How hard is LeetCode 294. Flip Game II?

LeetCode 294. Flip Game II is rated Medium on LeetCode.

What topics does LeetCode 294. Flip Game II cover?

LeetCode 294. Flip Game II is tagged Memoization, Math, Dynamic Programming, Backtracking and Game Theory on LeetCode.

Is LeetCode 294. Flip Game II a premium problem?

Yes. LeetCode 294. Flip Game II is a LeetCode Premium problem, so the full statement and test cases require a paid LeetCode subscription.

MediumNo. 0294LeetCode Premium

Flip Game II — LeetCode 294 Python Solution

Pattern: Backtracking
Topics: Memoization, Math, Dynamic Programming, Backtracking, Game Theory
Reading time: 3 min
Source: leetcode.com

The problem

You are playing a Flip Game with your friend. You are given a string currentState that contains only '+' and '-'.

Example

Input: currentState = "++++"
Output: true
Explanation: The starting player can guarantee a win by flipping the middle "++" to become "+--+".

Python solution

PythonSolution

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)

Complexity

Time: O(n·m) (typical)
Space · auxiliary: O(n·m)or optimized

Pattern: Backtracking

Build candidates one choice at a time and abandon a branch the moment it cannot work. LeetCode 294. Flip Game II is filed here because LeetCode tags it Backtracking, which is the vocabulary this hub collects.

The backtracking guide has the Python template for the pattern and the 105 LeetCode problems that use it.

Related problems
No.	Problem	Difficulty
0464	Can I Win	Medium
0913	Cat and Mouse	Hard
1728	Cat and Mouse II	Hard
0070	Climbing Stairs	Easy
0241	Different Ways to Add Parentheses	Medium
0357	Count Numbers with Unique Digits	Medium

Frequently asked questions

How hard is LeetCode 294. Flip Game II?: LeetCode 294. Flip Game II is rated Medium on LeetCode.
What topics does LeetCode 294. Flip Game II cover?: LeetCode 294. Flip Game II is tagged Memoization, Math, Dynamic Programming, Backtracking and Game Theory on LeetCode.
Is LeetCode 294. Flip Game II a premium problem?: Yes. LeetCode 294. Flip Game II is a LeetCode Premium problem, so the full statement and test cases require a paid LeetCode subscription.

Stuck on problems like this in a live interview?

Stealth Interview is a desktop app for macOS and Windows. It reads the problem off your screen and returns a working solution with a step-by-step explanation and its time and space complexity, invisible to screen sharing.

Download for Mac Download for Windows

Free to download. Check it runs on your machine, then pick a plan.

90%Subscription

python

Please analyze this context and help me answer the question or solve the problem.

Transcript

"Okay, next one. Given a string, return the length of the longest substring that has no repeating characters. Walk me through your approach before you code it."

Technical interview · 45 minsolution.pyShared editor · Python 3Longest substring without repeatsMediumGiven a string s, return the length of the longest contiguous substring in which every character is unique.s = "abcabcbb"
→ 31def length_of_longest_substring(s: str) -> int:2 last_seen = {}3 left = best = 04 for right, ch in enumerate(s):5 if ch in last_seen and last_seen[ch] >= left:

python

def length_of_longest_substring(s: str) -> int:    last_seen = {}                    # char -> index    left = best = 0    for right, ch in enumerate(s):        if ch in last_seen and last_seen[ch] >= left:            left = last_seen[ch] + 1  # jump past it        last_seen[ch] = right        best = max(best, right - left + 1)    return best    last_seen = {}                    # char -> index

Approach

At each index r, the window's left edge is:

left r = max (left r -1, last[s r] + 1)

Time O(n): each character enters and leaves the window once.
Space O(min(n, k)): k is the size of the character set.

Type your message...

GPT-6.1 Terra

Transcribing...

90%Subscription

python

Please analyze this context and help me answer the question or solve the problem.

Transcript

"Okay, next one. Given a string, return the length of the longest substring that has no repeating characters. Walk me through your approach before you code it."

python

def length_of_longest_substring(s: str) -> int:    last_seen = {}                    # char -> index    left = best = 0    for right, ch in enumerate(s):        if ch in last_seen and last_seen[ch] >= left:            left = last_seen[ch] + 1  # jump past it        last_seen[ch] = right        best = max(best, right - left + 1)    return best    last_seen = {}                    # char -> index

Approach

At each index r, the window's left edge is:

left r = max (left r -1, last[s r] + 1)

Time O(n): each character enters and leaves the window once.
Space O(min(n, k)): k is the size of the character set.

Type your message...

GPT-6.1 TerraClear chat

Transcribing...

QuitEscHide+QCapture+PCrop+OSubmit+EnterPress to get answerLeft/Right+Up/Down+

90%Subscription

python

Please analyze this context and help me answer the question or solve the problem.

Transcript

"Okay, next one. Given a string, return the length of the longest substring that has no repeating characters. Walk me through your approach before you code it."

python

def length_of_longest_substring(s: str) -> int:    last_seen = {}                    # char -> index    left = best = 0    for right, ch in enumerate(s):        if ch in last_seen and last_seen[ch] >= left:            left = last_seen[ch] + 1  # jump past it        last_seen[ch] = right        best = max(best, right - left + 1)    return best    last_seen = {}                    # char -> index

Approach

At each index r, the window's left edge is:

left r = max (left r -1, last[s r] + 1)

Time O(n): each character enters and leaves the window once.
Space O(min(n, k)): k is the size of the character set.

Type your message...

GPT-6.1 TerraClear chat

Transcribing...

Primary display

QuitEscHide+QCapture+PCrop+OSubmit+EnterPress to get answerLeft/Right+Up/Down+

The problem

Example

Python solution

Complexity

Pattern: Backtracking

Related problems

Frequently asked questions

Stuck on problems like this in a live interview?

Approach

Approach

Approach