Description
You are playing a Flip Game with your friend.
You are given a string currentState that contains only '+' and '-'. You and your friend take turns to flip two consecutive "++" into "--". The game ends when a person can no longer make a move, and therefore the other person will be the winner.
Return true if the starting player can guarantee a win, and false otherwise.
Example 1:
Input: currentState = "++++"
Output: true
Explanation: The starting player can guarantee a win by flipping the middle "++" to become "+--+".
Example 2:
Input: currentState = "+"
Output: false
Constraints:
1 <= currentState.length <= 60currentState[i] is either '+' or '-'.- There cannot be more than 20 consecutive
'+'.
Follow up: Derive your algorithm's runtime complexity.
Solutions
Solution 1
PythonJavaC++Go
class Solution:
def canWin(self, currentState: str) -> bool:
@cache
def dfs(mask):
for i in range(n - 1):
if (mask & (1 << i)) == 0 or (mask & (1 << (i + 1)) == 0):
continue
if dfs(mask ^ (1 << i) ^ (1 << (i + 1))):
continue
return True
return False
mask, n = 0, len(currentState)
for i, c in enumerate(currentState):
if c == '+':
mask |= 1 << i
return dfs(mask)(code-box)
class Solution {
private int n;
private Map<Long, Boolean> memo = new HashMap<>();
public boolean canWin(String currentState) {
long mask = 0;
n = currentState.length();
for (int i = 0; i < n; ++i) {
if (currentState.charAt(i) == '+') {
mask |= 1 << i;
}
}
return dfs(mask);
}
private boolean dfs(long mask) {
if (memo.containsKey(mask)) {
return memo.get(mask);
}
for (int i = 0; i < n - 1; ++i) {
if ((mask & (1 << i)) == 0 || (mask & (1 << (i + 1))) == 0) {
continue;
}
if (dfs(mask ^ (1 << i) ^ (1 << (i + 1)))) {
continue;
}
memo.put(mask, true);
return true;
}
memo.put(mask, false);
return false;
}
}(code-box)
using ll = long long;
class Solution {
public:
int n;
unordered_map<ll, bool> memo;
bool canWin(string currentState) {
n = currentState.size();
ll mask = 0;
for (int i = 0; i < n; ++i)
if (currentState[i] == '+') mask |= 1ll << i;
return dfs(mask);
}
bool dfs(ll mask) {
if (memo.count(mask)) return memo[mask];
for (int i = 0; i < n - 1; ++i) {
if ((mask & (1ll << i)) == 0 || (mask & (1ll << (i + 1))) == 0) continue;
if (dfs(mask ^ (1ll << i) ^ (1ll << (i + 1)))) continue;
memo[mask] = true;
return true;
}
memo[mask] = false;
return false;
}
};(code-box)
func canWin(currentState string) bool {
n := len(currentState)
memo := map[int]bool{}
mask := 0
for i, c := range currentState {
if c == '+' {
mask |= 1 << i
}
}
var dfs func(int) bool
dfs = func(mask int) bool {
if v, ok := memo[mask]; ok {
return v
}
for i := 0; i < n-1; i++ {
if (mask&(1<<i)) == 0 || (mask&(1<<(i+1))) == 0 {
continue
}
if dfs(mask ^ (1 << i) ^ (1 << (i + 1))) {
continue
}
memo[mask] = true
return true
}
memo[mask] = false
return false
}
return dfs(mask)
}(code-box)
Solution 2
PythonJavaC++Go
class Solution:
def canWin(self, currentState: str) -> bool:
def win(i):
if sg[i] != -1:
return sg[i]
vis = [False] * n
for j in range(i - 1):
vis[win(j) ^ win(i - j - 2)] = True
for j in range(n):
if not vis[j]:
sg[i] = j
return j
return 0
n = len(currentState)
sg = [-1] * (n + 1)
sg[0] = sg[1] = 0
ans = i = 0
while i < n:
j = i
while j < n and currentState[j] == '+':
j += 1
ans ^= win(j - i)
i = j + 1
return ans > 0(code-box)
class Solution {
private int n;
private int[] sg;
public boolean canWin(String currentState) {
n = currentState.length();
sg = new int[n + 1];
Arrays.fill(sg, -1);
int i = 0;
int ans = 0;
while (i < n) {
int j = i;
while (j < n && currentState.charAt(j) == '+') {
++j;
}
ans ^= win(j - i);
i = j + 1;
}
return ans > 0;
}
private int win(int i) {
if (sg[i] != -1) {
return sg[i];
}
boolean[] vis = new boolean[n];
for (int j = 0; j < i - 1; ++j) {
vis[win(j) ^ win(i - j - 2)] = true;
}
for (int j = 0; j < n; ++j) {
if (!vis[j]) {
sg[i] = j;
return j;
}
}
return 0;
}
}(code-box)
class Solution {
public:
bool canWin(string currentState) {
int n = currentState.size();
vector<int> sg(n + 1, -1);
sg[0] = 0, sg[1] = 0;
function<int(int)> win = [&](int i) {
if (sg[i] != -1) return sg[i];
vector<bool> vis(n);
for (int j = 0; j < i - 1; ++j) vis[win(j) ^ win(i - j - 2)] = true;
for (int j = 0; j < n; ++j)
if (!vis[j]) return sg[i] = j;
return 0;
};
int ans = 0, i = 0;
while (i < n) {
int j = i;
while (j < n && currentState[j] == '+') ++j;
ans ^= win(j - i);
i = j + 1;
}
return ans > 0;
}
};(code-box)
func canWin(currentState string) bool {
n := len(currentState)
sg := make([]int, n+1)
for i := range sg {
sg[i] = -1
}
var win func(i int) int
win = func(i int) int {
if sg[i] != -1 {
return sg[i]
}
vis := make([]bool, n)
for j := 0; j < i-1; j++ {
vis[win(j)^win(i-j-2)] = true
}
for j := 0; j < n; j++ {
if !vis[j] {
sg[i] = j
return j
}
}
return 0
}
ans, i := 0, 0
for i < n {
j := i
for j < n && currentState[j] == '+' {
j++
}
ans ^= win(j - i)
i = j + 1
}
return ans > 0
}(code-box)
