1, 0, 0],
[2, 0, 1, 2, 1, 0, 1, 0, 0], [0, 2, 1, 2, 1, 0, 1, 0, 0], [2, 2, 1, 2, 1, 0, 1, 0, 0],
[1, 2, 0, 1, 2, 0, 1, 0, 0], [1, 0, 2, 1, 2, 0, 1, 0, 0], [1, 2, 2, 1, 2, 0, 1, 0, 0],
[2, 2, 2, 0, 0, 1, 1, 0, 0]
*/

This is a java equivalent code sample
package testit;
public class TicTacToe {
public static void main(String[] args) {
// TODO Auto-generated method stub
// 0 1 2
// 3 4 5
// 6 7 8
char[] pc = {' ' ,'o', 'x' };
char[] c = new char[9];
// initialize c
for (int i = 0; i < 9; i++)
c[i] = pc[0];
for (int i = 0; i < 3; i++) {
c[0] = pc[i];
for (int j = 0; j < 3; j++) {
c[1] = pc[j];
for (int k = 0; k < 3; k++) {
c[2] = pc[k];
for (int l = 0; l < 3; l++) {
c[3] = pc[l];
for (int m = 0; m < 3; m++) {
c[4] = pc[m];
for (int n = 0; n < 3; n++) {
c[5] = pc[n];
for (int o = 0; o < 3; o++) {
c[6] = pc[o];
for (int p = 0; p < 3; p++) {
c[7] = pc[p];
for (int q = 0; q < 3; q++) {
c[8] = pc[q];
int countx = 0;
int county = 0;
for(int r = 0 ; r<9 ; r++){
if(c[r] == 'x'){
countx = countx + 1;
}
else if(c[r] == 'o'){
county = county + 1;
}
}
if(Math.abs(countx - county) < 2){
if(won(c, pc[2])+won(c, pc[1]) == 1 ){
System.out.println(c[0] + " " + c[1] + " " + c[2]);
System.out.println(c[3] + " " + c[4] + " " + c[5]);
System.out.println(c[6] + " " + c[7] + " " + c[8]);
System.out.println("*******************************************");
}
}
}
}
}
}
}
}
}
}
}
}
public static int won(char[] c, char n) {
if ((c[0] == n) && (c[1] == n) && (c[2] == n))
return 1;
else if ((c[3] == n) && (c[4] == n) && (c[5] == n))
return 1;
else if ((c[6] == n) && (c[7] == n) && (c[8] == n))
return 1;
else if ((c[0] == n) && (c[3] == n) && (c[6] == n))
return 1;
else if ((c[1] == n) && (c[4] == n) && (c[7] == n))
return 1;
else if ((c[2] == n) && (c[5] == n) && (c[8] == n))
return 1;
else if ((c[0] == n) && (c[4] == n) && (c[8] == n))
return 1;
else if ((c[2] == n) && (c[4] == n) && (c[6] == n))
return 1;
else
return 0;
}
}
`

Below Solution generates all possible combinations using recursion
It has eliminated impossible combinations and returned 888 Combinations
Below is a working code Possible winning combinations of the TIC TAC TOE game
const players = ['X', 'O'];
let gameBoard = Array.from({ length: 9 });
const winningCombination = [
[ 0, 1, 2 ],
[ 3, 4, 5 ],
[ 6, 7, 8 ],
[ 0, 3, 6 ],
[ 1, 4, 7 ],
[ 2, 5, 8 ],
[ 0, 4, 8 ],
[ 2, 4, 6 ],
];
const isWinningCombination = (board)=> {
if((Math.abs(board.filter(a => a === players[0]).length -
board.filter(a => a === players[1]).length)) > 1) {
return false
}
let winningComb = 0;
players.forEach( player => {
winningCombination.forEach( combinations => {
if (combinations.every(combination => board[combination] === player )) {
winningComb++;
}
});
});
return winningComb === 1;
}
const getCombinations = (board) => {
let currentBoard = [...board];
const firstEmptySquare = board.indexOf(undefined)
if (firstEmptySquare === -1) {
return isWinningCombination(board) ? [board] : [];
} else {
return [...players, ''].reduce((prev, next) => {
currentBoard[firstEmptySquare] = next;
if(next !== '' && board.filter(a => a === next).length > (gameBoard.length / players.length)) {
return [...prev]
}
return [board, ...prev, ...getCombinations(currentBoard)]
}, [])
}
}
const startApp = () => {
let combination = getCombinations(gameBoard).filter(board =>
board.every(item => !(item === undefined)) && isWinningCombination(board)
)
printCombination(combination)
}
const printCombination = (combination)=> {
const ulElement = document.querySelector('.combinations');
combination.forEach(comb => {
let node = document.createElement("li");
let nodePre = document.createElement("pre");
let textnode = document.createTextNode(JSON.stringify(comb));
nodePre.appendChild(textnode);
node.appendChild(nodePre);
ulElement.appendChild(node);
})
}
startApp();

This discovers all possible combinations for tic tac toe (255,168) -- written in JavaScript using recursion. It is not optimized, but gets you what you need.
const [EMPTY, O, X] = [0, 4, 1]
let count = 0
let coordinate = [
EMPTY, EMPTY, EMPTY,
EMPTY, EMPTY, EMPTY,
EMPTY, EMPTY, EMPTY
]
function reducer(arr, sumOne, sumTwo = null) {
let func = arr.reduce((sum, a) => sum + a, 0)
if((func === sumOne) || (func === sumTwo)) return true
}
function checkResult() {
let [a1, a2, a3, b1, b2, b3, c1, c2, c3] = coordinate
if(reducer([a1,a2,a3], 3, 12)) return true
if(reducer([a1,b2,c3], 3, 12)) return true
if(reducer([b1,b2,b3], 3, 12)) return true
if(reducer([c1,c2,c3], 3, 12)) return true
if(reducer([a3,b2,c1], 3, 12)) return true
if(reducer([a1,b1,c1], 3, 12)) return true
if(reducer([a2,b2,c2], 3, 12)) return true
if(reducer([a3,b3,c3], 3, 12)) return true
if(reducer([a1,a2,a3,b1,b2,b3,c1,c2,c3], 21)) return true
return false
}
function nextPiece() {
let [countX, countO] = [0, 0]
for(let i = 0; i < coordinate.length; i++) {
if(coordinate[i] === X) countX++
if(coordinate[i] === O) countO++
}
return countX === countO ? X : O
}
function countGames() {
if (checkResult()) {
count++
}else {
for (let i = 0; i < 9; i++) {
if (coordinate[i] === EMPTY) {
coordinate[i] = nextPiece()
countGames()
coordinate[i] = EMPTY
}
}
}
}
countGames()
console.log(count)
I separated out the checkResult returns in case you want to output various win conditions.

Could be solved with brute force but keep in mind the corner cases like player2 can't move when player1 has won and vice versa. Also remember Difference between moves of player1 and player can't be greater than 1 and less than 0.
I have written code for validating whether provided combination is valid or not, might soon post on github.

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