02. Remove Colored Pieces if Both Neighbors are the Same Color

The problem can be found at the following link: Question Link

My Approach

1. Initialize scores for Alice and Bob (aliceScore and bobScore) to zero.

2. Iterate through the colors (excluding edge pieces) using a loop. For each piece in the iteration:

   • Check if it's A and its neighboring pieces are also A. If yes, increment aliceScore as Alice can remove the current piece.

   • Check if it's B and its neighboring pieces are also B. If yes, increment bobScore as Bob can remove the current piece.

3. Return true if Alice's score is greater than Bob's score; otherwise, return false.

Time and Auxiliary Space Complexity

• Time Complexity: O(n)

• Auxiliary Space Complexity: O(1)

Code (C++)

class Solution {
public:
    bool winnerOfGame(string colors) {
        int alice=0;
        int bob=0;
        for(int i=1;i<colors.size()-1;i++)
        {
            if(colors[i-1]=='A' && colors[i+1]=='A' && colors[i]=='A')
            {
                alice++;
            }
            else if(colors[i-1]=='B' && colors[i+1]=='B' && colors[i]=='B')
            {
                bob++;
            }
        }
        cout<<alice<<" "<<bob<<endl;
        return alice>bob;
    }
};

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