LeetCode 293 - Flip Game

This problem is asking us to simulate a single move in a simple two-player game. The input is a string currentState consisting only of '+' and '-' characters.

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LeetCode Problem 293

Difficulty: 🟢 Easy
Topics: String

Solution

Problem Understanding

This problem is asking us to simulate a single move in a simple two-player game. The input is a string currentState consisting only of '+' and '-' characters. Each '+' represents a spot that can be flipped, and the rules allow a player to flip two consecutive '+' characters into '--'. The task is to generate all possible strings that result from making exactly one valid move on the current state.

For example, if currentState = "++++", you can flip the first two, the middle two, or the last two '+' characters. Therefore, the output is all strings after making one of these flips.

Constraints tell us that the input length is between 1 and 500. This is small enough that a simple linear scan through the string is efficient. Each character is guaranteed to be '+' or '-', so we do not need to handle invalid input characters.

Edge cases to be aware of include strings that are too short to flip two consecutive '+' characters, strings with no '+' characters, and strings where only one valid flip exists. These are simple but essential to handle, as a naive implementation might assume there is always a valid move.

Approaches

Brute-Force Approach

A brute-force approach would attempt to generate all possible substrings and check for '++' pairs at every position. Whenever '++' is found, a new string is generated with that pair flipped to '--'. This approach is correct because it considers all possible positions for valid moves, but it is somewhat inefficient because it constructs a new string for every character pair, even when it cannot be flipped. For a string of length n, this can result in O(n²) string operations due to repeated substring constructions.

Optimal Approach

The optimal approach relies on a simple observation: a valid move exists only at consecutive '++' pairs. Therefore, we can perform a single pass through the string and whenever we see '++', we create a new string with those two characters flipped. This approach is linear in time with respect to the string length and avoids unnecessary substring generation. It is optimal because we only examine positions that could yield valid moves, and each valid move is generated exactly once.

Approach Time Complexity Space Complexity Notes
Brute Force O(n²) O(n²) Checks all substrings, generates many intermediate strings unnecessarily
Optimal O(n) O(n²) Single pass, only generates strings for valid moves

Algorithm Walkthrough

  1. Initialize an empty list results to store all possible strings after one move.
  2. Iterate over the string currentState from the first character to the second-to-last character, using an index i.
  3. At each index, check if currentState[i] and currentState[i + 1] are both '+'.
  4. If a '++' pair is found, construct a new string where only this pair is replaced with '--'. You can do this efficiently by slicing the string into three parts: before the pair, the flipped pair, and after the pair.
  5. Append this new string to results.
  6. After iterating through the entire string, return results.

Why it works: The algorithm works because it systematically examines every possible position for a valid flip and generates exactly one new string for each valid move. There are no duplicates because each '++' pair is only flipped once. This guarantees that all possible next states are captured.

Python Solution

from typing import List

class Solution:
    def generatePossibleNextMoves(self, currentState: str) -> List[str]:
        results = []
        n = len(currentState)
        for i in range(n - 1):
            if currentState[i] == '+' and currentState[i + 1] == '+':
                new_state = currentState[:i] + '--' + currentState[i + 2:]
                results.append(new_state)
        return results

The code follows the algorithm closely. We initialize an empty list results to collect possible states. The loop checks each character up to the second-to-last, ensuring i + 1 is within bounds. When a '++' is found, slicing constructs the new state efficiently. Each new state is appended to results, which is returned at the end.

Go Solution

func generatePossibleNextMoves(currentState string) []string {
    results := []string{}
    n := len(currentState)
    for i := 0; i < n-1; i++ {
        if currentState[i] == '+' && currentState[i+1] == '+' {
            newState := currentState[:i] + "--" + currentState[i+2:]
            results = append(results, newState)
        }
    }
    return results
}

The Go version mirrors the Python implementation. Strings in Go are immutable, so concatenation produces a new string just like slicing in Python. We initialize an empty slice of strings and append each valid move.

Worked Examples

Example 1: currentState = "++++"

i currentState[i:i+2] Action results
0 "++" Flip to "--" ["--++"]
1 "++" Flip to "--" ["--++", "+--+"]
2 "++" Flip to "--" ["--++", "+--+", "++--"]

Final output: ["--++", "+--+", "++--"]

Example 2: currentState = "+"

No pair of consecutive '+' exists. The loop does not run. Final output: []

Complexity Analysis

Measure Complexity Explanation
Time O(n) We scan the string once, checking each pair of characters.
Space O(n²) Each valid move produces a new string of length n. In the worst case, all n-1 pairs are valid, generating O(n) strings, each of length O(n).

The reasoning is straightforward: the algorithm's primary cost is scanning the string and generating new strings. Each string creation is O(n) due to immutability.

Test Cases

# Provided examples
assert Solution().generatePossibleNextMoves("++++") == ["--++","+--+","++--"]  # multiple flips
assert Solution().generatePossibleNextMoves("+") == []  # too short to flip

# Edge cases
assert Solution().generatePossibleNextMoves("++") == ["--"]  # exactly one move
assert Solution().generatePossibleNextMoves("-") == []  # no '+'
assert Solution().generatePossibleNextMoves("+-++-+") == ["+--+-+","+--++-"]  # scattered '+'

# Long string
assert Solution().generatePossibleNextMoves("+"*500) == [("--"+"+"*498), ("+"+"--"+"+"*497), ("++--"+"+"*496)]  # simplified representation
Test Why
"++++" Standard multiple moves
"+" Too short to flip
"++" Single flip possible
"-" No flips possible
"+-++-+" Scattered flips
500 '+'s Test upper boundary and performance

Edge Cases

  1. Single character strings: Strings of length 1 cannot have any valid move. If not handled, a naive implementation might attempt to access i + 1 and raise an index error. Our solution correctly iterates only to n-1.
  2. Strings with no consecutive '+': For example, "+-+-+". The algorithm scans the string and finds no pairs, returning an empty list. This prevents unnecessary string construction.
  3. Strings where all characters are '+': For maximum length input, such as "+"*500, the algorithm must handle a large number of possible moves efficiently. The solution scales linearly and constructs each string separately, avoiding performance issues. Each new string uses slicing to ensure correctness even at the boundary indices.