Python - Algorithm #32

Python Algorithm

LeetCode

3019. Number of Changing Keys

• You are given a 0-indexed string s typed by a user. Changing a key is defined as using a key different from the last used key. For example, s = "ab" has a change of a key while s = "bBBb" does not have any.
• Return the number of times the user had to change the key.
• Note: Modifiers like shift or caps lock won't be counted in changing the key that is if a user typed the letter 'a' and then the letter 'A' then it will not be considered as a changing of key.
• Constraints
  • 1 ≤ s.length ≤ 100
  • s consists of only upper case and lower case English letters.

class Solution:
    def countKeyChanges(self, s: str) -> int:
        answer = 0
        # 문자열 lower, for문 사용해서 앞에 문자와 뒤에 문자 비교
        s = s.lower()
        for i in range(len(s) - 1):
            if s[i] != s[i+1]:
                answer += 1
        return answer

class Solution:
    def countKeyChanges(self, s: str) -> int:
        answer = 0
        s = s.lower()
        for i in range(len(s)-1):
            if s[i] != s[i+1]:
                answer += 1
        return answer

---

3238. Find the Number of Winning Players

• You are given an integer n representing the number of players in a game and a 2D array [pick] where pick[i] = [xi, yi] represents that the player xᵢ picked a ball of color yᵢ.
• Player i wins the game if they pick strictly more than i balls of the same color. In other words,
  • Player 0 wins if they pick any ball.
  • Player 1 wins if they pick at least two balls of the same color.
  • ...
  • Player i wins if they pick at leasti + 1 balls of the same color.
• Return the number of players who win the game.
• Note that multiple players can win the game.
• Constraints
  
  • 2 ≤ n ≤ 10
  • 1 ≤ pick.length ≤ 100
  • pick[i].length == 2
  • 0 ≤ xᵢ ≤ n - 1
  • 0 ≤ yᵢ ≤ 10

class Solution:
    def winningPlayerCount(self, n: int, pick: List[List[int]]) -> int:
        answer = 0
        # key: 플레이어 번호, value: 색깔별 공 개수
        d = defaultdict(Counter)
        for player, ball_color in pick:
            d[player][ball_color] += 1
        for player, ball_counter in d.items():
            if player + 1 <= max(ball_counter.values()):
                answer += 1
        return answer

class Solution:
    def winningPlayerCount(self, n: int, pick: List[List[int]]) -> int:
        answer = 0
        d = defaultdict(Counter)
        for p, c in pick:
            d[p][c] += 1
        for key, value in d.items():
            for i in value:
                if value[i] >= key + 1:
                    answer += 1
                    break
        return answer

---

https://leetcode.com/problems/number-of-changing-keys/

https://leetcode.com/problems/find-the-number-of-winning-players/