# Tic Tac Toe - Amazon Top Interview Questions

### Problem Statement :

```Implement the tic-tac-toe game with the following methods:

TicTacToe(int n) which instantiates an n x n game board. A player wins a game if their pieces either form a horizontal, vertical, or diagonal line of length n.
int move(int r, int c, boolean me) which places the next move at row r and column c. me indicates whether it's my move (me = true) or it's your opponent's (me = false) move. If this move makes you win, return 1, if your opponent wins, return -1, and otherwise return 0.
Constraints

1 ≤ n ≤ 100,000

0 ≤ m ≤ 100,000 where m is the number of calls to move

Example 1

Input

methods = ["constructor", "move", "move", "move", "move", "move"]

arguments = [[3], [0, 0, True], [2, 0, False], [0, 1, True], [2, 1, False], [0, 2, True]]`

Output

[None, 0, 0, 0, 0, 1]

Explanation

t = TicTacToe(3)

t.move(0, 0, True) == 0 # I place piece (0, 0)

t.move(2, 0, False) == 0 # Opponent places piece (2, 0)

t.move(0, 1, True) == 0 # I place piece (0, 1)

t.move(2, 1, False) == 0 # Opponent places piece (2, 1)

t.move(0, 2, True) == 1 # I place piece (0, 2) to win```

### Solution :

```                        ```Solution in C++ :

class TicTacToe {
int n;
unordered_map<int, int> row[2], col[2], diag[2], cross[2];

public:
TicTacToe(int n) : n(n) {
}

int move(int r, int c, bool me) {
int winner = 2 * me - 1;
if (++row[me][r] == n) return winner;
if (++col[me][c] == n) return winner;
if (++diag[me][r + c] == n) return winner;
if (++cross[me][r - c] == n) return winner;
return 0;
}
};```
```

```                        ```Solution in Java :

import java.util.*;

class TicTacToe {
int n;
int[] rowSum;
int[] colSum;
int diag;
int revDiag;

public TicTacToe(int n) {
this.n = n;
rowSum = new int[n];
colSum = new int[n];
diag = 0;
revDiag = 0;
}

public int move(int r, int c, boolean me) {
// Step 1) Make a move
int val = me ? 1 : -1;
rowSum[r] += val;
colSum[c] += val;

if (r == c)
diag += val;
if (r == n - c - 1)
revDiag += val;

// Step 2) Check if game over
boolean gameOver = Math.abs(rowSum[r]) == n || Math.abs(colSum[c]) == n
|| Math.abs(diag) == n || Math.abs(revDiag) == n;

if (gameOver)
return me ? 1 : -1;
return 0;
}
}```
```

```                        ```Solution in Python :

class TicTacToe:
def __init__(self, n):
self.n = n
self.row_sums = [0] * n
self.col_sums = [0] * n
self.diag = self.inv_diag = 0

def move(self, r, c, me):
val = 1 if me else -1

# update values
self.row_sums[r] += val
self.col_sums[c] += val
if r == c:
self.diag += val
if r == self.n - c - 1:
self.inv_diag += val

# check wins
if self.n in list(map(abs, [self.row_sums[r], self.col_sums[c], self.diag, self.inv_diag])):
return val
return 0```
```

## Print in Reverse

Given a pointer to the head of a singly-linked list, print each data value from the reversed list. If the given list is empty, do not print anything. Example head* refers to the linked list with data values 1->2->3->Null Print the following: 3 2 1 Function Description: Complete the reversePrint function in the editor below. reversePrint has the following parameters: Sing

Given the pointer to the head node of a linked list, change the next pointers of the nodes so that their order is reversed. The head pointer given may be null meaning that the initial list is empty. Example: head references the list 1->2->3->Null. Manipulate the next pointers of each node in place and return head, now referencing the head of the list 3->2->1->Null. Function Descriptio

You’re given the pointer to the head nodes of two linked lists. Compare the data in the nodes of the linked lists to check if they are equal. If all data attributes are equal and the lists are the same length, return 1. Otherwise, return 0. Example: list1=1->2->3->Null list2=1->2->3->4->Null The two lists have equal data attributes for the first 3 nodes. list2 is longer, though, so the lis

## Merge two sorted linked lists

This challenge is part of a tutorial track by MyCodeSchool Given pointers to the heads of two sorted linked lists, merge them into a single, sorted linked list. Either head pointer may be null meaning that the corresponding list is empty. Example headA refers to 1 -> 3 -> 7 -> NULL headB refers to 1 -> 2 -> NULL The new list is 1 -> 1 -> 2 -> 3 -> 7 -> NULL. Function Description C

## Get Node Value

This challenge is part of a tutorial track by MyCodeSchool Given a pointer to the head of a linked list and a specific position, determine the data value at that position. Count backwards from the tail node. The tail is at postion 0, its parent is at 1 and so on. Example head refers to 3 -> 2 -> 1 -> 0 -> NULL positionFromTail = 2 Each of the data values matches its distance from the t

## Delete duplicate-value nodes from a sorted linked list

This challenge is part of a tutorial track by MyCodeSchool You are given the pointer to the head node of a sorted linked list, where the data in the nodes is in ascending order. Delete nodes and return a sorted list with each distinct value in the original list. The given head pointer may be null indicating that the list is empty. Example head refers to the first node in the list 1 -> 2 -