leetcode - 2038. Remove Colored Pieces if Both Neighbors are the Same Color

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.

Example 1:

Input: colors = "AAABABB"
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.
1
2
3
4
5
6
7
8
9
10

Example 2:

Input: colors = "AA"
Output: false
Explanation:
Alice has her turn first.
There are only two 'A's and both are on the edge of the line, so she cannot move on her turn.
Thus, Bob wins, so return false.
1
2
3
4
5
6

Example 3:

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.
1
2
3
4
5
6
7
8
9
10
11
12
13

Constraints:

1 <= colors.length <= 10^5
colors consists of only the letters 'A' and 'B'
1
2

Solution

Solved after hints…

The number of removing alphabets doesn’t depend on the other player, but simply depends on the number of consecutive A or B. The number Alice could remove A is the number of consecutive A - 2, so does Bob.

Calculate the number of moving alphabets, then compare.

Time complexity: $o(n)$
Space complexity: $o(1)$

Code

class Solution:
    def winnerOfGame(self, colors: str) -> bool:
        alice, bob = 0, 0
        consec_a, consec_b = -2, -2
        for i in colors:
            if i == 'A':
                consec_b = -2
                consec_a += 1
                if consec_a > 0:
                    alice += 1
            elif i == 'B':
                consec_a = -2
                consec_b += 1
                if consec_b > 0:
                    bob += 1
        return alice > bob
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16

Also a recursive and TLE version.

class Solution:
    def winnerOfGame(self, colors: str) -> bool:
        memo = {}
        def helper(colors: str, cur_char: str) -> bool:
            if (colors, cur_char) in memo:
                return memo[(colors, cur_char)]
            if cur_char == 'A':
                for i in range(1, len(colors) - 1):
                    if colors[i - 1] == colors[i] == colors[i + 1] == 'A':
                        next_game = helper(colors[:i] + colors[i + 1:], 'B')
                        if next_game == True:
                            memo[(colors, cur_char)] = True
                if (colors, cur_char) not in memo:
                    memo[(colors, cur_char)] = False
            elif cur_char == 'B':
                for i in range(1, len(colors) - 1):
                    if colors[i - 1] == colors[i] == colors[i + 1] == 'B':
                        next_game = helper(colors[:i] + colors[i + 1:], 'A')
                        if next_game == False:
                            memo[(colors, cur_char)] = False
                if (colors, cur_char) not in memo:
                    memo[(colors, cur_char)] = True
            return memo[(colors, cur_char)]
        return helper(colors, 'A')
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24

Another prev version of counting:

class Solution:
    def winnerOfGame(self, colors: str) -> bool:
        cons_a, cons_b = 0, 0
        prev_ch = None
        cnt = 0
        for each_ch in colors:
            if prev_ch != each_ch:
                if prev_ch:
                    if prev_ch == 'A':
                        cons_a += max(0, cnt - 2)
                    elif prev_ch == 'B':
                        cons_b += max(0, cnt - 2)
                prev_ch = each_ch
                cnt = 0
            cnt += 1
        if prev_ch == 'A':
            cons_a += max(0, cnt - 2)
        elif prev_ch == 'B':
            cons_b += max(0, cnt - 2)
        return cons_a > cons_b
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20