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Remove Colored Pieces if Both Neighbors are the Same Color

Problem Description​

There are n pieces arranged in a line, and each piece is colored either by 'A' or by 'B'. You are given a string colors of length n where colors[i] is the color of the ith piece.

Alice and Bob are playing a game where they take alternating turns removing pieces from the line. In this game, Alice moves first.

Alice is only allowed to remove a piece colored 'A' if both its neighbors are also colored 'A'. She is not allowed to remove pieces that are colored 'B'. Bob is only allowed to remove a piece colored 'B' if both its neighbors are also colored 'B'. He is not allowed to remove pieces that are colored 'A'. Alice and Bob cannot remove pieces from the edge of the line. If a player cannot make a move on their turn, that player loses and the other player wins. Assuming Alice and Bob play optimally, return true if Alice wins, or return false if Bob wins.

Examples​

Example 1:

Output: true
Explanation:
AAABABB -> AABABB
Alice moves first.
She removes the second 'A' from the left since that is the only 'A' whose neighbors are both 'A'.

Now it's Bob's turn.
Bob cannot make a move on his turn since there are no 'B's whose neighbors are both 'B'.
Thus, Alice wins, so return true.

Example 2:

Input: colors = "ABBBBBBBAAA"
Output: false
Explanation:
ABBBBBBBAAA -> ABBBBBBBAA
Alice moves first.
Her only option is to remove the second to last 'A' from the right.

ABBBBBBBAA -> ABBBBBBAA
Next is Bob's turn.
He has many options for which 'B' piece to remove. He can pick any.

On Alice's second turn, she has no more pieces that she can remove.
Thus, Bob wins, so return false.

Constraints​

  • 1 <= colors.length <= 10^5
  • colors consists of only the letters 'A' and 'B'

Solution for Remove Colored Pieces if Both Neighbors are the Same Color​

Approach​

  1. Initialization: Start with two counters, countA and countB, initialized to zero. These will keep track of the number of triplets 'AAA' and 'BBB' found in the string colors.

  2. Iteration: Loop through the string colors from the beginning to the third last character (n - 2), where n is the length of colors.

  3. Counting Triplets:

    • Check for sequences of three consecutive 'A's ('AAA'). If found, increment countA.
    • Check for sequences of three consecutive 'B's ('BBB'). If found, increment countB.

  4. Comparison: After iterating through the string:

    • Compare countA and countB.
    • Return true if countA (number of 'AAA' sequences) is greater than countB (number of 'BBB' sequences), indicating player 'A' has more winning sequences.
    • Otherwise, return false.

Implementation​

Live Editor
function Solution(arr) {
   
  function winnerOfGame(colors) {
let countA = 0, countB = 0;
const n = colors.length;
for (let i = 0; i < n - 2; i++) {
    if (colors[i] === 'A' && colors[i + 1] === 'A' && colors[i + 2] === 'A') {
        countA++;
    }
    if (colors[i] === 'B' && colors[i + 1] === 'B' && colors[i + 2] === 'B') {
        countB++;
    }
}
return countA > countB;
}

  const input = "AAABABB"

  const output =winnerOfGame(input)
  return (
    <div>
      <p>
        <b>Input: </b>
        {JSON.stringify(input)}
      </p>
      <p>
        <b>Output:</b> {output.toString()}
      </p>
    </div>
  );
}
Result
Loading...

Complexity Analysis​

  • Time Complexity: O(n)O(n)
  • Space Complexity: O(1) O(1)

Code in Different Languages​

Written by @hiteshgahanolia
function winnerOfGame(colors) {
 let countA = 0, countB = 0;
 const n = colors.length;
 for (let i = 0; i < n - 2; i++) {
     if (colors[i] === 'A' && colors[i + 1] === 'A' && colors[i + 2] === 'A') {
         countA++;
     }
     if (colors[i] === 'B' && colors[i + 1] === 'B' && colors[i + 2] === 'B') {
         countB++;
     }
 }
 return countA > countB;
}

References​