Description
Alice and Bob take turns playing a game, with Alice starting first.
Initially, there are n stones in a pile. On each player's turn, that player makes a move consisting of removing any non-zero square number of stones in the pile.
Also, if a player cannot make a move, he/she loses the game.
Given a positive integer n, return true if and only if Alice wins the game otherwise return false, assuming both players play optimally.
Example 1:
Input: n = 1
Output: true
Explanation: Alice can remove 1 stone winning the game because Bob doesn't have any moves.
Example 2:
Input: n = 2
Output: false
Explanation: Alice can only remove 1 stone, after that Bob removes the last one winning the game (2 -> 1 -> 0).
Example 3:
Input: n = 4
Output: true
Explanation: n is already a perfect square, Alice can win with one move, removing 4 stones (4 -> 0).
Constraints:
Solutions
Solution 1
PythonJavaC++GoTypeScript
class Solution:
def winnerSquareGame(self, n: int) -> bool:
@cache
def dfs(i: int) -> bool:
if i == 0:
return False
j = 1
while j * j <= i:
if not dfs(i - j * j):
return True
j += 1
return False
return dfs(n)(code-box)
class Solution {
private Boolean[] f;
public boolean winnerSquareGame(int n) {
f = new Boolean[n + 1];
return dfs(n);
}
private boolean dfs(int i) {
if (i <= 0) {
return false;
}
if (f[i] != null) {
return f[i];
}
for (int j = 1; j <= i / j; ++j) {
if (!dfs(i - j * j)) {
return f[i] = true;
}
}
return f[i] = false;
}
}(code-box)
class Solution {
public:
bool winnerSquareGame(int n) {
int f[n + 1];
memset(f, 0, sizeof(f));
function<bool(int)> dfs = [&](int i) -> bool {
if (i <= 0) {
return false;
}
if (f[i] != 0) {
return f[i] == 1;
}
for (int j = 1; j <= i / j; ++j) {
if (!dfs(i - j * j)) {
f[i] = 1;
return true;
}
}
f[i] = -1;
return false;
};
return dfs(n);
}
};(code-box)
func winnerSquareGame(n int) bool {
f := make([]int, n+1)
var dfs func(int) bool
dfs = func(i int) bool {
if i <= 0 {
return false
}
if f[i] != 0 {
return f[i] == 1
}
for j := 1; j <= i/j; j++ {
if !dfs(i - j*j) {
f[i] = 1
return true
}
}
f[i] = -1
return false
}
return dfs(n)
}(code-box)
function winnerSquareGame(n: number): boolean {
const f: number[] = new Array(n + 1).fill(0);
const dfs = (i: number): boolean => {
if (i <= 0) {
return false;
}
if (f[i] !== 0) {
return f[i] === 1;
}
for (let j = 1; j * j <= i; ++j) {
if (!dfs(i - j * j)) {
f[i] = 1;
return true;
}
}
f[i] = -1;
return false;
};
return dfs(n);
}(code-box)
Solution 2
PythonJavaC++GoTypeScript
class Solution:
def winnerSquareGame(self, n: int) -> bool:
f = [False] * (n + 1)
for i in range(1, n + 1):
j = 1
while j <= i // j:
if not f[i - j * j]:
f[i] = True
break
j += 1
return f[n](code-box)
class Solution {
public boolean winnerSquareGame(int n) {
boolean[] f = new boolean[n + 1];
for (int i = 1; i <= n; ++i) {
for (int j = 1; j <= i / j; ++j) {
if (!f[i - j * j]) {
f[i] = true;
break;
}
}
}
return f[n];
}
}(code-box)
class Solution {
public:
bool winnerSquareGame(int n) {
bool f[n + 1];
memset(f, false, sizeof(f));
for (int i = 1; i <= n; ++i) {
for (int j = 1; j <= i / j; ++j) {
if (!f[i - j * j]) {
f[i] = true;
break;
}
}
}
return f[n];
}
};(code-box)
func winnerSquareGame(n int) bool {
f := make([]bool, n+1)
for i := 1; i <= n; i++ {
for j := 1; j <= i/j; j++ {
if !f[i-j*j] {
f[i] = true
break
}
}
}
return f[n]
}(code-box)
function winnerSquareGame(n: number): boolean {
const f: boolean[] = new Array(n + 1).fill(false);
for (let i = 1; i <= n; ++i) {
for (let j = 1; j * j <= i; ++j) {
if (!f[i - j * j]) {
f[i] = true;
break;
}
}
}
return f[n];
}(code-box)
